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laiz [17]
2 years ago
7

Please help and thank you

Physics
1 answer:
Mashutka [201]2 years ago
8 0

Answer:

B. C3H8

Chemical Formula for propane is C3H8

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The mass of the boy is 35 kilograms. What is the force of friction that slowed him down?
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Answer:

Newton's F=ma, which means the force (F) acting on an object is equal to the mass (m) of an object times its acceleration (a)

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2 years ago
At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr
svp [43]
The acceleration of the particle at time t is:
a(t)=24t^2 ft/s^2
The velocity of the particle at time t is given by the integral of the acceleration a(t):
v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s
and the position of the particle at time t is given by the integral of the velocity v(t):
x(t)=\int v(t) = \int (8t^3)=8  \frac{t^4}{4}=2t^4 ft

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
x(2 s)=2 t^4 = 2 (2s)^4=32 ft
5 0
3 years ago
Which term defines the energy of motion?
Korvikt [17]
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Why do electric fish need to force electric charges to move?
Anastasy [175]
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3 0
3 years ago
A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h
Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    \tau = I \times \alpha

and,       I = \sum mr^{2}

So,      I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}

                       = 4 kg m^{2}

      \tau_{1} = 4 kg m^{2} \times \alpha_{1}

     \tau_{2} = I_{2} \alpha_{2}

So,      I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}

                     = 1 kg m^{2}

 as \tau_{2} = I_{2} \alpha_{2}

                   = 1 kg m^{2} \times \alpha_{2}        

Hence,     \tau_{1} = \tau_{2}

                  4 \alpha_{1} = \alpha_{2}

            \alpha_{1} = \frac{1}{4} \alpha_{2}

Thus, we can conclude that the new rotation is \frac{1}{4} times that of the first rotation rate.

8 0
3 years ago
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