YO YO I NEED HELP!!!

You make the following measurements of an object: 42kg and 22m³. What would the objects density be? Show all work please and the

Tresset [83]

The formula for density is mass divided by volume
So 42kg divided by 22m3
= 1.909 kilogram/cubic meter
The object density would be 1.9
If you round it, the answer would be 1.91

3 0

4 years ago

carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr

denis23 [38]

The half-life equation $m=m_{0} (\frac{1}{2})^n$ in which n is equal to the number of half-lives that have passed can be altered to solve for n.

$n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}$

$\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2$

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

2 * 5700 = 11400 yr

3 0

3 years ago

How many simple machines are in a wine opener? The kind with the bottle opener at the head and two "arms."

Blababa [14]

It uses the corkscrew to anchor it to the cork and a lever to pull the cork out

Hope this helps buddy:D

7 0

3 years ago

Read 2 more answers

I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s.

AfilCa [17]

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

$v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s$

2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2) v$

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

$\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s$

3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

$I=\Delta p$ (1)

The impulse can be rewritten as product between force and time of collision:

$I=F \Delta t$

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

$\Delta p = m\Delta v = m(v_f -v_i)$

So, eq.(1) becomes

$F \Delta t = m(v_f -v_i)$

where:

F = ? is the unknown force

$\Delta t = 0.002 s$ is the duration of the impact

m = 0.16 kg is the mass of the ball

$v_f = 44 m/s$ is the final velocity of the ball

$v_i = -38 m/s$ is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

$F=\frac{m (v_f -v_i)}{\Delta t}=\frac{(0.16 kg)(44 m/s-(-38 m/s))}{0.002 s}=6560 N$

7 0

3 years ago

A white pool ball of mass 1.0 kg moving at 10 m/s collides with

solniwko [45]

The velocity of the red ball after the collision is 5.8 m/s

Explanation:

In absence of external forces on the system, we can apply the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$