Answer:
m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J
Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.
Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer
Explanation:
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.
What is friction?Friction is the force that prevents one hard material from scooting or rolling over the other.
Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.
We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.
Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.
In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.
Thus, the correct option is B.
For more details regarding friction, visit:
https://brainly.com/question/28356847
#SPJ2
How much time will elapse if a radioisotope with a half-life of 88 seconds decays to one-sixteenth of its original mass?
Answer:
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
Explanation:
The decay of radioisotopes are represented by the following ordinary differential equation:
[tex]\frac{dm}{dt} = -\frac{t}{\tau}[/tex]
Where:
[tex]t[/tex] - Time, measured in seconds.
[tex]\tau[/tex] - Time constant, measured in seconds.
[tex]m[/tex] - Mass of the radioisotope, measured in grams.
The solution of this expression is:
[tex]m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }[/tex]
Where [tex]m_{o}[/tex] is the initial mass of the radioisotope, measured in kilograms.
The ratio of current mass to initial mass is:
[tex]\frac{m(t)}{m_{o}} = e^{-\frac{t}{\tau} }[/tex]
The time constant is now calculated in terms of half-life:
[tex]\tau = \frac{t_{1/2}}{\ln2}[/tex]
Where [tex]t_{1/2}[/tex] is the half-life of the radioisotope, measured in seconds.
Given that [tex]t_{1/2} = 88\,s[/tex], the time constant of the radioisotope is:
[tex]\tau = \frac{88\,s}{\ln 2}[/tex]
[tex]\tau \approx 126.957\,s[/tex]
Now, if [tex]\frac{m(t)}{m_{o}(t)} = \frac{1}{16}[/tex] and [tex]\tau \approx 126.957\,s[/tex], the time is:
[tex]t = -\tau \cdot \ln\frac{m(t)}{m_{o}}[/tex]
[tex]t = -(126.957\,s)\cdot \ln \frac{1}{16}[/tex]
[tex]t \approx 352\,s[/tex]
352 seconds are needed for the radioisotope to decay to one-sixteenth of its original mass.
A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?
Answer:
The amplitude [tex]A(5) = 1 \ m[/tex]
Explanation:
From the question we are told that
The spring constant is [tex]k = 100 \ N/m[/tex]
The damping constant is [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]
The mass is [tex]m = 0.050 \ kg[/tex]
The maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]
The time considered is t = 5.0 s
Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as
[tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]
substituting values
[tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]
[tex]A(5) = 1 \ m[/tex]
A 2-slit arrangement with 60.3 μm separation between the slits is illuminated with 482.0 nm light. Assuming that a viewing screen is located 2.14 m from the slits, find the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side. A. 24.1 mm B. 34.2 mm C. 68.4 mm D. 51.3 mm
Answer:
The distance is [tex]y = 0.03425 \ m[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 60.3 \mu m= 60.3 *10^{-6}\ m[/tex]
The wavelength is [tex]\lambda = 482.0 \ nm = 482.0 *10^{-9} \ m[/tex]
The distance of the screen is [tex]D = 2.14 \ m[/tex]
Generally the distance of a fringe from the central maxima is mathematically represented as
[tex]y = [m + \frac{1}{2} ] * \frac{\lambda * D}{d}[/tex]
For the first dark fringe m = 0
[tex]y_1 = [0 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_1 = 0.00855 \ m[/tex]
For the second dark fringe m = 1
[tex]y_2 = [1 + \frac{1}{2} ] * \frac{482*10^{-9} * 2.14}{ 60.3*10^{-6}}[/tex]
[tex]y_2 = 0.0257 \ m[/tex]
So the distance from the first dark fringe on one side of the central maximum to the second dark fringe on the other side is
[tex]y = y_1 + y_2[/tex]
[tex]y = 0.00855 + 0.0257[/tex]
[tex]y = 0.03425 \ m[/tex]
A laboratory electromagnet produces a magnetic field of magnitude 1.38 T. A proton moves through this field with a speed of 5.86 times 10^6 m/s.
a. Find the magnitude of the maximum magnetic force that could be exerted on the proton.
b. What is the magnitude of the maximum acceleration of the proton?
c. Would the field exert the same magnetic force on an electron moving through the field with the same speed? (Assume that the electron is moving in the direction as the proton.)
