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2 years ago

4. What is the acceleration of the car in each section? b с d a

Physics

2 answers:

PilotLPTM [1.2K]2 years ago

4 0

Acceleration is the slope of a velocity time graph.

Units are all m/s ^2

a) 3
b) 0
c) -3
d) 1 - doesn’t LOOK like a slope of 1, but if you look carefully at the Y axis, you can see it rises by 4 rather than 2.

makvit [3.9K]2 years ago

3 0

Answer:

0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.

Explanation:

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What potential difference is needed to give a helium nucleus (q=2e) 85.0 kev of kinetic energy?

Vilka [71]

The kinetic energy K given to the helium nucleus is equal to its potential energy, which is
$E=q \Delta V$
where q=2e is the charge of the helium nucleus, and $\Delta V$ is the potential difference applied to it.
Since we know the kinetic energy, we have
$E=K=85~keV=q \Delta V$
and from this we can find the potential difference:
$\Delta V = \frac{K}{q}= \frac{85~keV}{2e}=42.5~kV$

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3 years ago

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3 years ago

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The Earth and Moon are separated by about 4.0 10^6 m. Suppose Mars is [{MathJax fullWidth='false' 2.9 \times 10^{11} }] m from E

Shalnov [3]

Answer:

a) α=7.9x10^-4 rad

b) θ=1.12x10^-4 rad

c) The Earth and the Moon cannot be seen without a telescope.

Explanation:

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3 years ago

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AURORKA [14]

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3 years ago

A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the

Serjik [45]

The elastic potential energy of a spring is given by
$U= \frac{1}{2}kx^2$
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
$W=-\Delta U = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$

In our problem, initially the spring is uncompressed, so $x_i=0$ . Therefore, the work done by the spring when it is compressed until $x_f$ is
$W=- \frac{1}{2}kx_f^2$
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.

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3 years ago