Answer:
The car travels the distance of 225m before coming to rest.
Explanation:
v = 45m/s
t = 5s
Therefore,
d = v*t
= 45*5
= 225m
Help pleaseeeeeeeeeeeeee
Answer:
hello the answer is 47m/s
A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.
Calculate:
Area of the largest face
Answer:
1.5
x 1.0
1.50
x 0.5
075.00
answer: 75.00m
Explanation:
I hope this help
Please show steps as to how to solve this problem
Thank you!
Explanation:
Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:
[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]
[tex] -m_1gx + m_2gd_2 = 0[/tex]
[tex]m_1x = m_2d_2[/tex]
Solving for x,
[tex]x = \dfrac{m_2}{m_1}d_2[/tex]
[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]
11. An object moves in circular path with constant speed
a. Is the object's velocity constant? Explain.
b. Is its acceleration constant? Explain.
Answer:
B. Is its acceleration constant
Explanation:
Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.
the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is
Answer:
The work done on the block by the spring as it accelerates the block is 4kx².
Explanation:
Let initial distance is x.
It was compressed three times farther and then the block is released, new distance is 3x.
The work done in compressing the spring is given by :
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]
[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]
So, the work done on the block by the spring as it accelerates the block is 4kx².
The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?
First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.
This was something from logic.
The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.
How are Newton’s 1 and 2 law related?
Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitude.
Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Explanation:
Let us assume that
[tex]q_{1} = q_{2} = +q[/tex]
[tex]q_{3} = -q[/tex]
As [tex]q_{3}[/tex] is the negative charge and placed midway between two equal positive charges ([tex]q_{1}[/tex] and [tex]q_{2}[/tex]).
Total distance between [tex]q_{1}[/tex] and [tex]q_{2}[/tex] is 2r. This means that the distance between [tex]q_{1}[/tex] and [tex]q_{3}[/tex], [tex]q_{2}[/tex] and [tex]q_{3}[/tex] = d = r
Now, force action on charge [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})[/tex]
where,
k = electrostatic constant = [tex]9 \times 10^{9} Nm^{2}/C^{2}[/tex]
Substitute the values into above formula as follows.
[tex]F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})[/tex] ... (1)
Similarly, force acting on [tex]q_{3}[/tex] due to [tex]q_{1}[/tex] is as follows.
[tex]F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\[/tex] ... (2)
As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge [tex]q_{3}[/tex] is zero.
Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.
Una pelota se lanza verticalmente hacia arriba desde la azotea de un edificio con una velocidad inicial de 35 m/s. Si se detiene en el aire a 200 m del suelo, ¿Cuál es la altura del edificio?
a. 138,8 m
b. 51.2 m
c. 71,2 m
d. 45,0 m
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons are pushed the same distance into the same electric field (but far enough apart that they don't effect eachother). What is the electrical potential of one of the electrons now?
Answer:
0.5 V
Explanation:
The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.58 g
Answer:
w = 1,066 rad / s
Explanation:
For this exercise we use Newton's second law
F = m a
the centripetal acceleration is
a = w² r
indicate that the force is the mass of the body times the acceleration
F = m 0.58g = m 0.58 9.8
F = 5.684 m
we substitute
5.684 m = m w² r
w = [tex]\sqrt{5.684/r}[/tex]
To finish the calculation we must suppose a cylinder radius, suppose it has r = 5 m
w = [tex]\sqrt{ 5.684/5}[/tex]
w = 1,066 rad / s
Distillation is the separation of multiple Choose... components based on their different Choose... . As the mixture is heated and the first component Choose... , its Choose... form travels through the distillation set-up and Choose... into a different container.
Answer:
Explanation:
Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container
Which one of the following is not an example of convection? An eagle soars on an updraft of wind. A person gets a suntan on a beach. An electric heater warms a room. Smoke rises above a fire. Spaghetti is cooked in water.
Answer: The statement that is not an example of convection is (A person gets a suntan on a beach).
Explanation:
There are different modes of heat energy transfer which includes:
--> conduction
--> Radiation and
--> Convection
CONVECTION is a process by which heat energy is transferred in a fluid or air by the actual movement of the heated molecules. The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed.
Convection in gases is very common and gas expands more than liquid when subjected to high temperature.
--> it is used in bringing about the circulation of fresh air in the room in a process known as ventilation.Here, cool air is constantly being replaced with denser air ( warm air).
-->An electric heater warms a room and Smoke rises above a fire are typical example of convection in gases.
-->Spaghetti is cooked in water: As the water close to the burner warms, it rises to the top and boils. At the same time, cooler water on top moves downward to replace the rising hot water.
--> also the eagle uses convection current to stay afloat in the sky without flapping its wings to conserve energy.
But the option (A person gets a suntan on a beach) is an example of heat transfer through radiation. This is because the sun emits it's rays from the sky down to earth without any material medium unlike others. Therefore, this option is the ODD one out.