1. Yes
2. No
.Answer;
Using Fmax=qVB
F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)
ANS=1.29*10^-12 N
2. Using Amax=Fmax/ m
Amax =(1.29*10^-12 N) / (1.67*10^-27 kg)
ANS=1.93*10^15 m/s^2*
3. No, the acceleration wouldn't be the same. Since The magnitude of the electron is equal to that of the proton, but the direction would be in the opposite direction and also Since an electron has a smaller mass than a proton
A fan rotating with an initial angular velocity of 1500 rev/min is switched off. In 2.5 seconds, the angular velocity decreases to 400 rev/min. Assuming the angular acceleration is constant, answer the following questions.
How many revolutions does the blade undergo during this time?
A) 10
B) 20
C) 100
D) 125
E) 1200
Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
Changing the speed of a synchronous generator changes A) the frequency and amplitude of the output voltage. B) only the frequency of the output voltage. C) only the amplitude of the output voltage. D) only the phase of the output voltage.
Answer:
A) the frequency and amplitude of the output voltag
Explanation:
Changing the speed of a synchronous generator changes both the output voltage (amplitude of the wave) and frequency as they tend to increase.
Changing the speed regulator will change the engine throttle setting to maintain the speed.
While the power, torque, current, fuel flow rate and torque angle will have decreased.
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire
Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
The molecules in Tyler are composed of carbon and other atoms that share one or more electrons between two atoms, forming what is known as a(n) _____ bond.
Answer:
covalent
Explanation:
covalent bonds share electrons
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.2 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.8 cm .
Required:
What is the magnitude of the charge (in nC) on each bead?
Answer:
The magnitude of the charge is 54.9 nC.
Explanation:
The charge on each bead can be found using Coulomb's law:
[tex] F_{e} = \frac{k*q_{1}q_{2}}{r^{2}} [/tex]
Where:
q₁ and q₂ are the charges, q₁ = q₂
r: is the distance of spring stretching = 4.8x10⁻² m
[tex]F_{e}[/tex]: is the electrostatic force
[tex] F_{e} = \frac{k*q^{2}}{r^{2}} \rightarrow q = \sqrt{\frac{F_{e}}{k}}*r [/tex]
Now, we need to find [tex]F_{e}[/tex]. To do that we have that Fe is equal to the spring force ([tex]F_{k}[/tex]):
[tex] F_{e} = F_{k} = -kx [/tex]
Where:
k is the spring constant
x is the distance of the spring = 4.8 - 4.0 = 0.8 cm
The spring constant can be found by equaling the sping force and the weight force:
[tex] F_{k} = -W [/tex]
[tex] -k*x = -m*g [/tex]
where x is 5.2 - 4.0 = 1.2 cm, m = 1.8 g and g = 9.81 m/s²
[tex] k = \frac{mg}{x} = \frac{1.8 \cdot 10^{-3} kg*9.81 m/s^{2}}{1.2 \cdot 10^{-2} m} = 1.47 N/m [/tex]
Now, we can find the electrostatic force:
[tex] F_{e} = F_{k} = -kx = -1.47 N/m*0.8 \cdot 10^{-2} m = -0.0118 N [/tex]
And with the magnitude of the electrostatic force we can find the charge:
[tex]q = \sqrt{\frac{F_{e}}{k}}*r = \sqrt{\frac{0.0118 N}{9 \cdot 10^{9} Nm^{2}/C^{2}}}*4.8 \cdot 10^{-2} m = 54.9 \cdot 10^{-9} C = 54.9 nC[/tex]
Therefore, the magnitude of the charge is 54.9 nC.
I hope it helps you!
The magnitude of the charge (in nC) on each bead is equal to 55.21 nC.
