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ra1l [238]
3 years ago
11

Starting from rest and moving in a straight line, a cheetah

Physics
1 answer:
UNO [17]3 years ago
3 0

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of  motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

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Determine the nuclear radius (in fm) for each of the following nuclei.
tekilochka [14]

Answer:

(a) 2.75 fm

(b) 2.89 fm

(c) 4.70 fm

(d) 7.12 fm

Explanation:

For a given element, the radius r of its nuclei is given by;

r = r₀A^{(1/3)}

Where;

A = Atomic mass of the element

r₀ = 1.2 x 10⁻¹⁵m = 1.2fm

Now let's solve for the given elements

(a) ¹²₆C

Carbon element => This has an atomic mass number of 12

Therefore its radius is given by;

r = 1.2  x 12^{1/3}

r = 1.2 x 2.29

r = 2.75 fm

(b) ¹⁴₇N

Nitrogen element => This has an atomic mass number of 14

Therefore its radius is given by;

r = 1.2  x 14^{1/3}

r = 1.2 x 2.41

r = 2.89 fm

(c) ⁶⁰₂₇Co

Cobalt element => This has an atomic mass number of 60

Therefore its radius is given by;

r = 1.2  x 60^{1/3}

r = 1.2 x 3.92

r = 4.70 fm

(d) ²⁰⁸₈₂Pb

Lead element => This has an atomic mass number of 208

Therefore its radius is given by;

r = 1.2  x 208^{1/3}

r = 1.2 x 5.93

r = 7.12 fm

4 0
2 years ago
You are an industrial engineer with a shipping company. As part of the package- handling system, a small box with mass 1.60 kg i
Kisachek [45]

Answer:

v = 0.84m/s, v(max)= 0.997m/s

Explanation:

Initial work done by the spring, where c is the compression = 0.28m:

W_s = \frac{1}{2}kc^2

Work lost to friction:

W_f =\mu mg(c-x)

Energy:

E = W_s-W_f=\frac{1}{2}kc^2 -\mu mg(c-x)=\frac{1}{2}mv^2+\frac{1}{2}kx^2

(a) Solve for v:

v=\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}

(b) Solve \frac{dv}{dx}=0 for x:

\frac{dv}{dx}=\frac{\mu g-\frac{k}{m}x}{\sqrt{\frac{k}{m}(c^2-x^2)-2\mu g(c-x)}}

\frac{dv}{dx}=0 if:

\mu g-\frac{k}{m}x = 0

x_{max} = \frac{\mu gm}{k}

v_{max}=\sqrt{\frac{k}{m}c^2-2\mu gc+(\mu g)^2\frac{m}{k}}

6 0
3 years ago
What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
EleoNora [17]

Below is an attachment containing the solution.

4 0
2 years ago
A student lifts a physical science book off the table and above their head. Is there work being done?
DiKsa [7]

Answer: A force must cause a displacement in order for work to be done. A book falls off a table and free falls to the ground. Yes. This is an example of work.

Explanation:

7 0
2 years ago
An empty 230 kg elevator accelerates upward
elena-s [515]

Answer:

7.2 as used in the equation

3 0
3 years ago
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