Una cuerda horizontal tiene una longitud de 5 m y masa de 0,00145 kg. Si sobre esta cuerda se da un pulso generando una longitud de onda de 0,6 m y una frecuencia de 120 Hz. La tensión a la cual está sometida la cuerda es:
a. 1,5 N
b. 15,0 N
c. 3,1 N
d. 5,2 N
Answer:
Option (A) is correct.
Explanation:
A horizontal rope has a length of 5 m and a mass of 0.00145 kg. If a pulse occurs on this string, generating a wavelength of 0.6 m and a frequency of 120 Hz. The tension to which the string is subjected is
mass of string, m = 0.00145 kg
Frequency, f = 120 Hz
wavelength = 0.6 m
Speed = frequency x wavelength
speed = 120 x 0.6 = 72 m/s
Let the tension is T.
Use the formula
[tex]v =\sqrt\frac{T L}{m}\\\\72 = \sqrt\frac{T\times 5}{0.00145}\\\\T = 1.5 N[/tex]
Option (A) is correct.
Find the current in the thin straight wire if the magnetic field strength is equal to 0.00005 T at distance 5 cm.
Answer:
Answer
Correct option is
A
5×10
−6
tesla
I=5A
x=0.2m
Magnetic field at a distance 0.2 m away from the wire.
B=
2πx
μ
0
I
=
2π×0.2
4π×10
−7
×5
=10×5×10
−7
=5×10
−6
tesla
A caris initially at rest starts moving with a constant acceleration of 0.5 m/s2 and travels a distance of 5 m. Find
(i) Final velocity
(ii)The time taken
Answer:
(I)
[tex] { \bf{ {v}^{2} = {u}^{2} - 2as }} \\ {v}^{2} = {0}^{2} - (2 \times 0.5 \times 5) \\ {v}^{2} = 5 \\ { \tt{final \: velocity = 2.24 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{v = u + at}} \\ 2.24 = 0 + (0.5t) \\ { \tt{time = 4.48 \: seconds}}[/tex]
Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 412 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 412 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
Required:
Find ÎEint, 1, Q1, and W1 for the process at constant volume.
Answer:
ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
Explanation:
Given the data in the question;
T[tex]_i[/tex] = 300 K, T[tex]_f[/tex] = 412 K, n = 0.7 mol
since helium is monoatomic;
Cv = (3/2)R, Cp = (5/2)R
W₁ = 0 J [ at constant volume or ΔV = 0]
Now for the first cylinder; from the first law of thermodynamics;
Q₁ = ΔE[tex]_{int[/tex],₁ + W₁
Q₁ = ΔE[tex]_{int[/tex],₁ = n × Cv × ΔT
we substitute
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × ( 3/2 )8.314 × ( 412 - 300 )
Q₁ = ΔE[tex]_{int[/tex],₁ = 0.7 × 12.471 × 112
Q₁ = ΔE[tex]_{int[/tex],₁ = 977.7 J
Therefore, ΔE[tex]_{int[/tex],₁ = 977.7 J , Q₁ = 977.7 J and W₁ = 0 J
true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields
Answer:
True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field
b) Two skaters collide and grab on to each other on a frictionless ice. One of them, of mass 80 kg, is moving to the right at 5.0 m/s, while the other of mass 70 kg is moving to the left at 2.0 m/s. What are the magnitude and direction of the two skaters just after they collide
Answer:
The two skaters move with a speed of 1.73 m/s after the collision in the right direction.
Explanation:
Given that,
The mas of skater 1, m₁ = 80 kg
The speed of skater 1, u₁ = 5 m/s (right)
The mass of skater 2, m₂ = 70 kg
The speed of skater 2, u₂ = -2 m/s (left)
Let v is the magnitude of the two skaters just after they collide. They must have a common speed. So, using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}[/tex]
Put all the values,
[tex]v=\dfrac{80(5)+70(-2)}{(80+70)}\\\\=1.73m /s[/tex]
So, the two skaters move with a speed of 1.73 m/s after the collision in the right direction.
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigger than the object and the screen is 5 m away from the mirror as shown in fig 5.2, calculate the focal length of the mirror.
Answer:
f = 1 m
Explanation:
The magnification of the lens is given by the formula:
[tex]M = \frac{q}{p}[/tex]
where,
M = Magnification = 4
q = image distance = 5 m
p = object distance = ?
Therefore,
[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]
Now using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]
f = 1 m
Which image illustrates reflection?
A
B
с
D
Answer: I beleive A
Explanation:
Answer:
A
Explanation:
We can see the light being reflected off the mirror.
A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).
Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2
Answer:
The pressure is 2505 Pa.
Explanation:
Height, h = 1.5 m
density of blood, d = 1000 kg/cubic meter
Gravity, g = 1.67 m/s^2
let the pressure is P.
The pressure due to the fluid is given by
P = h d g
P = 1.5 x 1000 x 1.67
P = 2505 Pa
10 A turning pork creates sound cares
with
Frequency of 170Hz: To the
speed of sound in is in 340mls
calculate the wave
wave length
of
in air is
the sound wales.
