Find the recoil velocity of a 65kg ice hockey goalie who catches a 0.15kg hockey puck slapped at him at a velocity of 50m/s. Assume that the goalie is at rest before catching the puck, and friction between the ice and the puck-goalie system is negligible.

Answers

Answer 1

Answer:

0.12m/s

Explanation:

Given data

mass m1=65kg

velocity v1=?

mass m2=0.15kg

v2=50m/s

We know that the expression for the conservation of momentum is given as

0=m1v1+m2v2

we want to find v1

v1=(m2v2)/m1

v1=0.15*50/65

v1=7.5/65

v1=0.115

v1=0.12m/s

Therefore the recoil velocity is 0.12m/s


Related Questions

It takes an airplane nearly 3/4 of a mile to stop. Which law is being used?
A) Newton's first law
B) Newton's second law
C) Newton's third law
D) Law of conservation of momentum
E) Law of universal gravitation

Answers

Answer:

I think it is D

Explanation:

might be wrong

A conveyor belt dumps 2500 yd3 of gravel to form a cone-shaped pile. How high could this pile of gravel be, and what could the circumference of the pile of gravel be at ground level

Answers

Answer:

Explanation:

Volume of the cone = 1/3πr²h

r is the radius

h is the height

2500 = 1/3πr²h

Criss multiply

r²h = 2500 × 3π

Rewrite the expression

r²h = 50² × 3π

Compare values

r² = 50²

r = √50²

r = 50yd

The radius is 50yards

Also;

h = 3π yards

h = 3(3.14)

h = 9.42yards

Hence the pile is 9.42yards high

Circumference of the pile = 2πr

Given r = 50

Substitute

Circumference = 2π(50)

Circumference = 100π

Circumference = 100(3.14)

Circumference = 314yards

Hence the circumference of the pile at the ground level is 314yards

how much tension (fm) must be supplied at the triceps to stabalize the arm against an external force (fe) of 200 N given dm

Answers

The missing part of the question is :

given dm-2 cm and de 25 cm

Answer:

The correct answer is = 2500 N

Explanation:

Given:

External force (fe) = 200 N

dm- 2 cm = 0.02 m

de = 25 cm = 0.25 m

Balancing torque about elbow

Fm*dm = Fe*de

By putting value

fm*0.02 = 200*0.25

fm= 200*0.25/0.02

fm = 50/0.02

fm = 2500 N

Thus, the correct answer is = 2500 N

I need help with this science work.

Answers

Answer:

1. Radiation

2. Conduction

3. Conduction

4. Convection

5. Convection

6. Convection

7. Radiation

I think that is correct

I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object? And what direction?

Answers

Answer:

20N to the North

Explanation:

the west and east forces cancel out, that leaves us with 50 to the north, and 30 to the south, the difference would be 20 N to the North

A pencil is dropped from rest from a height of 1.8 meters above the ground. How much time does it take the pencil to hit the ground?

Answers

Answer:

0.60s

Explanation:

Using the motion equation as follows:

S = ut + 1/2gt²

Where;

S = distance travelled (m)

u = initial velocity (m/s)

t = time (s)

g = acceleration due to gravity (m/s²)

Based on the information provided, S = 1.8m, u = 0m/s, t = ?

S = ut + 1/2gt²

1.8 = (0×t) + 1/2 (9.8t²)

1.8 = 0 + 4.9t²

1.8 = 4.9t²

t² = 1.8/4.9

t² = 0.3673

t = √0.3673

t = 0.6060

t = 0.60s

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.54 m/s, and its maximum acceleration is 8.13 m/s2. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum

Answers

Answer:

Explanation:

The spring will move under SHM .

Maximum speed is at the middle equilibrium position . Maximum acceleration is when the particle is at one of the extreme position . So time gap between the two position is one fourth of time period of oscillation .

=  T / 4

If A be the amplitude and ω be angular velocity

ω A =  maximum velocity = 1.54

ω² A = maximum acceleration = 8.13

dividing

ω = 5.28

2π / T = 5.28

T = 1.189

T / 4 = 0.3 s.

Calculate the rms speed of helium atoms near the surface of the Sun at a temperature of about 5200 K.

