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3 years ago

Knowing Newton’s 2nd Law, how would you rearrange it to solve for acceleration.

Physics

2 answers:

AysviL [449]3 years ago

8 0

Answer:

Force / mass

Explanation:

Divide mass on both sides to get acceleration by itself leaving you with mass below force hence divide force by mass

mezya [45]3 years ago

4 0

Newton’s 2nd : Force = rate of change of momentum
= (mv - mu) / t
= m (v-u) / t
= m a

( since a = (v-u)/t )

From F = m a then a= F/m

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rodikova [14]

Answer:

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Explanation:

7 0

3 years ago

Read 2 more answers

A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric

vichka [17]

Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

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3 years ago

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A little boy is standing at the edge of a cliff 1000 m high. He throws a ball straight downward at an initial speed of 20 m/s, a

marin [14]

Answer:

The ball will be at 700 m above the ground.

Explanation:

We can use the following kinematic equation

$y(t) = \ y_0 \ + \ v_0 \ t \ + \frac{1}{2} \ a \ t^2$ .

where y(t) represent the height from the ground. For our problem, the initial height will be:

$y_0 \ = \ 1000 m$ .

The initial velocity:

$v_0 = - 20 \frac{m}{s}$ ,

take into consideration the minus sign, that appears cause the ball its thrown down. The same minus appears for the acceleration:

$a=-10\frac{m}{s}$

So, the equation for our problem its:

$y(t) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ t \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ t^2$ .

Taking t=6 s:

$y(6 \ s) = \ 1000 m \ - \ 20 \ \frac{m}{s} \ * \ 6 \ s \ - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ (6 s)^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - \frac{1}{2} \ 10 \frac{m}{s^2} \ * \ 36 s^2$ .

$y(6 \ s) = \ 1000 m \ - 120 m - 180 m$ .

$y(6 \ s) = \ 1000 m \ - 300 m$ .

$y(6 \ s) = \ 700 m$ .

So this its the height of the ball 6 seconds after being thrown.

6 0

3 years ago

Determine the amount of potential energy of a 5N book that is 1.5m high on a shelf.

Alex73 [517]

Answer:

Potential energy of book = 7.5 J

Explanation:

Given:

Weight of book = 5 N

Height of shelf = 1.5 meter

Find:

Potential energy of book

Computation:

Weight = Mass x Acceleration of gravity

Mass x Acceleration of gravity = 5 N

Potential energy = Mass x Acceleration of gravity x Height

Potential energy of book = Mass x Acceleration of gravity x Height

We know that;

Mass x Acceleration of gravity = 5 N

So,

Potential energy of book = 5 x 1.5

Potential energy of book = 7.5 J

3 0

3 years ago

A packet is dropped from a stationary helicopter, hovering at a height 'h' from the ground level, reaches the ground in 12s. Cal

Ksju [112]

Use kinematic equations to solve:

1) yf = yo + vo*t + 1/2at²

yf = final height
yo = initial height
vo = initial velocity
a = acceleration
t = time

yf - yo = vo*t + 1/2at²

yf - yo = h

vo = 0

Thus,

h = 1/2at²

h = 1/2(9.8)(12)² = 705.6 m

2) vf = vo + at

vo = 0

Thus,

vf = at

vf = (9.8)(12) = 117.6 m/s

4 0

3 years ago