All categories

3 years ago

PLEASE ANSWER ILL GIVE YOU BRAIN!!!!

Physics

1 answer:

solniwko [45]3 years ago

3 0

Answer:

I think the first two one is 70 the second on is 14

I hope this helps if I’m wrong I’m sorry

Explanation:

You might be interested in

a car with a mass of 1200 kilograms is moving around a circular curve at a uniform velocity of 20 meters per second. the centrip

qaws [65]

Well, first of all, a car moving around a circular curve is not moving
with uniform velocity. The direction of motion is part of velocity, and
the direction is constantly changing on a curve.

The centripetal force that keeps an object moving in a circle is

   Force = (mass of the object) · (speed)² / (radius of the circle)

   F = m s² / r

We want to know the radius, to rearrange the formula to give us
the radius as a function of everything else.

      F   = m s² / r

Multiply each side by 'r': F· r = m · s²

Divide each side by 'F':    r = m · s² / F

We know all the numbers on the right side,
so we can pluggum in:

r =       m    · s² /    F

      r = (1200 kg) · (20 m/s)² / (6000 N) .

I'm pretty sure you can finish it up from here.

5 0

3 years ago

A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

7 0

3 years ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

3 years ago

Question 4 (2 points)

Igoryamba

The answer is a) Teres Major Muscle

7 0

2 years ago

Read 2 more answers

Earth's gravitational force just got three times stronger! What happens to your weight?

kicyunya [14]

Your "weight" is the name you give to that gravitational force.
So your question actually says:

"Your weight just got three times stronger !
What happens to your weight ?"

8 0

3 years ago