Answers

Answer: Root mean square speed, Vrms=5,694m/s

Explanation:

Root mean square speed  is defined as the  the square root of the average of the square of the velocity and given as [tex]\sqrt 3RT/M[/tex]

Where R =8.3145J/mol.k

M molar mass of the molecule in kg/mol

Helium= He=4g/mol =  0.004kg/mol

T= Temperature = 5200k

Vrms=  [tex]\sqrt 3RT/M[/tex]

=[tex]\sqrt 3 x 8.3145 x 5200/0.004[/tex]

=[tex]\sqrt32426550[/tex]

Vrms =5,694m/s

At room temperature, most metals are
Select one:
a. gas
O b. liquid
O c. an alloy
O d. solid

Answers

most metals are solid at a room temperature

Science Motion Problem pelase elp

Answers

Answer:120

Vav = S / t = (120 + 45) m / (4 + 60) sec = 2.6 m/s

A biker pushes on the pedal and accelerates her 50 kg bike from 0 m/s to 10 m/s. What is the the impulse?

Answers

Answer:

3.7

Explanation:

How do mass and the type of material affect thermal energy transfer?

Answers

Answer:

Warmer objects have faster particles and higher temperatures. If two objects have the same mass, the object with the higher temperature has greater thermal energy. ... Heat is the transfer of thermal energy between objects that have different temperatures.

9. A 0.25-kg mass is attached to a string and swung in a vertical circle whose radius is 0.75 m. At the bottom of the circle, the mass is observed to have a speed of 10 m/s. What is the magnitude of the tension in the string at that point

Answers

Answer:

T =  33.34 N

Explanation:

Given that,

Mass attached to the spring, m = 0.25 kg

Radius of the vertical circle, r = 0.75 m

Speed of the mass, v = 10 m/s

We need to find the magnitude of the tension in the string at that point. We know that the tension in the string is equal to the centripetal force acting on it. It can be given by

[tex]T=\dfrac{mv^2}{r}\\\\T=\dfrac{0.25\times (10)^2}{0.75}\\\\T=33.34\ N[/tex]

So, the tension in the string is 33.34 N.

the whale shark is the largest of all fish and can have the mass of 3 adult elephants. suppose that a crane is lifting a whale shark into a tank for delivery to an aquarium. The crane must exert a force of 25000 N to lift the shark from rest. if the sharks acceleration equals 1.25m/s^2, what is the sharks mass ?

Answers

Answer: The whale shark is the largest type of fish in the world. Its mass can be as large as 2.00 x 10^4kg, which is the equivalent mass of three average adult elephants. Suppose a crane lifts a net with a 2.00 x 10^4kg whale shark off the ground. The net is steadily accelerated from rest over an interval of 2.5s until the net reaches a speed of 1.0 m/s.

Explanation:

PLEASE HELPP WITH QUESTIONS :(

Answers

Answer:no

Explanation: lol

In theory, if the airspeed of an airplane is doubled while in level flight, parasite drag will become

Answers

Answer:

Four times greater

Explanation:

Drag is comprised of parasite drag and induced drag. The Parasite drag is the drag that acts on an object when the said object happens to be moving through a fluid. Parasite drag is also a combination of two forms of drag. These two forms of drag are called form drag and skin friction drag.

As the speed of an airplane increases, the parasite drag also increases while the induced drag of the airplane decreases. The parasite drag of an airplane in theory becomes four times greater, when it's airspeed is doubled.

hello what is momentum?

Answers

Answer:

Momentum is vector quantity and it is defined as the product of mass of the object and velocity of the object.

As it is a vector quantity it has both direction and magnitude, and momentum is written as

p = m × v

p = momentum

m = mass of the object

v = velocity

Two kinds of momentum are there and they are:

Linear MomentumAngular Momentum

QUESTION 1 During the time of Pangaea most of the dry land on Earth was joined into one huge landmass that covered nearly a third of the planet's surface. The giant ocean that surrounded the continent is known as what?
QUESTION 2Pangaea existed during what geological time periods, which were times of great change?
QUESTION 3 Most distributions of rocks within Earth's crust, including minerals, fossil fuels, and energy resources, are a direct result of the history of plate motions and collisions and the corresponding changes in the configurations of the continents and ocean basins. Research and explain how plate tectonics could account for these resources exactly.

PLS ANSWeR

Answers

Answer:

Pangea existed between about 299 million years ago (at the start of the Permian Period of geological time) to about 180 million years ago (during the Jurassic Period). It remained in its fully assembled state for some 100 million years before it began to break up.

Explanation:

GOOGLE

How can humans detect infrared waves?
А
They see them as color.
B
They feel them as heat.
с
They hear them as sound.
D
They feel them as vibrations.

Answers

The answer is B. They feel them as heat.
I’m not good with these type of Stuff so don’t pick my answer but I think the answer may be D

All of the following would motivate someone toward attaining a career goal except
A expectation of money
B. low self-esteem
C. belief in success
D. financial stability

Answers

B
......................

