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Mandarinka [93]
3 years ago
6

How long would this acceleration last

Physics
1 answer:
Aleksandr-060686 [28]3 years ago
6 0
It would last as long as the applied force continued, or until the accelerating object hit something.
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What are various systems of unit?​
Naya [18.7K]

Answer:

hlw its jess bregoli

your answer is here

SI (International System of Units) (meter-kilogram-second-ampere-kelvin-mole-candela)

FPS (foot-pound-second)

MKS (meter-kilogram-second)

CGS (centimeter-gram-second)

EMU (Electromagnetic) (centimeter-gram-second-abampere)

ESU (Electrostatic) (centimeter-gram-second-abcoulomb)

Atomic (bohr-electron mass-atomic second-electron)

MTS (meter-tonne-second)

Explanation:

hope it may help you !!

6 0
2 years ago
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Alexandra is dragging her 37.8 kg golden retriever cross the wooden floor by applying a horizontal force. What force must be app
Mila [183]

Answer:

F = 385.56 N

Explanation:

Given that,

Mass, m = 37.8 kg

He applies a horizontal force and cross the wooden floor.

We need to find the force that must be applied to move a dog with a constant speed of 1 m/s.

The coefficient of kinetic friction between the dog in the floor is 1.02.

The net force acting on it is given by :

F = f- μmg

f is force due to constant speed, f = 0 (since, a = 0)

F = μmg

= 1.02 × 37.8 × 10

= 385.56 N

Hence, the required force is 385.56 N.

6 0
3 years ago
What are 90 percent of the stars in space called?
Naddik [55]
Yes there called stars because when a star in space goes nova it shines in the sky like a little twinkle because it far away the sun is also a star too.<span />
6 0
3 years ago
Two boxes sit side by side on a smooth horizontal surface. The lighter box 5.2 kg, the heavier box has a mass of 7.4kg (a) find
Paladinen [302]

OK.  So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.

-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.

-- You're pushing it with 5.0N of force.

-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)

-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.

The force on it to make it accelerate is (mass) x (acceleration) =

                                                              (7.4 kg) x (5/12.6 m/s²) =  <em>2.936 N.</em>

The only force on the big box comes from the small box, pushing it from behind. 
So that same  <em>2.936N</em>  must be the contact force between the boxes.

8 0
3 years ago
If a truck loses 5610 J of energy as it slows down due to an external force of 425 N, what distance will it have moved after the
lara31 [8.8K]

Answer:

13.2m

Explanation:

Step one:

given data

Energy= 5610J

Force F= 425N

Required

The distance traveled

Step two:

We know that work done is given as

WD= force* distance

so

5610=425*d

divide both sides by 425

d= 5610/425

d=13.2m

3 0
3 years ago
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