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kompoz [17]
4 years ago
14

Which of the following best describes to use for non-renewable resource?

Physics
1 answer:
iogann1982 [59]4 years ago
3 0

Answer:

the answer is the last one

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4 years ago
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Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd
Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0
4 years ago
A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoi
Pavlova-9 [17]

Answer:

v(1.5)=0.7648\ m/s

Explanation:

<u>Dynamics</u>

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

\displaystyle a=\frac{F}{m}

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

F=2cos1.1t

The variable acceleration is calculated by:

\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}

a=0.8439cos1.1t

The instant velocity is the integral of the acceleration:

\displaystyle v(t)=\int_{t_o}^{t_1}a.dt

\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt

Integrating

\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}

\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)

\boxed{v(1.5)=0.7648\ m/s}

3 0
3 years ago
PLEASE HELP PLEASE HELP ME PLEASE HELP ME ​
klemol [59]

Answer:24

Explanation:

7 0
2 years ago
Consider two thin disks, of negligible thickness, of radiusR oriented perpendicular to thex axis such that the x axis runs throu
Veseljchak [2.6K]

Complete question

The complete question is shown on the first and second uploaded image

Answer:∈

Answer to first question is shown on the second uploaded image.

Part B the Answer is:

The ratio  \frac{R}{a} is evaluated to be 49.99

Explanation:

The explanation is shown on the third ,fourth and fifth image.

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