Answers
Answer:
hello the answer is 47m/s
Related Questions
A TV satellite dish is designed to receive radio waves of wavelength
0.0644 meters. What is the frequency of the waves it receives? _______GHz
Give your answer in gigahertz (GHz). 1 GHz = 10^9 Hz.
Give your answer to the nearest tenth of a GHz (one place after the decimal). Just enter the number; do NOT use scientific notation.
Answers
Answer:
4.7 GHz
Explanation:
Applying,
v = λf................. Equation 1
Where v = velocity of the radio wave, λ = wavelength, f = frequency
make f the subject of the equation
f = v/λ.............. Equation 2
Note: A radio wave is an electromagnetic wave, as such it moves with a velocity of 3.00 x 10⁸ m/s
From the question,
Given: λ = 0.0644 meters
Constant: v = 3.00 x 10⁸ m/s
Substitute these values into equation 2
f = (3.00 x 10⁸)/0.0644
f = 4.66×10⁹ Hz
f = 4.7 GHz
g How much buoyancy force, in N, a person with a mass of 70 kg experiences by just standing in air
Answers
Answer:
686.7N
Explanation:
Given data
Mass= 70kg
We know that the buoyant force experienced by the person is equal to the weight of the person
Hence the weight is
Weight = mass* Acceleration
Weight= 70*9.81
Weight= 686.7N
Therefore the weight is 686.7N
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn't skid, what is the force exerted on it by static friction?
Answers
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. Projectile A is
fired with a speed of 19.5 m/s at a launch angle 30° of while projectile B is fired with a speed of 19.5 m/s at a launch angle of 60°. How long
it takes one projectile to be directly above the other?
Answers
Let the point where A is launched act as the origin, so that the horizontal positions at time t of the respective projectiles are
• A : x = (19.5 m/s) cos(30°) t
• B : x = 62.2 m + (19.5 m/s) cos(60°) t
These positions are the same at the moment one projectile is directly above the other, which happens for time t such that
(19.5 m/s) cos(30°) t = 62.2 m + (19.5 m/s) cos(60°) t
Solve for t :
(19.5 m/s) (cos(30°) - cos(60°)) t = 62.2 m
t = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))
t ≈ 8.71 s
You are holding one end of a horizontal stretched string. Flicking your wrist will send a pulse down the string. Which actions will make the pulse travel faster
Answers
Answer:
Use a lighter string of the same length, under the same tension.
Stretch the string tighter to increase the tension
Explanation:
The wave speed depends on propertices of the medium, not on how you generate the wave. For a string
Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.
A pulse will be sent down a horizontal extended thread if a person flicks the wrist. To raise the tension, pull the string tighter.
Define pulse.Pulse is the same thing as monitoring heartbeat. The pulse rate could also be measured via auscultation, which includes hearing to the heart rhythm using a stethoscope then counting it for one minute.
A pulse will be sent down a horizontal extended thread if a person holds one end of the rope and flicks their wrist.
To raise the tension, pull the string tighter.
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An object moving along a horizontal track collides with and compresses a light spring (which obeys Hooke's Law) located at the end of the track. The spring constant is 52.1 N/m, the mass of the object 0.250 kg and the speed of the object is 1.70 m/s immediately before the collision.
(a) Determine the spring's maximum compression if the track is frictionless.
?? m
(b) If the track is not frictionless and has a coefficient of kinetic friction of 0.120, determine the spring's maximum compression.
??m
Answers
(a) As it gets compressed by a distance x, the spring does
W = - 1/2 (52.1 N/m) x ²
of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so
W = ∆K
- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.118 m
(b) Taking friction into account, the only difference is that more work is done on the object.
By Newton's second law, the net vertical force on the object is
∑ F = n - mg = 0
where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives
n = mg = 2.45 N
and from this we get the magnitude of kinetic friction,
f = µn = 0.120 (2.45 N) = 0.294 N
Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:
W (friction) = - (0.294 N) x
W (spring) = - 1/2 (52.1 N/m) x ²
==> W (total) = W (friction) + W (spring)
Solve for x :
- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²
==> x ≈ 0.112 m
For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:
a) The spring's maximum compression when the track is frictionless is 0.118 m.
b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.
a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:
[tex] E_{i} = E_{f} [/tex]
[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex] (1)
Where:
[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg
[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s
k: is the spring constant = 52.1 N/m
x: is the distance of compression
After solving equation (1) for x, we have:
[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]
Hence, the spring's maximum compression is 0.118 m.
b) When the track is not frictionless, we can calculate the spring's compression by work definition:
[tex] W = \Delta E = E_{f} - E_{i} [/tex]
[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex] (2)
Work is also equal to:
[tex] W = F*d = F*x [/tex] (3)
Where:
F: is the force
d: is the displacement = x (distance of spring's compression)
The force acting on the object is given by the friction force:
[tex] F = -\mu N = -\mu m_{o}g [/tex] (4)
Where:
N: is the normal force = m₀g
μ: is the coefficient of kinetic friction = 0.120
g: is the acceleration due to gravity = 9.81 m/s²
The minus sign is because the friction force is in the opposite direction of motion.
