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Alik [6]
2 years ago
11

A 0.2 kg hockey puck moving at 80 m/s is caught by a 40 kg goalie at rest. With what speed does the goalie slide on the slide?

Physics
1 answer:
Lady_Fox [76]2 years ago
5 0

Answer:

V = 0.39 m/s

Explanation:

Given that,

Mass of hockey puck, m = 0.2 kg

Mass of goalie = 40 kg

Speed of hockey puck, v = 80 m/s

We need to find the speed with which the goalie slide on the slide. Let V be the speed. Using the conservation of momentum as follows :

mv=(M+m)V\\\\V=\dfrac{mv}{M+m}\\\\V=\dfrac{0.2\times 80}{0.2+40}\\\\V=0.39\ m/s

So, the required speed is 0.39 m/s.

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Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram
bagirrra123 [75]

Speed of car A is given as

v_a = 70 mph

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

v_a = 70 *\frac{1609}{3600} = 31.3 m/s

now its distance from Bambi is given as

d_a = 350 m

time taken by it to hit the Bambi

t = \frac{d}{v}

t = \frac{350}{31.3}

t = 11.2 s

Now other car is moving at speed 50 mph

so its speed in SI unit will be

v_b = 50* \frac{1609}{3600}

v_b = 22.35 m/s

now its distance from Bambi is given as

d_b = 590 feet

as we know that 1 feet = 0.3048 m

d_b = 590*0.3048 = 179.83 m

now the time to hit the other car is

t_2 = \frac{179.83}{22.35}

t_b = 8.05 s

So Car B will hit the Bambi first

7 0
3 years ago
Suppose there is a large amount of (weakly interacting) dark matter between us and a distant galaxy. How will this affect our vi
leonid [27]

Answer:

Dark matter does not affect our view, humans can see through them.

Explanation:

They do not affect our view because we can see right through the (weakly interacting) dark matter, as they do not interact or interfere with electromagnetic force.

Dark matter are often invisible substances and are difficult to spot because they don't absorb or reflect light.

7 0
3 years ago
A. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6
steposvetlana [31]

Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle.  There are two forces on the object: tension force T pulling up and weight force mg pulling down.

Sum the forces in the radial (+y) direction:

∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle.  There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

T = (1.52 kg) (4.21 m/s)² / (0.676 m) −  (1.52 kg) (9.8 m/s²)

T = 24.9 N

6 0
3 years ago
suppose you have a 69.0-kg wooden crate resting on a wood floor. what maximum force can you exert horizontally on the crate with
marusya05 [52]

You've got a 69.0-kg wooden crate on a wooden floor. The box can withstand a force of up to 338N in a horizontal direction without being moved. Following this, the wooden creates moving stats.

In order to calculate the friction coefficient, divide the force pushing two objects together by the force acting between them. friction coefficient might be 0 or one. They can be split into two categories: friction coefficient that is static. Kinetic friction coefficient (also known as sliding coefficient of friction).

the acceleration brought on by the gravitational pull of large masses generally, gravitational , often known as the acceleration brought on by the Earth's gravitational pull and centrifugal force,

F= friction coefficient *M*g

F= 0.5*69*9.8

F=338N

Learn more about gravitational here

brainly.com/question/3009841

#SPJ4

7 0
1 year ago
show the trajectory of a body projected with an initial velocity vat an angle of departure thita is a paraboler​
Jobisdone [24]

Answer:

Let me look up a couple of things regarding this question.

Explanation:

Then I will get back to you.

5 0
2 years ago
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