What is the displacement of the object after 3 seconds?

a person does 100 joules of work in pulling back the string of a bow. what will be the initial speed of a 0.5 kg arrow when it i

eduard

If all the energy she put into bending the bow is completely
transmitted to the arrow, then the arrow has the 100 joules
of kinetic energy when it leaves the bow.

   Kinetic energy = (1/2) (mass) (speed)²

   100 J = (1/2) (0.5 kg) (speed²)

Divide each side by 0.25 kg: 100 J / 0.25 kg = speed²

[ joule ] = [ newton-meter ] = kg-m²/sec²

100 kg-m²/sec² / 0.25 kg = speed²

   400 m²/sec² = speed²

Take the square root of each side: speed = √400 m/s

    20 m/s

      (about 44.7 mph)

3 0

2 years ago

A young hockey player stands at rest on the ice holding a 1.3-kg helmet. The player tosses the helmet directly in front of him w

navik [9.2K]

Answer

given,

initial speed of hockey player= 0 m/s

mass of the helmet, m = 1.3 Kg

initial speed of the helmet, u = 0 m/s

final speed of the helmet, v = 6 m/s

recoil speed of the hockey player, v' = 0.25 m/s

we need to calculate the mass of the hockey player, M = ?

using conservation of momentum

m u + M u' = M v' + m v

initial speed of ice skater is zero

1.3 x 0 + M x 0 = M x (-0.25) + 1.3 x 6

negative sign is taken because recoil velocity is in opposite direction

0 = -0.25 M + 7.8

0.25 M = 7.8

M = 31.2 Kg

Hence, the mass of the young hockey player is equal to 31.2 Kg

5 0

2 years ago

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

dem82 [27]

Answer:

Explanation:

a )

Reaction force of the ground

R = mg

= 160 N

Maximum friction force possible

= μ x R

= μ x 160

= .4 x 160

= 64 N .

b )

160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,

Taking moment about top point of ladder

160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3

240 + 444 + 4f = 2700

f = 504 N

c )

Let x be the required distance.

Taking moment about top point of ladder

160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )

240 + 444 x + 1440 = 2700

x = 2.3 m

so man can go upto 2.3 at which maximum friction acts .

8 0

3 years ago

A tennis player swings her 1000 g racket with a speed of 11 m/s. She hits a 60 g tennis ball that was approaching her at a speed

shusha [124]

Answer:

- 3.72 Ns.

9.44 m/s

Explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

final velocity of ball, v = - 44 m/s

Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

= - 2.64 - 1.08 = - 3.72 Ns

Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 = 1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .

3 0

3 years ago

A gas is compressed from an initial volume of 5.75 L to a final volume of 1.23 L by an external pressure of 1.00 atm. During the

nataly862011 [7]

Answer:

<em>The internal energy change is 330.01 J</em>

Explanation:

Given

$v_{1}$ the initial volume = 5.75 L

$v_{2}$ the final volume = 1.23 L

$P_{e}$ is the external pressure = 1.00 atm

q the heat energy removed = -128 J (since is removed from the system)

expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;

W = - $P_{e}$ ΔV

W = -1.00 x(1.23 - 5.75)

W = -1.00 x -4.52

W = 4.52 L atm

converting to joules we have

W = 4.52 L atm x 101.33 J/ L atm = 458.01 J

The internal energy change during compression can be calculated thus;

ΔU = q + W

ΔU = -128 J + 458.01 J

ΔU = 330.01 J

Therefore the internal energy change is 330.01 J

8 0

2 years ago