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insens350 [35]
3 years ago
14

IF YOU ANSWER THESE 3 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 17 POINTS!!!

Physics
1 answer:
gogolik [260]3 years ago
3 0

Answer:

1. a

2. a [im iffy on this but 95% positive its this]

3. b [walking is a form of aerobics, so i would say b]

Explanation:

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B What is SI and CGs unit of force​
choli [55]

Answer:

The relation between SI and CGS unit of force is the CGS unit of force is equal to SI unit of force. The force is the product of mass and acceleration. its SI unit is Newton and CGS unit is dyne. The relation between SI and CGS unit of force is.  

HOPE THIS HELPS!!!!!

Explanation:

6 0
3 years ago
The amount of electrons that an atom loses, shares or gains is the ________________.
mr Goodwill [35]
It’s known as valency
4 0
3 years ago
The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the poo
Yuri [45]

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

4 0
3 years ago
*An inductor is capable of dissipating 50W of heat energy when a current 0.8A flows through it at a certain frequency. Calculate
ale4655 [162]
I think that your answer would be D
8 0
3 years ago
A 3.00-m rod is pivoted about its left end. A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from t
dalvyx [7]

Answer:

The net torque about the pivot is and the answer is 'c'

c. T_{net}=8.58

Explanation:

T=F*d

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

T_1=F_1*d_1

T_1=7.8N*1.6m=12.48N*m

T_2=F_2*d_2

T_2=2.60N**cos(30)*3.0m

T_2= - 3.9 N*m

Now to find the net Torque is the summation of both torques

T_{net}=T_1+T_2

T_{net}=12.48N-3.9N=8.58N

8 0
3 years ago
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