The Sex Equity in Education Act requires educational institutions to distribute a sexual harassment policy to:

The diagram shows a ramp with a toy car at the bottom. A string attached to the front of the car and the string goes over a pull

Annette [7]

The answer would be Gravity.

Gravity is pulling the weight down, which is pulling the car up the ramp.

3 0

4 years ago

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40

Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$

So total energy at the top of the hill $=mgh+0=mgh$

At the bottom of the hill kinetic energy is equal to $\frac{1}{2}mv^2$ and potential energy will be 0

So total energy at the bottom of the hill is equal to $0+\frac{1}{2}mv^2$

Form energy conservation $mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$ , for v = 20 m/sec

$20=\sqrt{2\times 9.8\times h}$

Squaring both side

$19.6h=400$

h = 20.408 m

Now if velocity is 0 m/sec

$40=\sqrt{2gh}$

$19.6h=1600$

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

5 0

3 years ago

Two objects of equal mass collide on a horizontal frictionless surface. Before the collision, object A is at rest while object B

fenix001 [56]

Answer: 6m/s

Explanation:

Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.

After collision, the two objects will move at the same velocity (v).

Let mA and mB be the mass of the two objects

uA and uB be their velocities before collision.

v be their velocity after collision

Since the two objects has the same mass, mA= mB= m

Also since object A is at rest, its velocity = 0m/s

Velocity of object B = 12m/s

Mathematically,

mAuA + mBuB = (mA+mB )v

m(0) + m(12) = (m+m)v

0+12m = (2m)v

12m = 2mv

12 = 2v

v = 6m/s

Therefore the speed of the composite body (A B) after the collision is 6m/s

7 0

4 years ago

The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. Which comparison is best supported by this

Vesnalui [34]

The atomic number of beryllium (Be) is 4, and the atomic number of barium (Ba) is 56. the <span>comparison is best supported by this information is that beryllium has a lower atomic radius than Barium</span>

8 0

3 years ago

Read 2 more answers

Can someone help me with this

yKpoI14uk [10]

bonded pairs of electrons, lone pairs of electrons.

4 0

3 years ago