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solniwko [45]
3 years ago
13

How much larger is the acceleration with two people pushing on one side of the cart compared to only one person pushing the cart

?
Physics
1 answer:
il63 [147K]3 years ago
6 0
As we said to the two people name A and the single person nameB.as every person make an acceleration 60ms2.so group A has more acceleration then group B. because in A there are two person so we add 60+60=120ms2. so group A can produce 120ms2 while groupB can produce 60ms2. i hope it will help you.

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Which is a key criterion for the design of an automobile bumper?
Mrac [35]

Answer: The answer is A

Explanation: The bumper is the first part of an automobile to be impacted when in a head-on accident

3 0
3 years ago
If a stereo has a dial that changes the volume of sounds the stereo makes what is the dial doing
vredina [299]
The volume dial is acting as a voltage divider or a variable resistor. 
4 0
4 years ago
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A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, caus
babunello [35]

Answer:

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

Explanation:

The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:

I = m\cdot (\vec{v}_{2} - \vec{v_{1}}) (1)

Where:

I - Impulse, in kilogram-meters per second.

m - Mass, in kilograms.

\vec{v_{1}} - Initial velocity of the hockey park, in meters per second.

\vec{v_{2}} - Final velocity of the hockey park, in meters per second.

If we know that m = 0.2\,kg, \vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right] and \vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right], then the impulse applied by the stick to the park is approximately:

I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]

I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]

The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.  

8 0
3 years ago
Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air
frosja888 [35]

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

f=\frac{v}{\lambda}

where v is the wave's speed and \lambda is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

f=\frac{400 m/s}{2 m}=200 Hz

- underwater, the frequency of the wave is:

f=\frac{1600 m/s}{8 m}=200 Hz

So, the frequency has not changed.

3 0
3 years ago
Read 2 more answers
Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w
amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

1)

  • Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
  • There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

        v = \lambda * f (1)

  • Replacing by the givens, and solving for f, we get:

       f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)

⇒     f = 8.6 Ghz (with two significative figures)

2)

  • Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

       v =\frac{d}{t} (3)

       where v= c= speed of light in vacuum = 3*10⁸ m/s

       d= distance between mountaintops = 52 km = 52*10³ m

  • Solving for t, we get:

       t = \frac{d}{c} = \frac{52e3m}{3e8m/s} = 17.3e-5 sec = 0.173e-3 sec = 0.173 ms (4)

       ⇒  t = 0.2 ms (with two significative figures)

6 0
3 years ago
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