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3 years ago

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If the mass of a portion is 1.67 x 10-27kg and the mass of an electron is 9.11 x 10-31kg, calculate the force of gravitation bet

ween: i. a proton and an electronii. two electronsiii. two protons. Take G= 6.67 x 10-11 Nm2kg-2, distance between the protons = 4.0m., distance between the electrons = 2 x 10-2m, distance between the proton and the electron = 5.4 x 10-11m

1 answer:

emmainna [20.7K]3 years ago

7 0

Answer:

1.38 into 10 power minus 58 Newton is the answer

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What is the relation between the gravitational force acting between any two bodies object with their masses and the distance bet

LenaWriter [7]

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

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3 years ago

A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an

3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

$a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2$

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

$F_s = am = 1.31*16 = 20.92 N$

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

$F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N$

So the maximum acceleration on the block is

$a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2$

4)As the box slides, it is now subjected to kinetic friction, which is

$F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N$

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0

3 years ago

A truck moves 60 kilometers east from point A to point B. At point B, it turns back west and stops 15 kilometers away from point

iogann1982 [59]

Answer:

Distance: 75 km

Displacement: 45 km

Explanation:

- Distance is a scalar quantity that refers to the total space covered by an object. It is calculated as the sum of the distances covered in each motion, regardless of their direction. therefore in this case:

distance = 60 km + 15 km = 75 km

- Displacement is a vector quantity whose magnitude is equal to the difference between the final point and the starting point of the motion, so it also takes into account the direction of each motion. In this case, the truck moves 60 km east, and then 15 km west: if we call '0' the starting point, the final point will be then

$p_B = 60 km - 15 km=45 km$

And so the displacement is

$d=p_B - p_A=45 km - 0 = 45 km$

6 0

3 years ago

Read 2 more answers

4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two

Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=mV_{o}+MU_{o}$ (2)

$p_{f}=(m+M)V_{f}$ (3)

$m=1200 kg$ is the mass of the first car

$V_{o}=20 m/s$ is the velocity of the first car, to the North

$M=1400 kg$ is the mass of the second car

$U_{o}=-22 m/s$ is the mass of the second car, to the South

$V_{f}$ is the final velocity of both cars after the collision

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg}$ (6)

Finally:

$V_{f}=-2.61 m/s$ (7) This is the resulting velocity of the wreckage, to the south

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4 years ago

The diagram below represents the rock cycle.

lara31 [8.8K]

Answer:

C. Heat and Pressure

Explanation:

The arrow which is labeled A points from igneous rock to metamorphic rock.

There are three types of Rocks:

1. Igneous Rock

2. Metamorphic Rock

3. Sedimentary Rock

Rock cycle:

Rock cycle is the process that describes the transition between these three types of rocks. Each type has its own form and its own equilibrium condition. The rock type alters when it is pushed out of its equilibrium conditions.

Transition of Igneous rock to Metamorphic rock:

Igneous rock forms when magma cools down. The transition of Igneous Rock to Metamorphic Rock is a result of a process called Metamorphism. Metamorphism is the alteration in the structure of rock as a result of certain heat and pressure conditions. Inside Earth heat comes from pressure. Heat with pressure does not melt the rock but it bakes the rock. Baking is not melting but it changes the shape of the rock while it is still solid. It actually forms crystals. Because the rock changes its structure, it is called Metamorphic Rock.

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3 years ago

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