All categories

3 years ago

What is the net force acting on a 0.15 hockey puck accelerating at 12 m/s/s?

Physics

1 answer:

svetoff [14.1K]3 years ago

3 0

1.8N

Explanation:

Given parameters:

Mass of the hockey puck = 0.15kg

Acceleration = 12m/s²

Unknown:

Net force = ?

Solution:

According newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Force = mass x acceleration

Input the variables;

Net force = 0.15 x 12 = 1.8N

learn more:

Newton's law of motion brainly.com/question/11411375

#learnwithBrainly

You might be interested in

A cyclist intends to cycle up a 7.70o hill whose vertical height is 126m. Assuming the mass of bicycle plus person is 75.0kg, ca

Ksivusya [100]

Answer:

92704.5 J

596.44737 N

Explanation:

m = Mass of person + bicycle = 75 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Vertical height = 126 m

$\theta$ = Angle = 7.7°

d = Diameter = 0.388 m

Work done against gravity is given by

$P=mgh\\\Rightarrow P=75\times 9.81\times 126\\\Rightarrow P=92704.5\ J$

Work done is 92704.5 J

Force required is given by

$F=\dfrac{mgrsin\theta}{\pi d}\\\Rightarrow F=\dfrac{75\times 9.81\times sin7.7\times 5.12}{\pi\times 0.388}\\\Rightarrow F=596.44737\ N$

The force is 596.44737 N

7 0

3 years ago

Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2

jek_recluse [69]

The orbital radius is: $r=\frac{GM}{v^2}$

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

$\frac{GMm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the satellite

r is the radius of the orbit

v is the speed of the satellite

Re-arranging the equation, we find:

$\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

7 0

3 years ago

Read 2 more answers

What is (a) the x component and (b) the y component of the net electric field at the square's center

Sav [38]

Answer:

What is (a) the x component and (b) the y component of the net electric field at the square's center

8 0

3 years ago

In her hand, a softball pitcher swings a ball of mass 0.245 kg around a vertical circular path of radius 59.8 cm before releasin

GuDViN [60]

Answer:

The velocity is $v_b = 20.17 \ m/s$

Explanation:

From the question we are told that

The mass of the ball is $m = 0.245 \ kg$

The radius is $r = 59.8 \ cm = 0.598 \ m$

The force is $F = 30.9 \ N$

The speed of the ball is $v = 16.0 \ m/s.$

Generally the kinetic energy at the top of the circle is mathematically represented as

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

=> $K_t = \frac{1}{2} * 0.245 * 16.0 ^2$

=> $K_t = 31.36 \ J$

Generally the work done by the force applied on the ball from the top to the bottom is mathematically represented as

$W = F * d$

Here d is the length of a semi - circular arc which is mathematically represented as

$d = \pi * r$

$W = 30.9 * 0.598$

$W = 18.48 \ J$

Generally the kinetic energy at the bottom is mathematically represented as

$K_b = \frac{1}{2} * m * v_b^2$

=> $K_b = \frac{1}{2} * 0.245 * v_b^2$

=> $K_b = 0.1225 * v_b^2$

From the law of energy conservation

$K_t + W =K_b$

=> $31.36+ 18.48 = 0.1225 * v_b^2$

=> $v_b = 20.17 \ m/s$

3 0

3 years ago

Light travels at a speed of about 3.0 108 m/s. (a) how many miles does a pulse of light travel in a time interval of 0.1 s, whic

babunello [35]

Speed of light is given as

$c = 3 * 10^8 m/s$

time interval is given as

$\Delta t = 0.1 s$

so the distance covered by the light is given as

$d = v * \Delta t$

$d = 3 * 10^8 * 0.1 = 3 * 10^7 meter$

now as we know that

$1 mile = 1609 meter$

so the distance moved by light is

$d = \frac{3 * 10^7}{1609} = 18645.12 miles$

now for the comparision of this distance with diameter of earth

as we know that radius of earth is

$R = 6.38 * 10^6 m$

so the diameter of earth will be

$d = 2R = 12.76 * 10^6 m$

now the ratio of diameter with the distance that light move will be

$\frac{distance}{diameter} = \frac{3 * 10^7}{12.76 * 10^6}$

$\frac{distance}{diameter} = 2.35$

<em>so it is 2.35 times more than the diameter of earth</em>

4 0

3 years ago