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Svetach [21]
4 years ago
7

How does kinetic energy affect the stopping distance of small vehicle compared to a large vehicle

Physics
1 answer:
Nadusha1986 [10]4 years ago
4 0
The kinetic energy for a large vehicle is different from that of a smaller vehicle, assuming that the vehicles are travelling at the same speed and stopping the same distance. This is because for a larger vehicle the kinetic energy is higher, as the mass for a larger vehicle, is more than the smaller vehicle.
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Can you help me answer this?
Pavel [41]
The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.
5 0
4 years ago
The maximum current output of a 60 ω circuit is 11 A. What is the rms voltage of the circuit?
bezimeni [28]

Answer:

660V

Explanation:

V=IR

V=11×60

=660V

hope this helps

please like and Mark as brainliest

8 0
4 years ago
Saturn orbits the Sun at a radius of about 1.4×109 km. At this distance the force of gravity on Saturn due to the Sun is 3.7×102
Afina-wow [57]

Explanation:Saturn is the second largest planet of the solar system in mass and size and the sixth nearest planet in distance to the Sun.

In the night sky Saturn is easily visible to the unaided eye as a non twinkling point of light.

8 0
3 years ago
An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i
Diano4ka-milaya [45]

Answer:

It is real, inverted, and smaller than the object.

Explanation:

First of all, we can use the lens equation to find the location of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = 15 cm is the focal length (positive for a converging lens)

p = 50 cm is the distance of the object from the lens

Solving the equation for q,

\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}\\q=\frac{1}{0.047 cm^{-1}}=21.3 cm

The distance of the image from the lens is positive, so we can already conclude that the image is real.

Now we can also write the magnification equation:

{h_i}=-h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Substituting p = 50 cm and q = 21.3 cm, we have

{h_i}=-h_o \frac{21.3 cm}{50 cm}=-0.43 h_o

So from this relationship we observe that:

|h_i| < |h_o| --> this means that the image is smaller than the object, and

h_i < 0 --> this means that the image is inverted

so, the correct answer is

It is real, inverted, and smaller than the object.

4 0
3 years ago
Which type of electromagnetic wave is used to transmit signals for
baherus [9]

Answer:

OB. Radio Waves

Explanation:

Radio Waves are wireless transmission of sound messages or info

8 0
2 years ago
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