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Nookie1986 [14]
3 years ago
6

What is the speed of a cheetah if it takes 20 sesonds to run 300 m

Physics
1 answer:
natta225 [31]3 years ago
8 0

the answer to your question is 15 :)

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Anon25 [30]

In several of the questions you've posted during the past day, we've already said that a wave with larger amplitude carries more energy.  That idea is easy to apply to this question.

8 0
4 years ago
Read 2 more answers
How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium
Minchanka [31]

The height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

<h3>Pressure and temperature at equilibrium </h3>

The relationship between pressure and temperature can be used to determine the height risen by the water.

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where;

  • V₁ = AL
  • V₂ = A(L - y)
  • P₁ = Pa
  • P₂ = Pa + ρgh
  • T₁ = 20⁰C = 293 K
  • T₂ = 10⁰ C = 283 k

\frac{PaAL}{T_1} = \frac{(P_a + \rho gh)A(L-y)}{T_2} \\\\\frac{PaL}{T_1} = \frac{(P_a + \rho gh)(L-y)}{T_2} \\\\L-y = \frac{PaLT_2}{T_1(P_a + \rho gh)} \\\\y = L (1 - \frac{PaT_2}{T_1(P_a + \rho gh)})\\\\y = 4.2(1 - \frac{101325 \times 283}{293(101325\  +\  1000 \times  9.8 \times  100)} )\\\\y = 3.8 \ m

Thus, the height risen by water in the bell after enough time has passed for the air to reach thermal equilibrium is 3.8 m.

The complete question is below:

A diving bell is a 4.2 m -tall cylinder closed at the upper end but open at the lower end. The temperature of the air in the bell is 20 °C. The bell is lowered into the ocean until its lower end is 100 m deep. The temperature at that depth is 10°C. How high does the water rise in the bell after enough time has passed for the air to reach thermal equilibrium?

Learn more about thermal equilibrium here: brainly.com/question/9459470

#SPJ4

3 0
2 years ago
A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is
Umnica [9.8K]

Answer:

<h2>Magnitude of the second charge is -4.17*10^{-7}C</h2>

Explanation:

According to columbs law;

F = kq1q2/r^{2}

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

0.03=9*10^{9}*  -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C

6 0
4 years ago
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A 4.0-kilogram ball moving at 8.0 m/s to the right collides with a 1.0-kilogram ball at rest. After the collision, the 4.0-kilog
Illusion [34]
Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
So overall momentum its 32kgm/s
Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball 
3 0
3 years ago
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Acceleration is defined as a change in velocity over time. If a runner increases velocity
Anton [14]

Answer:

a = 0,1[m/s^2]

Explanation:

First we need to indentify the initial data.

And using this kinematic equation we have:

v = 4[m/s]\\v_{0}= 2 [m/s] \\t = 20[s]\\\\v= v_{0}+a*t\\a=\frac{v-v_{0}}{t} \\a= \frac{4-2}{20} \\a=0.1[m/s^{2}]

8 0
4 years ago
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