All would motivate someone toward attaining a career goal except low self-esteem. The correct option is B.

What is self-esteem?

We tend to feel better about ourselves and about life in general when we have healthy self-esteem.

It improves our ability to deal with life's ups and downs. When we have low self-esteem, we tend to view ourselves and our lives in a more negative and critical light.

People's choices and decisions are heavily influenced by their self-esteem. In other words, self-esteem motivates people by making it more or less likely that they will take care of themselves and reach their full potential.

Except for low self-esteem, everyone would not motivate somebody to pursue a career goal.

Thus, the correct option is B as low self esteem cannot motivate anyone for career goal.

For more details regarding self-esteem, visit:

https://brainly.com/question/28768312

#SPJ5

find the current of 20 c of charge pass a piticular point in a circuit in 10 seconds

Answers

I = Q/t = 20C/10s = 2A

A student is standing on top of the school building and has a height of 44.1m. They kick a rock off the edge of the building with velocity of 6 m/s . how far away from the base of the building does the rock land?

Answers

At a height of 44.1m the time taken to fall is
s= 1/2 g t^2
44.1 = 1/2 x 9.8 x t^2
t = square root (9)= 3 seconds
The horizontal distance is 3x6 = 18 m away from the building base

PLS HELP PLS 10. This graph represents the velocity of an object over time. What is the average acceleration of the
object during the first 4 seconds?

Answers

Answer:

8 M/S

Explanation:

This is because the points match from 4 seconds (X) and it matches up to 8 (Y), therefore it went up 8 M/S in 4 seconds.

If George Washington had become "King of America," our government might have become?

Answers

Answer:

Monachry

Explanation:

Hope this helped!!!

what is the answer to life and everything in the universe?

Answers

The answer to life and the universe is 42

A 5.00 gg bullet is fired horizontally into a 1.20 kgkg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 mm along the surface before stopping.

Answers

Answer:

V= 295.2m/s

Explanation:

Given that mass of bullet = 5.0g = 0.005kg

Mass of wooden block = 1.20kg

The coefficient of kinetic friction between block and surface = 0.20

The force of friction of the block

Fk = (μ) mg

Fk = 0.2×1.2×9.8m/s

F = 2.352N

= Force × distance

The work done = 2.352× 0.390

= 0.91728J

The initial velocity of the block can be calculated

1/2mv^2

0.5×1.2×v^2 = 0.917J

0.6V^2 = 0.917J

V^2 = 0.917J/0.6

V^2 = 1.52

V = √1.52

V = 1.23m/s

Now we shall use conservation of momentum to find the velocity of the bullet

0.005kg×v = 1.2kg × 1.23m/s

0.005v = 1.476

V= 1.476/0.005

V= 295.2m/s

A man accelerates at 2.52 m/s2 for 3.73s. If his initial velocity was 1.75 m/s, what is his final velocity?

Answers

Answer:

Final Velocity = 11.15 m/s

Explanation:

By using the equation v = u + at we can find the answer.

v = final velocity

u = initial velocity

a = acceleration

t = time

Plug the numbers in and you get 11.15 m/s, hope this helps!

If the first stage provides a thrust of 5.25 mega-newtons [MN] and the space shuttle has a mass of 4,470,000 pound-mass [lbm], what is the acceleration of the spacecraft in miles per hour squared [mi/h2]

Answers

Answer:

20,861.65 mi/h²

Explanation:

We convert  4,470,000 pound-mass [lbm], to kg. Since  2.205 lbm = 1 kg, then 4,470,000 lbm = 4,470,000 lbm × 1 kg/2.205 = 2,027,210.88 kg

Since Force , F = ma where m = mass and a = acceleration, and our force of thrust , F = 5.25 MN = 5,250,000 N and or mass = mass of spacecraft = 2,027,210.88 kg, we then find the acceleration, a.

a = F/m = 5,250,000 N/2,027,210.88 kg = 2.59 m/s².

We now convert this acceleration into miles per hour. Since 1 mile = 1609 meters and 60 × 60 s = 1 hour ⇒ 3600 s = 1 hour, Our conversion factor for meter to mile is 1 mile/1609 m and that for second to hour is 3600 s/1 hour. We square the conversion factor for the time so we have (3600 s/1 hour)².

