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olasank [31]
3 years ago
8

Starting from rest, you accelerate at 6.0 m/s/s for 5 seconds. You then travel at constant speed for 10 seconds. What is the con

stant velocity you are moving at after 5 seconds? _______ m/s
Physics
1 answer:
choli [55]3 years ago
7 0

Answer:

6.0 m/s²(5 s) = 30 m/s

Explanation:

Distance traveled in 15 s is

½(6.0)5² + 6.0(5)(10) = 375 m

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Electromagnetic waves are used extensively in modern technology. Many devices are built to emit and/or receive EM waves at a ver
il63 [147K]

Answer: Option D and E

Explanation: Across the electromagnet spectrum, the frequency increases but the wavelength reduces.

This implies that radio waves have the longest wavelength but the smallest frequency, while gamma rays have the shortest wavelength but the highest frequency.

This implies that there is an inverse relationship between wavelength and frequency.

Of all the technologies listed above, wireless internet has the highest frequency hence making it have the shortest wavelength.

Also by comparing values from the data given to us above, it is possible for some wireless internet device to operate within the same frequency range of the microwave.

3 0
3 years ago
a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on
alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

4 0
3 years ago
A high school physics student with a mass of 68.18 KG is sitting in a seat reading this question the magnitude of the force with
love history [14]
The answer to the question is C
3 0
3 years ago
Calculate the change in internal energy of the following system: A balloon is cooled by removing 0.659 kJ of heat. It shrinks on
bixtya [17]
<h2>Answer:</h2>

-310J

<h2>Explanation:</h2>

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

<em>In this case, the system is the balloon and;</em>

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

<em>Substitute these values into equation (i) as follows;</em>

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>

<em />

<em />

5 0
3 years ago
What physics principle is used in radar guns to find the speeds of tennis balls and baseballs at sporting events?
Luda [366]
<h2>Answer: Doppler effect </h2>

Explanation:

A radar gun  (also known as a Doppler radar) uses the Doppler effect when measuring "return echoes" after having sent a microwave signal (a type of electromagnetic radiation).  

In this context the Doppler effect consists of the change in a wave perceived frequency when the emitter of the waves, and the observer move relative to each other.  

In the case of radars, a microwave signal is sent to a target (the tennis or baseball in this case) and then is reflected after "hitting" the target, so that the radar system measures this difference between the sent signal and the reflected signal.

6 0
3 years ago
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