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3 years ago

What is the best inference for the speed of the car after 3 seconds

Physics

1 answer:

SVEN [57.7K]3 years ago

8 0

Answer:

B. 17m/s

Explanation:

This question contains a graph that illustrates the relationship between the speed of a car over time. The graph shows that one can make an inference of the amount of time it takes for the car to cover a particular speed and vice versa.

In this case, after 3 seconds, the speed of the car will be 17 m/s. This inference was got by tracing the position of 3s in the x-axis to the value on the y-axis. Doing this, the best inference for the speed of the car after 3 seconds is 17m/s.

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A 2800 kg truck moving at 12 m/s to the right hits a stopped 1100 kg car. What is the combined velocity the moment they stick to

leva [86]

Answer:

The combined velocity is 8.61 m/s.

Explanation:

Given that,

The mass of a truck, m = 2800 kg

Initial speed of truck, u = 12 m/s

The mass of a car, m' = 1100 kg

Initial speed of the car, u' = 0

We need to find the combined velocity the moment they stick together. Let it is V. Using the conservation of momentum.

$m_1v_1+m_2v_2=(m_1+m_2)V\\\\V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}\\\\V=\dfrac{2800\times 12+0}{2800+1100}\\\\V=8.61\ m/s$

So, the combined velocity is 8.61 m/s.

5 0

3 years ago

If the force applied to an object is not greater than the starting friction, what will happen to the object?

bonufazy [111]

Answer:

Explanation:

the object will not move as the force exerted is not sufficient enough to overcome its force of friction

3 0

3 years ago

Read 2 more answers

A rocky space object of varying size

kow [346]

That would be an asteroid

3 0

4 years ago

An object with velocity 141 ft/s has a kinetic energy of 1558.71 ftâˆ™lbf, on a planet whose gravity is 31.5 ft/s2. What is its

Sidana [21]

Answer:

The mass of the object is 5.045 lbm.

Explanation:

Given;

kinetic energy of the object, K.E = 1558.71 ft.lbf

velocity of the object, V = 141 ft/s

The kinetic energy of the object is calculated as;

$K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }$

$m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm$

Therefore, the mass of the object is 5.045 lbm.

6 0

3 years ago

Why is it important to use placebos and a double-blind approach in some studies

viktelen [127]

Another name for these two words is "constant" and you want to have a "constant", because you want something to compare your experimental group to, to see whether data had changed or not. So you have placebos or a double- blind to compare your experimental group to it and also so you know you don't have a bias or anything in the study.

6 0

3 years ago