The mass of an object with 500j of kenetic energy moving with a velocity of 5 m/s is

In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired

Sauron [17]

They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)

4 0

3 years ago

When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0

lora16 [44]

Answer:

$d=0.165m$

Explanation:

Given

$m=0.25kg$ , $x_{1}=5cm*\frac{1m}{100cm}=0.05m$ , $x_{2}=4cm*\frac{1m}{100cm}=0.04m$ , $v=2\frac{rev}{s}$

The tension of the spring is

$F_{k}=K*x_{1}=m*g$

$K=\frac{m*g}{x_{1}}$

$K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m$

The force in the spring is equal to centripetal force so

$F_{c}=\frac{m*v^2}{r}$

$v=w*r=2\pi*r$

But Fc is also

Fc=KxΔr

$F_{c}=K*(r-x_{2})$

Replacing

$m*4\pi^2*r=K*(r-x_{2})$

$0.25kg*4\pi^2*r=49*(r-0.04m)$

$r=0.205m$

total distance is

$d=0.205-0.04=0.165m$

3 0

3 years ago

Find the Y component of 35m/s at 57 degrees from the X-axis.

rosijanka [135]

Answer:

35m/s[57o].

X = 35*Cos57 =

Y = 35*sin7 =Explanation:

learn man but there u go

3 0

3 years ago

A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A

Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2) Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

Em₀ = U = mg y

Final. Low point just before the crash

Emf = K = ½ m v²

Em₀ = Emf

m g y = ½ m v²

v = √ 2 g y

Let's calculate

v = √ (2 9.8 0.05)

v = 0.99 m / s

1) the moment before the crash is

p₀ = m v

p₀ = 0.221 0.99

p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

p = 0

2) change of momentum

Δp = p - p₀

Δp = 0- 0.219

Δp = -0.219 kg m / s

3) the reason is

Δp / p = 1

In percentage form it is 100%

3 0

3 years ago

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at

devlian [24]

Force acting during collision is internal so momentum is conserve so (initial momentum = final momentum) in both directions Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1150 kg and was approaching at 5.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west. Let Vx is and Vy are final velocities of car in +x and +y direction respectively. initial momentum in +ve x (east) direction = final momentum in +ve x direction (east)

- 750*25 + 1150*0 = (750+1150)
Vx initial momentum in +ve y (north) direction = final momentum in +ve y direction (north)

750*0 - 1150*5 = (750+1150)
Vy from here you can calculate Vx and Vy so final velocity V is

V=(√V2x+V2y)

and angle make from +ve x axis is

θ=tan−1(VyVx)

 kinetic energy loss in the collision = final KE - initial KE

5 0

3 years ago

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