Multiplying both conversion factors with our acceleration, we have

a = 2.59 m/s²

= 2.59 m/s² × 1 mile/1609 m × (3600 s/1 hour)²

= 33566440/1609 miles/hour²

= 20,861.65 mi/h²

= 20,861.65 miles per hour squared

Problem 6 (20 points) A ballistic pendulum is a device often used for measuring the velocity of a projectile. Assume to projectile fully embeds itself in the pendulum. If the projectile mass is 25 grams, the pendulum mass is 2.5 kg, and the vertical displacement of the pendulum after the collision is 12 cm, find the velocity of the projectile just prior to the collision.

Answers

Answer:

The velocity of the projectile just prior to the collision is 154.934 meters per second.

Explanation:

Let consider the projectile-pendulum system as our system of study, from statement we know that projectile gives kinetic energy to the pendulum and turns into gravitational potential energy. By Principle of Energy Conservation we construct the following expression:

[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}[/tex] (Eq. 1)

Where:

[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energies, measured in joules.

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies, measured in joules.

By applying definitions of gravitational potential and kinetic energies, we expand and simplify the equation above:

[tex](m+M)\cdot g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (m+M)\cdot (v_{1}^{2}-v_{2}^{2})[/tex]

[tex]g\cdot (y_{2}-y_{1})=\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex]

And we clear the initial velocity of the pendulum-projectile system:

[tex]v_{1}=\sqrt{v_{2}^{2}+2\cdot g\cdot (y_{2}-y_{1})}[/tex] (Eq. 2)

Where:

[tex]m[/tex] - Mass of the projectile, measured in kilograms.

[tex]M[/tex] - Mass of the ballistic pendulum, measured in kilograms.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speeds of the pendulum-projectile system, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and vertical position of the pendulum-projectile system, measured in meters.

Under the consideration of inellastic collision, we find the velocity of the projectile prior to the collision by Principle of Linear Momentum Conservation:

[tex]m\cdot v_{o, P} + M\cdot v_{o,B} = (m+M)\cdot v_{1}[/tex]

[tex]v_{o,P} = \frac{(m+M)\cdot v_{1}-M\cdot v_{o,B}}{m}[/tex] (Eq. 2)

Where:

[tex]v_{o, P}[/tex] - Velocity of the project prior to the collision, measured in meters per second.

[tex]v_{o,B}[/tex] - Velocity of the ballistic pendulum prior to the collision, measured in meters per second.

If we know that [tex]v_{2} = 0\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{2}-y_{1} = 0.12\,m[/tex], [tex]m = 0.025\,kg[/tex], [tex]M = 2.5\,kg[/tex] and [tex]v_{o,B} = 0\,\frac{m}{s}[/tex], then the velocity just prior to the collision is:

By (Eq. 1):

[tex]v_{1} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.12\,m)}[/tex]

[tex]v_{1} \approx 1.534\,\frac{m}{s}[/tex]

By (Eq. 2):

[tex]v_{o,P} = \frac{(0.025\,kg+2.5\,kg)\cdot \left(1.534\,\frac{m}{s} \right)-(2.5\,kg)\cdot \left(0\,\frac{m}{s} \right)}{0.025\,kg}[/tex]

[tex]v_{o,P} = 154.934\,\frac{m}{s}[/tex]

The velocity of the projectile just prior to the collision is 154.934 meters per second.

Neil travelled 36km at a speed of 8km/h Grant travelled 48km at a speed of 10km/h whose journey was quickest​

Answers

Answer:

[tex]\boxed {\tt Neil's \ journey}[/tex]

Explanation:

We want to find who travelled faster, so we must find the time for each person's journey. Time can be found by dividing distance by speed.

Neil

[tex]t=\frac{d}{s}[/tex]

Neil travelled 36 kilometers at a speed of 8 kilometers per hour.

[tex]d= 36 \ km \\s= 8 \ km/hr[/tex]

[tex]d=\frac{36 \ km}{8 \ km/hr}[/tex]

Divide (the kilometers or "km" will cancel each other out).

[tex]d=4.5 \ hr[/tex]

Grant

[tex]t=\frac{d}{s}[/tex]

Grant travelled 48 kilometers at a speed of 10 kilometers per hour.

[tex]d= 48 \ km \\s= 10 \ km/hr[/tex]

[tex]d=\frac{48 \ km} { 10 \ km/hr}[/tex]

Divide (the kilometers or "km" will cancel each other out).  

[tex]d=4.8 \ hr[/tex]

Neil's journey took 4.5 hours and Grant's took 4.8 hours. We want to find the journey that was quicker and 4.5 is less than 4.8. So Neil's journey was quicker.

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