Given the following data:
Original length = 4.0 cm to m = 0.04 mMass = 1.8 grams to kg = 0.0018New length = 5.2 cm to m = 0.052.Final length = 4.8 cm to m = 0.048 m.Extension, e = [tex]0.052 - 0.048[/tex] = 0.012 m
Scientific data:
Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]Coulomb's constant = [tex]8.99 \times 10^9\; Nm^2/C^2[/tex]To calculate the magnitude of the charge (in nC) on each bead, we would apply Coulomb's law:
First of all, we would determine the spring constant of this lightweight spring by using this formula:
[tex]W = mg = Ke \\\\K=\frac{mg}{e} \\\\K=\frac{0.0018 \times 9.8}{0.012} \\\\K=\frac{0.01764}{0.012}[/tex]
Spring constant, K = 1.47 N/m.
For the electrostatic force:
[tex]F = ke\\\\F = 1.47 \times 0.08[/tex]
F = 0.01176 Newton.
Coulomb's law of electrostatic force.
Mathematically, the charge in an electric field is given by this formula:
[tex]q = \sqrt{\frac{F}{k} } \times r[/tex]
Substituting the given parameters into the formula, we have;
[tex]q = \sqrt{\frac{0.01176 }{8.99 \times 10^9} } \times 0.048\\\\q=\sqrt{1.3228 \times 10^{-12}} \times 0.048\\\\q=1.1502 \times 10^{-6} \times 0.048\\\\q= 5.521 \times 10^{-8}\;C[/tex]
Note: 1 nC = [tex]1 \times 10^{-9}\;C[/tex]
Charge, q = 55.21 nC.
Read more on electric field here: https://brainly.com/question/14372859
If an electron is accelerated from rest through a potential difference of 1.60 x 102V, what is its de Broglie wavelength
Answer:
0.09 x10^-10m
Explanation:
Using wavelength=( 12.27 A)/√V
= 12.27 x 10^-10/ √1.6x10^2
= 0.09x10^-10m
Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 A. The relative permeability of the core is 80.
(a) What is the magnetic field in the core?
(b) What part of the magnetic field is due to atomic currents?
Answer:
A) 0.0267 T
B) 0.0263 T
Explanation:
Given that
The number of turns, N = 400
Radius of the wire, r = 6 cm = 0.06 m
Current in the wire, I = 0.25 A
Relative permeability, K(m) = 80
See the attached picture for the calculation
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
Answer:
A) the moment of inertia of the system decreases and the angular speed increases.
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
[tex]I_{1} w_{1} = I_{2} w_{2}[/tex] ....1
where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.
and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
[tex]I = mr^{2}[/tex] ....2
where [tex]m[/tex] is the mass of the rotating body,
and [tex]r[/tex] is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
Difference between matter and energy
Answer:
Energy is the strength and vitality required for sustained physical or mental activity.
Matter occupies space and possesses rest mass, especially as distinct from energy.
Hope this helps! （づ￣3￣）づ╭❤～
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Required:
How far away did the impact occur?
Answer:
The distance is [tex]d = 193.6 \ m[/tex]
Explanation:
From the question we are told that
The time interval between the sounds is k[tex]t_1 = k + t_2[/tex] = 0.50 s
The speed of sound in air is [tex]v_s = 343 \ m/s[/tex]
The speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]
Generally the distance where the collision occurred is mathematically represented as
[tex]d = v * t[/tex]
Now from the question we see that d is the same for both sound waves
So
[tex]v_c t = v_s * t_1[/tex]
Now
So [tex]t_1 = k + t[/tex]
[tex]v_c t = v_s * (t+ k)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]t = 0.0645 \ s[/tex]
So
[tex]d = 3000 * 0.0645[/tex]
[tex]d = 193.6 \ m[/tex]
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal:_____.
