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Over [174]
2 years ago
15

Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball

to hit the ground.(a)Define your coordinate system, be thorough.(b)Write down the given infor-mation, be sure to include hidden information.(c)State what physics principleis at play here. How do you know this?(d)Select an equation.(e)Solve forthe time it takes for the ball to hit the ground.
Physics
1 answer:
Nimfa-mama [501]2 years ago
6 0

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

     t = √(2 125 / 9.8)

     t = 5.05 s

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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
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Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
A police radar gun uses X-band microwave radiation at a frequency of 12.2 GHz. Microwaves travel at the speed of light, or 3x108
quester [9]

Answer:

A police radar gun uses X-band microwave radiation at a frequency of 13.1 GHz. Microwaves travel at the speed of light, or 3x108 m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car.

a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

ANSWER: 2533 Hz

Explanation:

6 0
3 years ago
This diagram shows two different forces acting on a skateboarder. The
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Answer:

B

Explanation:

7 0
3 years ago
Waves transport
Brilliant_brown [7]

Answer:

d)energy

Explanation:

Waves can transfer energy over distance without moving matter the entire distance. For example, an ocean wave can travel many kilometers without the water itself moving many kilometers. The water moves up and down—a motion known as a disturbance. It is the disturbance that travels in a wave, transferring energy.

5 0
2 years ago
A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron
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Answer:

259.62521 seconds

Explanation:

m_1 = Mass of astronaut = 87 kg

m_2 = Mass of wrench = 0.57 kg

v_1 = Velocity of astronaut

v_2 = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s

Time = Distance / Speed

Time=\frac{38.1}{0.14675}=259.62521\ s

The time taken to reach the ship is 259.62521 seconds

4 0
3 years ago
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