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3 years ago

A 0.15 kg baseball is pushed with 100 N force. what will its acceleration be?

Physics

1 answer:

sammy [17]3 years ago

8 0

Answer:

The acceleration of the ball is 666.67 m/s²

Explanation:

It is given that,

Mass of the baseball, m = 0.15 kg

Applied force to it, F = 100 N

We need to find the acceleration of the ball. It can be calculated using Newton's second law of motion as :

F = ma

$a=\dfrac{F}{m}$

$a=\dfrac{100\ N}{0.15\ kg}$

$a=666.67\ m/s^2$

So, the acceleration of the ball is 666.67 m/s². Hence, this is the required solution.

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Recall the topography of the Sierra Nevada Mountains east of California's central valley. Which of the following locations would

marysya [2.9K]

Answer:

3. western slope

Explanation:

If we investigate the physical features (such as natural and artificial features) of the Sierra Nevada Mountains located at the California's central valley eastern part, the western slope has the highest value of the precipitation level yearly. This is due to the location of the slope and the relevant conditions such as temperature and humidity.

7 0

3 years ago

A car with a mass of 2000 kg is moving around a circular curve at a uniform velocity of 25 m/s per second. The curve has a radiu

IrinaK [193]

The centripetal force is:
F = mv² / R
Where:
m: mass of the object
v: object speed
R: radius of the curve.
We have to:
m = 2000kg
v = 25 m / s
R = 80 meters.
Then the centripetal force acting on the vehicle is:
F = (2000kg * (25m / s) ²) / 80m
F = 15625 N

4 0

3 years ago

The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa

kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0

2 years ago

A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3>the ball is in flight before striking the ground for 2.26seconds.</h3>

b. How far from the building does the ball strike the ground?

Horizontal range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3>The ball strikes 18.63m far from building. </h3>

7 0

2 years ago

you have been called to testify as a as an expert witness in a trial involving a head-on collision Car A weighs 1515 pounds and

GrogVix [38]

Answer:

70.6 mph

Explanation:

Car A mass= 1515 lb

Car B mass=1125 lb

Speed of car B is 46 miles/h

Distance before locking, d=19.5 ft

Coefficient of kinetic friction is 0.75

Initial momentum of car B=mv where m is mass and v is velocity in ft/s

46 mph*1.46667=67.4666668 ft/s

$Momentum_B=1125*67.4666668 ft/s$

Initial momentum of car A is given by

$Momentum_A=1515v_a$ where $v_a$ is velocity of A

Taking East as positive and west as negative then the sum of initial momentum is

$1515v_a-(1125*67.4666668 ft/s)$

The common velocity is represented as $v_c$ hence after collision, the final momentum is

$Momentum_final=(m_a+m_b)v_c=(1515+1125)v_c=2640v_c$

From the law of conservation of linear momentum, sum of initial and final momentum equals each other hence

$1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

The acceleration of two cars $a=-\mug=-0.75*32.17=-24.1275 ft/s^{2}$

From kinematic equation

$v^{2}=u^{2}+2as$ hence

$v^{2}-u^{2}=2as$

$0^{2}-(v_c)^{2}=2*-24.1275*19.5$

$v_c=\sqrt{2*24.1275*19.5}=30.67 ft/s$

Substituting the value of $v_c$ in equation $1515v_a-(1125*67.4666668 ft/s)= 2640v_c$

$1515v_a-(1125*67.4666668 ft/s)= 2640*30.67$

$1515v_a=(1125*67.4666668 ft/s)+2640*30.67$

$v_a=\frac {156868.8}{1515}=103.5438 ft/s$

$\frac {103.5438}{1.46667}=70.59787 mph\approx 70.60 mph$

3 0

3 years ago