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Triss [41]
3 years ago
7

I NEEED WITH EXPLANATION PLEASE

Physics
1 answer:
Delicious77 [7]3 years ago
5 0

Answer:

20

Explanation:

u=0

a=2m/s^2

s=100m

V=?

v^2=u^2+2as

V^2=0^2+2(2*100)

V^2=400

v=root400

V=20

brainleast please! !!

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The fatty layer contains many nerves and blood vessels. True or False
mylen [45]

Answer:

False , The fatty layer do not contains many nerves and blood vessels. The fatty layer is related to skin layers.

Explanation:

The fatty layer is the most under a layer of skin. It made up of a system of collagen and fat cells. It helps maintain the body's temperature and shields the body from harm by serving as a shock absorber.

The dermis is the central layer of the skin. The dermis is maintained collectively by a protein termed collagen. This layer gives skin elasticity and power. The dermis also holds shock and feel receptors.

5 0
3 years ago
If a compound has a very low melting and boiling point, it is likely that the compound possesses mainly which type of intermolec
sergejj [24]
For the given question above, I think there is an associated choice of answer for it. However, the answer for this is London Dispersion Forces. <span>Dipole-dipole forces and hydrogen bonding are much stronger, leading to higher melting and boiling points.</span>
6 0
3 years ago
A bumblebee darts past at 3 m/s. The frequency of the hum made by its wings is 152 Hz. Assume the speed of sound to be 342 m/s.
Veronika [31]

It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.

Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.

After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.

8 0
3 years ago
Read 2 more answers
The smallest known galaxy, Segue 2, has an approximate radius of 1.05 × 1015 kilometers. Use the conversion factors 1 light-year
scoundrel [369]

( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =

(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =

(0.03388) x (psc) x (10³) =

33.88 parsecs


5 0
3 years ago
A 1.1 kg ball is attached to a ceiling by a 2.16 m long string. The height of the room is 5.97 m . The acceleration of gravity i
nydimaria [60]

1. -23.2 J

The gravitational potential energy of the ball is given by

U=mgh

where

m = 1.1 kg is the mass of the ball

g = 9.8 m/s^2 is the acceleration of gravity

h is the height of the ball, relative to the reference point chosen

In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is

h = -2.16 m

And the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(-2.16 m)=-23.2 J

2. 41.1 J

Again, the gravitational potential energy of the ball is given by

U=mgh

In this part of the problem, the reference point is the floor.

The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:

h = 5.97 m - 2.16 m = 3.81 m

And so the gravitational potential energy of the ball relative to the floor is

U=(1.1 kg)(9.8 m/s^2)(3.81 m)=41.1 J

3. 0 J

As before, the gravitational potential energy of the ball is given by

U=mgh

Here the reference point is a point at the same elevation of the ball.

This means that the heigth of the ball relative to that point is zero:

h = 0 m

And so the gravitational potential energy is

U=(1.1 kg)(9.8 m/s^2)(0 m)=0 J

4 0
3 years ago
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