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Lelu [443]
3 years ago
7

A slug traveling at 3 mm/h, East decided to race the slug next to him increasing his velocity to 5 mm/h, East in one hour. What

was the slug's acceleration?
Physics
1 answer:
N76 [4]3 years ago
7 0

The Acceleration of  slug  would be 2 mm/hr²

<u> Step by step Explanation:</u>

Given of the solution

Initial velocity of the slug,(we can represent as v_{i})  v_{i}= 3 mm/hr

Final velocity of the slug,(we can represent as v_{f} ) v_{f}= 5 mm/hr

Time, t = 1 hr

Acceleration, a = ?

We know,

a = \frac{(5-3)}{1} \\\\a = 2 mm/hr^2

Therefore, The Acceleration of  slug would be 2 mm/hr²

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What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.
Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0
3 years ago
A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.
Serga [27]

Answer:

P = 4000 [Pa]

Explanation:

Pressure is defined as the relationship between Force and the area where the body rests.

The support area is equal to:

A=50*20=1000[cm^{2} ]

But we must convert from square centimeters to square meters.

1000[cm^{2}]*\frac{1^{2}m^{2}  }{100^{2}m^{2}  }=0.1[m^{2} ]

And the pressure is:

P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]

6 0
2 years ago
One disadvantage of using proprietary licensed software is that
Vaselesa [24]

Answer:

its b

Explanation:

5 0
2 years ago
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At each corner of a square of side l there are point charges of magnitude Q, 2Q, 3Q, and 4Q.What is the magnitude and direction
bogdanovich [222]

Answer:

magnitude of force on charge 2Q  = \frac{KQ^{2} }{I^{2} }

Direction of force on charge = 61 ⁰

Explanation:

The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q  i.e x-component of the net force and the y-component of the net force

║F║ = \sqrt{f_{x}^{2} + f_{y}^{2}    }  =  after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW

magnitude of force acting on 2Q = \frac{KQ^{2} }{I^{2} }

The direction of the force on charge 2Q is calculated as

tan ∅ = \frac{f_{y} }{f_{x} } = 1.8284

therefore ∅ = tan^{-1}  1.8284

= 61⁰

3 0
3 years ago
What kind of image is formed by a convex lens when the object is less than a focal length away from the object?
maw [93]
Bigger image and can get clearer
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2 years ago
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