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2 years ago

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When a bat strikes a ball, the force exerted can send the ball deep into the outfield. Where is the equal and opposite force in

this case? Select one: A. bat's force on the batter B. ball's force on the bat C. ball's force on the air

1 answer:

Delvig [45]2 years ago

3 0

Answer:

The answer is option B: the ball´s force on the bat

Explanation:

Using Newton's third law (action and reaction) that says: If two bodies interact the force of one exerts on the other is of equal magnitude, direction and opposite direction

Therefore, if the bat hits the ball (action), the ball (reaction) with a force on the bat of equal magnitude, but opposite direction

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A 3 kg model airplane is traveling at a speed of 33 m/s. The operator then increases the speed up to 45 m/s in 2 seconds. How mu

nlexa [21]

Answer:

Force = 18N

Explanation:

Force = [ mass ( final velocity - initial velocity ) ] / time taken

using the formula, here mass is 3 kg, final velocity = 45 m/s , initial velocity = 45 m/s , time taken = 2 seconds

Force = [ 3 ( 45 - 33 ) ] / 2

Force = 18N

8 0

2 years ago

Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from

mojhsa [17]

Answer:

it is reduced four times.

Explanation:

By definition, the electric field is the force per unit charge created by a charge distribution.

If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.

So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.

8 0

3 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

Newton’s first law describes the tendency calles

beks73 [17]

The first law is about force or push and pull

5 0

3 years ago

Read 2 more answers

A bike rider approaches a hill with a speed of 8.5 m/s. The total mass of the bike rider is 91kg. What is the kinetic energy of

Anestetic [448]

a) The kinetic energy (KE) of an object is expressed as the product of half of the mass (m) of the object and the square of its velocity (v²):

$KE = \frac{1}{2}m* v^{2}$

It is given:

v = 8.5 m/s

m = 91 kg

So:

$KE= \frac{1}{2}*91*8.5^{2} =3,287.4J$

b) We can calculate height by using the formula for potential energy (PE):
PE = m*g*h

In this case, h is eight, and PE is the same as KE:
PE = KE = 3,287.4 J

m = 91 kg

g = 9.81 m/s² - gravitational acceleration

h = ? - height

Now, let's replace those:

3,287.4= 91 * 9.81 * h

⇒ h = 3,287.4/(91*9.81) = 3,287.4/892.7 = 3.7 m

3 0

3 years ago