Answer:
In an electromagnetic wave in free space, the ratio of the magnitudes of electric and magnetic field vectors E and B is equal: speed of light(c)
Explanation:
Generally the ratio of the E(electric field ) and the B(magnetic field ) is equal to the speed of the electromagnetic wave i.e the speed of light (c) the value is
[tex]c = 3.0 *10^{8} \ m/s[/tex]
A pulley 326 mm in diameter and rotating initially at 4.00 revolutions per second receives a constant angular acceleration of 2.25 radians per second squared by a drive belt. What is the linear velocity of the belt after 5.00 seconds
Answer:
The linear velocity, v = 5.93 m/s
Explanation:
To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using
ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' = = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s
ω' = ω + αt
= 25.13 rad/s + 2.25 rad/s² × 5.00 s
= 25.13 rad/s + 11.25 rad/s
= 36.38 rad/s
The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m
v = rω'
= 0.163 m × 36.38 rad/s
= 5.93 m/s
So, the linear velocity v = 5.93 m/s
Monochromatic light falls on two very narrow slits 0.047 mm apart. Successive fringes on a screen 6.60 m away are 8.9 cm apart near the center of the pattern.
Determine the wavelength and frequency of the light.
Answer::
[tex]\lambda = 634 nm[/tex]
[tex]f = 4.73 *10^{14} \ Hz[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.047 \ mm = 0.047 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 6.60 \ m[/tex]
The width of the fringe is [tex]y = 8.9 \ cm = 0.089 \ m[/tex]
Generally the width of the width of the fringes is mathematically represented as
[tex]y = \frac{\lambda * D }{d }[/tex]
=> [tex]\lambda = \frac{y * d }{D }[/tex]
=> [tex]\lambda = \frac{ 0.089 * (0.047 *10^{-3}) }{6.60 }[/tex]
=> [tex]\lambda = 634 *10^{-9}[/tex]
=> [tex]\lambda = 634 nm[/tex]
Generally the speed of light is mathematically represented as
[tex]c = f * \lambda[/tex]
=> [tex]f= \frac{ c}{\lambda }[/tex]
=> [tex]f= \frac{ 3.0 *10^{8}}{634 *10^{-9}}[/tex]
=> [tex]f = 4.73 *10^{14} \ Hz[/tex]
g When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 3.2 10-15 m before interacting. From this information, find the time interval required for the strong interaction to occur.
Answer:
Time, [tex]t=1.07\times 10^{-23}\ s[/tex]
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]
So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]
On a separate sheet of paper, tell why scientists in different countries can easily compare the amount of matter in similar objects in their countries
Answer: no u
Explanation: no u
A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?
Answer:
Pressure, P = 1250 Pa
Explanation:
Given that,
Weight of a brick, F = 50 N
Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm
We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.
A = 40 cm × 10 cm = 400 cm² = 0.04 m²
So,
Pressure,
[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]
So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.
The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
What is pressure?The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.
The given data in the problem is;
W is the weight of a brick = 50 N
The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm
A is the area,
The area is found as;
A=40 cm × 10 cm = 400 cm² = 0.04 m²
The pressure is the ratio of the force and area
[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]
Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.
To learn more about the pressure refer to the link;
https://brainly.com/question/356585
The Moon orbits Earth in a nearly circular orbit (mean distance is 378,000 km ). The moon Charon orbits Pluto in a nearly circular orbit as well (mean distance is 19,600 km ).
Earth Moon Pluto Charon
Mass (kg) 5.97 x 10^24 0.07342 x 10^24 0.0146 x 10^24 0.00162 x 10^24
Equatorial radius (km) 6378.1 1738.1 1185 604
Which object exhibits the longest orbital period? Hint: perform order of magnitude analysis.
a. Moon around Earth
b. Charon around Pluto
c. About the same for both
Answer:
a. Moon around Earth.
Explanation:
Charon orbit takes around 6.4 earth days to complete its orbit. Charon does not rises or sets, it hovers over same spot around the Pluto. The same side of Charon faces the Pluto, this is called Tidal Locking.
The moon orbit takes around 27 days to complete its orbit. The moon has different sides that are faced with sun which creates light or dark face of moon on the earth. Moon has 384,400 km distance from the earth.
The object that should exhibit the longest orbital period is option a. Moon around Earth.
