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Artyom0805 [142]
3 years ago
7

Consider a compact car that is being driven

Physics
1 answer:
tigry1 [53]3 years ago
6 0
The answer is 2847.830m
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If the horizontal component of a vector is 6 m/s and the vertical component is also 6 m/s, what is the resultant value of the ve
BlackZzzverrR [31]
Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.
6 0
3 years ago
When the force acting on the body equal to acceleration?
topjm [15]

Answer:

Acceleration and velocity Newton's second law says that when a constant force acts on a massive body, it causes it to accelerate, i.e., to change its velocity, at a constant rate. In the simplest case, a force applied to an object at rest causes it to accelerate in the direction of the force.

5 0
2 years ago
If a radio wave carries information by changing the amplitude of the transmitted wave, what kind of broadcast might it possibly
mariarad [96]
Option A because those are the correct ones
4 0
2 years ago
Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude
DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, v_p_i = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

W =K.E_f - K.E_i

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d  = (EQ) x d = EQd

K.E_f =EQd + \frac{1}{2}m_pv_p_i^2

m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C

K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5   \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\

K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0
3 years ago
A heat pump absorbs heat from the atmosphere at a rate of 30 kW. If work is being done to run this heat pump at a rate of 7.7 kW
Volgvan

Answer:

Option D 3.9

Explanation:

First, you need to use the correct equation which is the following:

COP = Q/W

Where:

Q = heat absorbed

W = work done by the pump

COP = coefficient of perfomance

We have all the data, so, all you need to do is replace in the above expression and you shoould get the correct result:

COP = 30 / 7.7

COP = 3.896

This result you can round it to 3.9. option D.

8 0
3 years ago
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