After entering equations (3) and (4) into (2), we have:
[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]
[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]
[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]
Solving the above quadratic equation for x
[tex] x = 0.112 m [/tex]
Therefore, the spring's compression is 0.112 m when the track is not frictionless.
Read more here:
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Question 4(Multiple Choice Worth 4 points)
(02.04 MC)
Which explanation justifies why the theory of evolution is a theory and not a law?
Predicts an organism's ability to adapt to its environment
It can be expressed as a simple mathematical statement
Explains the existence of diverse forms of life on Earth
O Additional evidence will change the theory into a law
Answers
Answer:
A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)
Explanation:
Stationary waves are
A) transverse waves
B) longitudinal waves
C) mechanical waves
Answers
Answer:
stationary waves are transverse waves
crushing chalk into powder is and irreversible change. is this example a physical or chemical change?Why?
Answers
Answer:
It is a example of physical change
Assuming the atmospheric pressure is 1 atm at sea level, determine the atmospheric pressure at Badwater (in Death Valley, California) where the elevation is 86.0 m below sea level.
Answers
Answer:
Atmospheric pressure at Badwater is 1.01022 atm
Explanation:
Data given:
1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa
Elevation (h) = 86m
gravity (g) = 9.8 m/s2
Density of air P = 1.225 kg/m3
Therefore pressure at bad water Pb = Pi + Pgh
Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)
Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa
hence:
Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the generator has an area of 0.016 m2. If the magnetic field used in the generator has a magnitude of 0.052 T, how many turns of wire are needed
Answers
Answer:
The number of turns of wire needed is 573.8 turns
Explanation:
Given;
maximum emf of the generator, = 190 V
angular speed of the generator, ω = 3800 rev/min =
area of the coil, A = 0.016 m²
magnetic field, B = 0.052 T
The number of turns of the generator is calculated as;
emf = NABω
where;
N is the number of turns
[tex]\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s[/tex]
[tex]N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns[/tex]
Therefore, the number of turns of wire needed is 573.8 turns
what is the average velocity if the initial velocity is at rest and the final velocity is 16 m/s
Answers
Answer:
8m/s
Explanation:
Vavg= 16-0/2=8m/s
12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen, what things could you do to help yourself (before calling me over to assist you?)
Answers
Answer:
See the answer below
Explanation:
After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem
1. Drop a few drops of immersion oil on the slide and view again under high the power objective.
2. If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.
An ice chest at a beach party contains 12 cans of soda at 3.78 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 6.48-kg watermelon at 29.4 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
Answers
Answer:
T = 13.25°C
Explanation:
From the law of conservation of energy:
Heat Lost by Watermelon = Heat Gained by Cans
[tex]m_wC_w\Delta T_w = m_cC_c\Delta T_c[/tex]
where,
[tex]m_w[/tex] = mass of watermelon = 6.48 kg
[tex]m_c[/tex] = mass of cans = (12)(0.35 kg) = 4.2 kg
[tex]C_w[/tex] = specific heat capacity of watermelon = 3800 J/kg.°C
[tex]C_c[/tex] = specific heat capacity of cans = 4200 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temprature of watermelon = 29.4°C - T
[tex]\Delta T_c[/tex] = Change in Temperature of cans = T - 3.78°C
T = final temperature = ?
Therefore,
[tex](4.2\ kg)(3800\ J/kg.^oC)(29.4^oC-T)=(6.48\ kg)(4200\ J/kg^oC)(T-3.78^oC)\\469224\ J-(15960\ J/^oC)T = (27216\ J/^oC)T-102876.48\ J\\469224\ J + 102876.48\ J = (27216\ J/^oC)T+(15960\ J/^oC)T\\\\T = \frac{572100.48\ J}{43176\ J/^oC}[/tex]
T = 13.25°C
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resistance. But driving slowly in dry sand is another story. If a 1500 kg car is driven in sand at 4.9 m/s , the coefficient of rolling friction is 0.060. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
Required:
a. What propulsion force is needed to keep the car moving forward at a constant speed?
b. What power is required for propulsion at 5.0 m/s?
c. If the car gets 15 mpg when driving on sand, what is the car's efficiency? One gasoline contains 1.4×10 ^8 J of chemical energy.
