Whats the mass of the sun times pi?

A 0.0382-kg bullet is fired horizontally into a 3.78-kg wooden block attached to one end of a massless, horizontal spring (k = 8

wel

Answer:

$280.87$ ms⁻¹

Explanation:

Consider the motion of the bullet-block combination after collision

$m$ = mass of the bullet = 0.0382 kg

$M$ = mass of wooden block = 3.78 kg

$V$ = velocity of the bullet-block combination after collision

$k$ = spring constant of the spring = 833 N m⁻¹

$A$ = Amplitude of oscillation = 0.190 m

Using conservation of energy

Kinetic energy of bullet-block combination after collision = Spring potential energy gained due to compression of spring

$(0.5)(m + M)V^{2} = (0.5)kA^{2}$

$(0.0382 + 3.78)V^{2} = (833)(0.190)^{2}$

$V = 2.81$ ms⁻¹

$v_{o}$ = initial velocity of the bullet before striking the block

Using conservation of momentum for the collision between bullet and block

$m v_{o} = (m + M) V$

$(0.0382) v_{o} = (0.0382 + 3.78) (2.81)$

$v_{o} = 280.87$ ms⁻¹

7 0

2 years ago

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

const2013 [10]

Answer: Fourth option. It increased by a factor of 3.

Solution:

m1=1.0 kg

Cylinder's gravitational potential energy: Ep=m*g*h

Ep1=(1.0 kg)*g*h

Ep1=g*h

m2=3.0 kg

Ep2=(3.0 kg)*g*h

Ep2=3*g*h

Replacing g*h by Ep1 in the equation above:

Ep2=3*Ep1

Then, the cylinder's gravitational potential energy increased by a factor of 3.

3 0

3 years ago

Read 2 more answers

a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?

Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !

The formula is D = 1/2 a T²

   Distance = (1/2 acceleration) x (time in seconds)²

   Acceleration = (change in speed) / (time for the change)
    = (18 m/s) / (12 sec)
    = 1.5 m/s² .

Distance = (1/2 x 1.5 m/s²) x (12 sec)²
      =     (0.75 m/s²) x (144 sec²) = 108 meters .

5 0

3 years ago

Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m

stepladder [879]

Answer: $F_{2}=\frac{3}{4}F_{1}$

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have two situations:

1) Two bags with masses $4M$ and $4M$ mutually exerting a gravitational attraction $F_{1}$ on each other:

$F_{1}=G\frac{(4M)(4M)}{r^2}$ (1)

$F_{1}=G\frac{16M^2}{r^2}$ (2)

$F_{1}=16\frac{GM^2}{r^2}$ (3)

2) Two bags with masses $2M$ and $6M$ mutually exerting a gravitational attraction $F_{2}$ on each other (assuming the distance between both bags is the same as situation 1):