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3 years ago

10

A car starts from rest and accelerates uniformly over a time of 10.21 seconds for a distance of 210 m. Determine the acceleratio

n of the car.

1 answer:

uysha [10]3 years ago

6 0

Answer:

4.03 m/s/s

Explanation:

d=1/2 a t^2

210 = 1/2 a (10.21)^2

210 = 52.1 a

a = 4.03 m/s/s

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Please help! A net force of 2.0 N acts on a 2.0-kg object for 10 seconds. What is the object’s kinetic energy after that 10 seco

padilas [110]

For Newton's second law, the force F applied to the object of mass m will cause an acceleration a of the body:
$F=ma$
So, the acceleration is
$a= \frac{F}{m}= \frac{2.0 N}{2.0 kg}=1 m/s^2$

The object undergoes through this acceleration for 10 seconds, t=10 s. Since it is an accelerated motion, we can find its final velocity after 10 seconds:
$v_f = v_i + at=0 m/s+(1m/s^2)(10 s)=10 m/s$
where $v_i$ is the initial velocity of the object, which is zero since it starts from rest.

Finally we can calculate the final kinetic energy of the object, which is given by
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3 years ago

Determine the magnitude of the resultant force acting on a 5 −kg particle at the instant t=2 s, if the particle is moving along

Alex787 [66]

Answer:

F = 296.7N

Explanation:

The x and y component of the position vector are given by:

$x(t) =rcos\phi, y(t) = rsin\phi\\ \frac{dx}{dt} =\frac{dr}{dt} cos\phi - rsin\phi\frac{d\phi}{dt} , \frac{dy}{dt}=\frac{dr}{dt}sin\phi + rcos\phi\frac{d\phi}{dt}\\ \frac{d^2x}{dt^2} = \frac{d^2r}{dt^2} cos\phi - \frac{dr}{dt} sin\phi\frac{d\phi}{dt} -\frac{dr}{dt} sin\phi\frac{d\phi}{dt} - r(cos\phi\frac{d^2\phi}{dt^2} + sin\phi\frac{d^2\phi}{dt^2})$

$\frac{d^2y}{dt^2} = \frac{d^2r}{dt^2} sin\phi + \frac{dr}{dt} cos\phi\frac{d\phi}{dt}+\frac{dr}{dt}cos\phi\frac{d\phi}{dt}+r(-sin\phi\frac{d^2\phi}{dt^2} +cos\phi\frac{d^2\phi}{dt^2})$

At t = 2s:

$\phi(2) = 1.5t^2-6t= -6\\ \frac{d\phi}{dt}(2) = 3t-6=0\\ \frac{d^2\phi}{dt^2}=3\\r(2)=2t+10=14\\ \frac{dr}{dt}=2\\\frac{d^2r}{dt^2} = 0$

Plugging in:

$\frac{d^2x}{dt^2}=-42(cos(-6) + sin(-6))=-52\frac{m}{s^2} \\\frac{d^2y}{dt^2} = 42(cos(-6)-sin(-6))=28.6\frac{m}{s^2}$

The resulting force F is:

$F = m\sqrt{(\frac{d^2x}{dt^2})^2 + (\frac{d^2y}{dt^2})^2}=296.7N$

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3 years ago

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This type of energy is called non mechanical energy so the answer is C

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3 years ago

A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?

docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

$v=\sqrt{2gh}$

g is acceleration due to gravity

$v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}$

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

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4 years ago