Answers
Answer:
a) [tex]F_p=882N[/tex]
b) [tex]P=4410W[/tex]
c) [tex]V_p'=24135[/tex] ,[tex]n=15.2\%[/tex]
Explanation:
From the question we are told that:
Mass [tex]M=1500kg[/tex]
Velocity [tex]v=4.9m/s[/tex]
Coefficient of Rolling Friction [tex]\mu=0.06[/tex]
a)
Generally the equation for The Propulsion Force is mathematically given by
[tex]F_p=\mu*mg[/tex]
[tex]F_p=0.06*1500*9.81[/tex]
[tex]F_p=882N[/tex]
b)
Therefore Power Required at
[tex]V_p=5.0m/s[/tex]
[tex]P=F_p*V_p[/tex]
[tex]P=882*5[/tex]
[tex]P=4410W[/tex]
c)
[tex]V_p' =15mpg[/tex]
[tex]V_p'=15*\frac{1609}[/tex]
[tex]V_p'=24135[/tex]
Generally the equation for Work-done is mathematically given by
[tex]W=F_p*V_p'[/tex]
[tex]W=882*15*1609[/tex]
[tex]W=2.13*10^7[/tex]
Therefore
Efficiency
[tex]n=\frac{W}{E}*100\%[/tex]
Since
Energy in one gallon of gas is
[tex]E=1.4*10^8J[/tex]
Therefore
[tex]n=\frac{2.1*10^7}{1.4*10^8}*100\%[/tex]
[tex]n=15.2\%[/tex]
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answers
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
The image of an object placed 30cm from a diverging lens is formed 10cm in front of the lens.
Calculate the focal length of the lens.
Answers
Answer:
15cm
Explanation:
Since the lens is a diverging lens, the image distance is negative (virtual)
v = -30cm
u = 10cm
Required
focal length f
Using the lens formula;
1/u + 1/v = 1/u
1/10 - 1/30 = 1/f
(3-1)/30 = 1/f
2/30 = 1/f
f = 30/2
f = 15cm
Hence the focal length of the lens is 15cm
Electrical charges are of two types. True False
Answers
Answer:
Electrical charges r of 2 types its true.they are positive and negative.hope it helps.stay safe healthy and happy..Answer: Think its true
If the cornea is reshaped (this can be surgically done or with contact lenses) to correct myopia, should its curvature be made greater or smaller? Explain. Also, explain how hyperopia can be corrected?
Answers
Answer:
Myopia curvature of the cornea if it is negative the curvatures are positive,
hypermetry,
Explanation:
Myopia is the visual defect that does not allow to see distant objects, which is why it is corrected with a divergent lens so that the image is formed on the retina, therefore, by reforming the curvature of the cornea if it is negative
therefore the curvature must decrease
To correct hypermetry, the curvatures are positive, consequently the curvature of the lens must increase
which vector best represents the net force acting on the +3 C charge
Answers
This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.
Two bar magnet with the north pole of one near the south pole of the second. field lines are leaving the north pole and bent away from the south pole of the other. Field lines are leaving the south pole of one and bending away from the north pole of the other.
Which change will correct the error in the diagram?
a)changing the N to S
b)reversing the arrows on the left to point toward the N
c)changing the S to N
d)reversing the arrows on the right to point toward the S
Answers
Answer:
changing the N to S. that's how the error will be corrected
Answer:
C is the correct answer
Explanation:
i took the test
Please show steps as to how to solve this problem
Thank you!
Answers
Answer:
Torques must balance
F1 * X1 = F2 * Y2
or M1 g X1 = M2 g X2
X2 = M1 / M2 * X1 = 130.4 / 62.3 * 10.7
X2 = 22.4 cm
Torque = F1 * X2 =
62.3 gm* 980 cm/sec^2 * 22.4 cm = 137,000 gm cm^2 / sec^2
Normally x cross y will be out of the page
r X F for F1 will be into the page so the torque must be negative
A rigid tank contains 10 lbm of air at 30 psia and 60 F. Find the volume of the tank in ft3. The tank is now heated until the pressure doubles. Find the heat transfer in Btu.