Answer:
2m
Explanation:
wavelength=speed/frequency
=340/170
=2m
Characteristics or properties of matter or energy that can be measured
Answer:
Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:
color (intensive)
density (intensive)
volume (extensive)
mass (extensive)
boiling point (intensive): the temperature at which a substance boils
melting point (intensive): the temperature at which a substance melts
Explanation:
Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle θ if the resultant force is directed vertically upward.
Answer:
how to solve this problem ???????
The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.
Resultant of the two forces
The resultant of the two forces is determined by resolving the force into x and y component as shown below;
[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]
where;
F1 = 500 NF2 = 600 NValue of Angle θThe value of Angle θ is determined from equation (1)
-500sinθ + 600sin(30) = 0
500sinθ = 600sin(30)
500sinθ = 300
sinθ = 3/5
θ = 36.87⁰
Resultant of the two forcesThe resultant of the forces is determined using the second equation;
500cosθ + 600cos(30) = R
500 x cos(36.87) + 600 x cos(30) = R
919.6 N = R
Learn more about resultant forces here: https://brainly.com/question/25239010
A 2090-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that its vertical velocity as a function of time is given by v(t) =At+Bt^2 , where A and B are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50m/s 2 at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m/s. (a) Determine A and B , including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket’s weight. (d) What was the initial thrust due to the fuel?
Answer:
a) A = 1.50 m / s², B = 1.33 m/s³, b) a = 12.1667 m / s²,
c) I = M (1.5 t + 1.333 t²) , d) ΔI = M 2.833 N
Explanation:
In this exercise give the expression for the speed of the rocket
v (t) = A t + B t²
and the initial conditions
a = 1.50 m / s² for t = 0 s
v = 2.00 m / s for t = 1.00 s
a) it is asked to determine the constants.
Let's look for acceleration with its definition
a = [tex]\frac{dv}{dt}[/tex]
a = A + 2B t
we apply the first condition t = 0 s
a = A
A = 1.50 m / s²
we apply the second condition t = 1.00 s
v = 1.5 1 + B 1²
2 = 1.5 + B
B = 2 / 1.5
B = 1.33 m/s³
the equation remains
v = 1.50 t + 1.333 t²
b) the acceleration for t = 4.00 s
a = 1.50 + 1.333 2t
a = 1.50 + 2.666 4
a = 12.1667 m / s²
c) The thrust
I = ∫ F dt = p_f - p₀
Newton's second law
F = M a
F = M (1.5 + 2 1.333 t) dt
we replace and integrate
I = M ∫ (1.5 + 2.666 t) dt
I = 1.5 t + 2.666 t²/2
I = M (1.5 t + 1.333 t²) + cte
in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant
cte = 0
I = M (1.5 t + 1.333 t²)
d) the initial push
For this we must assume some small time interval, for example between
t = 0 s and t = 1 s
ΔI = I_f - I₀
ΔI = M (1.5 1 + 1.333 1²)
ΔI = M 2.833 N
a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?
Answer:
Look at explanation
Explanation:
a)Only force acting on the object is gravity, so a=-g (consider up to be positive)
use: v^2=v0^2+2a(y-y0)
plug in givens, at max height v=0
0=400-19.6(H)
Solve for H
H= 20.41m
b) Use: y=y0+v0t+1/2at^2
Plug in givens
0=0+20t-4.9t^2
solve for t
t=4.08 seconds
c) v=v0+at
v=20-39.984= -19.984m/s
A 10-cm-long spring is attached to the ceiling. When a 2.0 kg mass is hung from it, the spring stretches to a length of 15 cm.What is the spring constant k?How long is the spring when a 4.0 kg mass is suspended from it?
As the spring is stretched, it exerts an upward restoring force f. At maximum extension, Newton's second law gives
∑ F = f - mg = 0 ==> f = (2.0 kg) (9.8 m/s²) = 19.6 N
By Hooke's law, if k is the spring constant, then
f = kx ==> k = f/x = (19.6 N) / (0.15 m) ≈ 130 N/m
A 4.0 kg mass would cause the spring to exert a force of
f = (4.0 kg) (9.8 m/s²) = 39.2 N
which would result in the spring stretching a distance x such that
39.2 N = (130 N/m) x ==> x = (39.2 N) / (130 N/m) ≈ 0.30 m ≈ 30 cm
If I could lift up to ten tons and I threw a ball the size of an orange but weighed a ton, to the ground, how big of an impact would it make? And could you also show me the equation to solve similar problems myself. Thank you.
Answer:
The impact force is 98000 N.
Explanation:
mass = 10 tons
The impact force is the weight of the object.
Weight =mass x gravity
W = 10 x 1000 x 9.8
W = 98000 N
The impact force is 98000 N.
If there are no other changes, explain what effect reducing the mass of the car will have on its acceleration when starting to move.
Answer:
when the mass of an object is decreased, the acceleration will increase
when mass is increased, acceleration decreases
