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Assoli18 [71]
2 years ago
14

An RV is traveling 60 km/h along a highway. A boy sitting near the driver of the RV throws a ball to another boy at the back end

of the RV with a speed of 25 km/h. What is the speed of the ball relative to the boys? km/h What is the speed of the ball relative to a stationary observer on the side of the road? km/h
Physics
1 answer:
alina1380 [7]2 years ago
5 0

Answer:

Speed of the ball relative to the boys: 25 km/h

Speed of the ball relative to a stationary observer: 35 km/h

Explanation:

The RV is travelling at a velocity of

v_{RV}=+60 km/h

Here we have taken the direction of motion of the RV as positive direction.

The boy sitting near the driver throws the ball back with speed of 25 km/h, so the velocity of the ball in the reference frame of the RV is

v_B = -25 km/h

with negative sign since it is travelling in the opposite direction relative to the RV. Therefore, this is the velocity measured by every observer in the reference frame of the RV: so the speed measured by the boys is

v = 25 km/h

Instead, a stationary observer outside the RV measures a velocity of the ball given by the algebraic sum of the two velocities:

v = +60 km/h + (-25 km/h) = +35 km/h

So, he/she measures a speed of 35 km/h.

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The wires in a household lamp cord are typically 3.5 mm apart center to center and carry equal currents in opposite directions.
zhuklara [117]

Explanation:

It is given that,

Distance between wires, d = 3.5 mm = 0.0035 m

Power of light bulb, P = 100 W

Potential difference, V = 120 V

(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :

\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}

Power, P = V × I

I=\dfrac{P}{V}=\dfrac{100}{120}=0.83\ A

This gives, \dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times (0.83)^2}{2\pi \times 0.0035}

\dfrac{F}{l}=0.0000393\ N/m

\dfrac{F}{l}=3.93\times 10^{-5}\ N/m

(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.

(c) This force is negligible.

Hence, this is the required solution.

8 0
3 years ago
A weather emergency siren is mounted on a tower, 105 m above the ground. On one hand, it would be a good idea to make the siren
Svetradugi [14.3K]

Answer:

Explanation:

101 dB = 10.1 B.

Maximum intensity of sound allowed = 10.1 B

Intensity of sound in terms of W/m² can be found as follows

log (I / I₀) = 10.1

I / I₀ = 10¹⁰°¹

I = I₀ X 10¹⁰°¹

= 10⁻¹² X  10¹⁰°¹

= 10⁻¹°⁹ W/m²

105 m above the ground the this intensity will be 105² times

intensity at source point = 10⁻¹°⁹ x 105²

= 138.79 W/m²

energy of sound from source

= 4π times

= 4 x 3.14 x 138.79

= 1743.28W/m²

To calculate in terms of decibel :

log 1743.28 / 10⁻¹²

= log 1743.28 +12

= 15.24 B

= 152.4 dB .

152.4 dB .

4 0
3 years ago
The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull
Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>
  • Mass of block 1 = m1
  • Mass of block 2, = m2
  • Height of block 1 above the ground, = h1
  • Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 )  - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 )  = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2

Thus, we can conclude the following before the block m2 reaches the ground;

  • For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
  • For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

5 0
2 years ago
An automobile is traveling away from Jill and towards Jack. The horn is honking, producing a sound wave.
wlad13 [49]

Answer:

a. The sound will travel at the speed to both Jill and Jack

b. Jack

Explanation:

a. Doppler effect describes how the frequency of a sound wave changes with regards to an observer that has relative motion to the sound source;

The Doppler effect is given by the following formula;

For a sound that is moving away, as observed by Jill, we have;

f' = \dfrac{(v - v_0)}{(v + v_s)} \cdot f =  \dfrac{v }{(v + v_s)} \cdot f

For approaching sound, as observed by Jack, we have;

f' = \dfrac{(v + v_0)}{(v - v_s)} \cdot f =  \dfrac{v }{(v - v_s)} \cdot f

Where;

f = The sound wave's actual frequency

f' = The frequency of the moving sound to the observer

v = The speed of the sound wave

v₀ = The observer's velocity = 0

v_s = The velocity of the source (the automobile honking) of the sound wave

From the above equation, we have that the speed of sound, 'v', is the same to both the source, and the observer although the frequency, and therefore, the wavelength of the sound alternatively increases or decreases

b. From the Doppler effect equation, the person who will hear the highest frequency is given by the formula for the frequency when the sound is approaching the observer, which has the lower denominator

Therefore, given that the automobile is travelling towards Jack, jack will hear the higher frequency

5 0
3 years ago
What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?
weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0
2 years ago
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