Answers
Answer:
Hence the amount of heat transfer is 918.75 Btu.
Explanation:
Now,
A magnetohydrodynamic (MHD) drive works by applying a magnetic field to a fluid which is carrying an electric current.
a. True
b. False
Answers
Answer:
True
Explanation:
A magnetohydrodynamic drive or MHD accelerator is a method which is used for propelling the vehicles using only by applying the electric and magnetic fields. It has no moving parts. It accelerates an electrically conductive propellant (liquid or gas) with magnetohydrodynamics.
Its working principle is same as an electric motor except that in an MHD drive, the moving rotor is replaced by the fluid acting directly as the propellant.
An MHD accelerator is reversible.
So, the statement is true.
Newspapers often talk about an energy crisis-about running out of certain energy sources in the not-so-distant future. About which kind of energy sources are they talking
Answers
Answer:
Nonrenewable energy
Explanation:
Renewable energy is also known as clean energy and it can be defined as a type of energy that are generated through natural sources or technology-based processes that are replenished constantly. Some examples of these natural sources are water (hydropower), wind (wind energy), sun (solar power), geothermal, biomass, waves etc.
Basically, a renewable energy source is sustainable and as such can not be exhausted.
On the other hand, a non-renewable energy refers to an energy source such as fossil fuels that takes a very long time to be created or their creation happened long ago and isn't likely to happen again e.g uranium.
For example, fossil fuels such as coal, oil, and natural gas, come from deep inside the Earth where they formed over millions of years ago.
In this scenario, the kind of energy the newspaper sources are talking about is a nonrenewable energy source because they are capable of being exhausted in the not-so-distant future.
One hazard of space travel is the debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a 0.100-mg chip of paint that strikes a spacecraft window at a relative speed of 4.00×10^3 m/s, given the collision lasts 6.00×10^8s.
Answers
Answer:
F = 6666.7 N
Explanation:
Given that,
Mass of a chip, m = 0.1 mg
Initial speed, u = 0
Final speed,[tex]v=4\times 10^{3}\ m/s[/tex]
Time of collision,[tex]t=6\times 10^{-8}\ s[/tex]
We know that,
Force, F = ma
Put all the values,
[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.1\times 10^{-6}\times (4\times 10^3-0)}{6\times 10^{-8}}\\\\F=6666.7\ N[/tex]
So, the required force is 6666.7 N.
A metre rule is used to measure the length of a piece of string in a certain experiment. It is found to be 20 cm long to the nearest millimeter. How should thisresult be recorded in a table of results? a. 0.2000m b. 0.200m c. 0.20m d. 0.2m
Answers
Answer:
C
Explanation:
20 cm = 0.2m
since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty
therefore it's 0.200m
helppp!!! what's the answer to this??
when an ideal capacitor is connected across an ac voltage supply of variable frequency, the current flowing
a) is in phase with voltage at all frequencies
b) leads the voltage with a phase independent of frequency
c) leads the voltage with a phase which depends on frequency
d) lags the voltage with a phase independent of frequency
what would be the correct option?
Answers
Answer:
(b)
Explanation:
The voltage always lags the current by 90°, regardless of the frequency.
two point charges two point charges are separated by 25 cm in the figure find The Net electric field these charges produced at point a and point b
Answers
The image is missing and so i have attached it.
Answer:
A) E = 8740 N/C
B) E = -6536 N/C
Explanation:
The formula for electric field is;
E = kq/r²
Where;
q is charge
k is a constant with value 8.99 x 10^(9) N•m²/C²
A) Now, to find the net electric field at point A, the formula would now be;
E = (kq1/(r1)²) - (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point A
r2 is distance from charge q2 to point A.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 25 cm - 10 cm = 15 cm = 0.15 m
r2 = 10 cm = 0.1 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.15^(2)) - ((-12.5 × 10^(-9))/0.1^(2))
E = 8740 N/C
B) similarly, electric field at point B;
E = (kq1/(r1)²) + (kq2/(r2)²)
Where;
r1 is distance from charge q1 to point B
r2 is distance from charge q2 to point B.
q1 = -6.25 nC = -6.25 × 10^(-9) C
q2 = -12.5 nC = -12 5 × 10^(-9) C
From the attached image, r1 = 10 cm = 0.1 m
r2 = 25cm + 10 cm = 35 cm = 0.35 m
Thus;
E = (8.99 x 10^(9)) × ((-6.25 × 10^(-9))/0.1^(2)) + ((-12.5 × 10^(-9))/0.35^(2))
E = -6536 N/C
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