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3 years ago

Robert is riding the Giant Drop at Great America. If

Physics

1 answer:

lora16 [44]3 years ago

7 0

Explanation:

i hope this helps, its not the same person but its the same equation.

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You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do

melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

$W = F_g .d$

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0

3 years ago

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

lord [1]

Answer:

The value is $F_f = 46.935 \ N$

Explanation:

From the question we are told that

The magnitude of the horizontal force is $F = 92.7 \ N$

The mass of the crate is $m = 40.5 \ kg$

The acceleration of the crate is $a = 1.13 \ m/s$

Generally the net force acting on the crate is mathematically represented as

$F_{net} = F - F_f = ma$

Here $F_f$ is force of kinetic friction (in N) acting on the crate

$92.7 - F_f = 40.5 * 1.13$

=> $F_f = 46.935 \ N$

5 0

3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

In the 1960s what dance was being created in the west coast

DerKrebs [107]

The dance that was being created was Hip Hop dance!

3 0

3 years ago

Read 2 more answers

If an object were released in space far away from planted or stars and given an initial momentum, describe what would happen to

mamaluj [8]

Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.

Have a look at Newton's first law of motion.

7 0

3 years ago

Read 2 more answers