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Readme [11.4K]
3 years ago
5

If a bat with a mass of 5 kg and acceleration of 2 m/s2 hits a ball whose mass is 0.5 kg in the forward direction, what is the r

eaction force of the ball on the bat?
10 N, forward
0 N, forward
0.5 N, upward
10 N, backward
Physics
2 answers:
ivanzaharov [21]3 years ago
8 0

Answer:

10 N , forward

Explanation:

F=m×a

F= 5×2

F=10 N

lina2011 [118]3 years ago
6 0

Answer:

10 N, backward

Explanation:

Force exerted on the ball by the bat can be calculated according to newtons second law,

Force = mass × acceleration

Given mass of the ball = 5kg

Acceleration of the ball = 2m/s²

Force exerted on the ball= 5×2

Force exerted on the ball by the bat = 10N

If the force of 10N is applied on the ball by the bat in the forward direction, the reaction force of the ball on the bat will also be 10N but in the opposite direction (backward) to the direction of the bat according to newton's third law which states that actions and reaction are equal and opposite.

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A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
charle [14.2K]

Answer:

<em>1.</em> <em>39068.07 N</em>

<em>2. 19534.036 N</em>

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

<em>height below tank surface = 10 - 0.01 = 9.99</em>

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>

surface area of plug = πr^{2} = 3.142 x 0.25^{2} = 0.196 m^{2}

<em>force required to lift left plug = pressure x area</em>

F =  199326.9 x 0.196 = <em>39068.07 N</em>

<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>

A = 0.196/2 = 0.098 m^{2}

force F required to lift right plug =  199326.9  x 0.098 =<em> 19534.036 N</em>

6 0
3 years ago
A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns
Katena32 [7]

Answer:

166 W

Explanation:

Power is the rate at which work is done.

\text{Power} = \dfrac{\text{Work done}}{\text{time}}

The work done by Jill is the product of the weight of the pail and the height it moves.

The weight is the product of the mass and acceleration of gravity, <em>g</em>. Taking <em>g</em> as 9.81 m/s², the weight is

<em>W</em> = (6.90 kg)(9.81 m/s²) = 67.689 N

Work done = (67.689 N)(27.0 m) = 1827.603 J

Power = (1827.603 J) ÷ (11.0 s) = 166 W

4 0
3 years ago
Mrs. Paul and Dr. Mykannen ride a tandem bike on the beach. They ride along the beach for about 455 meters. Their final velocity
algol [13]

Answer:

Time = 80.91 seconds

Explanation:

Given the following data;

Velocity = 5.50 m/s.

Distance = 445 meters

To find the time;

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

Velocity = \frac{distance}{time}

Substituting into the formula, we have;

5.5 = 445/time

Time = 445/5.5

Time = 80.91 seconds

5 0
3 years ago
Two waves overlap producing a new wave that exhibits destructive
suter [353]

Answer:B

Explanation:

8 0
3 years ago
Read 2 more answers
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

8 0
3 years ago
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