What is Charon's orbit?Charon's orbit takes around 6.4 earth days to finish its orbit. Charon does not rise or sets, it hovers over similar spot around Pluto. The same side of Charon faces Pluto, this we called Tidal Locking. Here the moon orbit should take approx 27 days to finish its orbit. The moon has various sides that are faced with the sun which developed the light or dark face of the moon on the earth. Also, Moon has 384,400 km distance from the earth.
learn more about orbit here: https://brainly.com/question/25404554
The earth has a vertical electric field at the surface,pointing down, that averages 102 N/C. This field is maintained by various atmosphericprocesses, including lightning.
What is the excess charge on the surface of the earth? inC
Answer:
[tex]q = -461532.5 \ C[/tex]
Explanation:
From the question we are told that
The electric filed is [tex]E = 102 \ N/C[/tex]
Generally according to Gauss law
=> [tex]E A = \frac{q}{\epsilon_o }[/tex]
Given that the electric field is pointing downward , the equation become
[tex]- E A = \frac{q}{\epsilon_o }[/tex]
Here [tex]q[/tex] is the excess charge on the surface of the earth
[tex]A[/tex] is the surface area of the of the earth which is mathematically represented as
[tex]A = 4\pi r^2[/tex]
Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]
substituting values
[tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]
[tex]A =5.1128 *10^{14} \ m^2[/tex]
So
[tex]q = -E * A * \epsilon _o[/tex]
Here [tex]\epsilon_o[/tex] s the permitivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]
[tex]q = -461532.5 \ C[/tex]
a car moves for 10 minutes and travels 5,280 meters .What is the average speed of the car?
Answer:use the formular distance over time i.e distance/time. Make sure to convert the distance from metres to kilometers and time from minutes to hours .
Explanation:
The average speed of the car is 31,680 meters per hour.
To calculate the average speed of the car, you need to divide the total distance traveled by the time it took to travel that distance.
Given:
Time taken (t) = 10 minutes = 10 minutes × (1 hour / 60 minutes) = 10/60 hours = 1/6 hours
Distance traveled (d) = 5,280 meters
Average Speed (v) = Distance (d) / Time (t)
Average Speed (v) = 5280 meters / (1/6) hours
To simplify, when you divide by a fraction, it's equivalent to multiplying by its reciprocal:
Average Speed (v) = 5280 meters × (6/1) hours
Average Speed (v) = 31,680 meters per hour
Hence, the average speed of the car is 31,680 meters per hour.
To know more about average speed here
https://brainly.com/question/17661499
#SPJ2
Which of the following statements is true vibrations ?
Answer:
C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.
Explanation:
The complete question is
Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.
Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.
Answer:
Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."
Explanation:
My options were:
A) Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.
B) Resonance occurs as a result of sympathetic vibrations.
C) A non-vibrating object can begin to vibrate as a result of forced vibrations.
D) Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.
A is correct
.
A cylinder rotating about its axis with a constant angular acceleration of 1.6 rad/s2 starts from rest at t = 0. At the instant when it has turned through 0.40 radian, what is the magnitude of the total linear acceleration of a point on the rim (radius = 13 cm)?
a. 0.31 m/s^2
b. 0.27 m/s^2
c. 0.35 m/s^2
d. 0.39 m/s^2
e. 0.45 m/s^2
Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;
[tex]a_t = ar[/tex]
where;
a is the angular acceleration and
r is the radius of the circular path
[tex]a_t = ar\\\\a_t = 1.6 *0.13\\\\a_t = 0.208 \ m/s^2[/tex]
Determine time of the rotation;
[tex]\theta = \frac{1}{2} at^2\\\\0.4 = \frac{1}{2} (1.6)t^2\\\\t^2 = 0.5\\\\t = \sqrt{0.5} \\\\t = 0.707 \ s\\\\[/tex]
Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration
[tex]a_r = \omega ^2r\\\\a_r = 1.131^2 (0.13)\\\\a_r = 0.166 \ m/s^2[/tex]
The magnitude of total linear acceleration is given by;
[tex]a = \sqrt{a_t^2 + a_r^2} \\\\a = \sqrt{0.208^2 + 0.166^2} \\\\a = 0.266 \ m/s^2\\\\a = 0.27 \ m/s^2[/tex]
Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
