You observe the moon rising at 6 p.m., around sunset. its phase is

A bird has a kinetic energy of 3 J and a potential energy of 25 J. What is the mechanical energy of the bird?

Ahat [919]

Answer:

28 J

Explanation:

Mechanical energy is the sum of kinetic and potential energy in an object that is used to do work. In other words, it is energy in an object due to its motion or position, or both.

In physics, mechanical energy (Emech) is the energy associated with the motion and position of an object usually in some force field.

Therefore:

ME = KE + PE = 25 + 3 = 28 J

8 0

3 years ago

Read 2 more answers

You lift your pet cat vertically by 2m and then you lower it vertically by 2m .during this exercise is the total work done by gr

rosijanka [135]

Zero, cause the positive and negative work cancel out :)

8 0

4 years ago

If a poison is ingested that prevents the release of acetylcholine, which one of the following would occur at a myoneural juncti

kirill115 [55]

Had to look for the options and here is my answer.
Given the situation that a poison is consumed and this prevented the acetylcholine release, the one that would most likely occur at a myoneural junction is that "sodium and potassium gates on the motor end plate will not open". Hope this helps.

4 0

3 years ago

Read 2 more answers

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

Velocity of sled

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Explanation:

As we know that here we we consider both people + sled as a system then there is no external force on it

So here we can use momentum conservation

since both people + sled is at rest initially so initial total momentum is zero

now when first people will jump with relative velocity "s" then let say the sled + other people will move off with speed v

so by momentum conservation we have

$0 = m_1(v - s) + (m_2 + M)v$

$v = \frac{m_1 s}{m_1 + m_2 + M}$

so velocity of the sled + other person is

$v = \frac{m_1 s}{m_1 + m_2 + M}$

velocity of first man who jump off

$v_1 = \frac{m_1 s}{m_1 + m_2 + M} - s$

$v_1 = -\frac{(m_2 + M) s}{m_1 + m_2 + M}$

Part b)

now when other man also jump off with same relative velocity

so let say the sled is now moving with speed vf

so by momentum conservation we have

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) = m_2(v_f - s) + Mv_f$

$(m_2 + M)(\frac{m_1 s}{m_1 + m_2 + M}) + m_2s = (m_2 + M)v_f$

Now we have

$v_f = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s$

Also the speed of second person is given as

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) + (\frac{m_2}{m_2 + M})s - s$

$v_2 = (\frac{m_1 s}{m_1 + m_2 + M}) - \frac{Ms}{m_2 + M}$

Part c)

change in kinetic energy of sled + two people is given as

$KE = \frac{1}{2}Mv_f^2 + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

here we know all values of speed as we found it in part a) and part b)

4 0

3 years ago

A project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow fo

Ksju [112]

Answer:

A. The project's energy costs will decrease

Explanation:

Since the project is located in an area with a demand-response program and on a site that has enough room for a wind-turbine to allow for on-site renewable energy.

Hence, the project's energy costs will decrease very well because it's implementing both of these strategies;

- Area with demand-response program.

- On-site renewable energy.

5 0

3 years ago