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3 years ago

“A 39.2 N object is pushed with 12 N of force but experiences 3 N of frictionWhat is the object's acceleration?”

Physics

1 answer:

Aleks [24]3 years ago

5 0

Answer:

The awnser would be 2.25m/s square

Explanation:

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Is India a rich country?

Norma-Jean [14]

Explanation:

India. Total wealth: $8.9 trillion | Wealth per capita: $6,440 | India, which is the fifth-largest economy in the world, is home to 3,57,000 HNWIs and 128 billionaires.

4 0

2 years ago

Read 2 more answers

If the car ha a mass of 1000 kilograms, what is its momentum (v=35m/s)

Black_prince [1.1K]

Answer:

<em>The momentum of the car is 35,000 kg.m/s</em>

Explanation:

<u>Momentum</u>

Momentum is often defined as <em>mass in motion.</em>

Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:

$\vec p = m\vec v$

If only magnitudes are considered:

p = mv

The car has a mass of m=1,000 kg and travels at v=35 m/s. Calculating its momentum:

p = 1,000 kg * 35 m/s

p = 35,000 kg.m/s

The momentum of the car is 35,000 kg.m/s

4 0

3 years ago

Use Snell's Law to solve the following:

azamat

Answer:

1.171

Explanation:

if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;

$n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705$

3 0

3 years ago

A jet plane is cruising at 300 m/s when suddenly the pilot turns the engines up to full throttle. After traveling 4.0 km, the je

jenyasd209 [6]

<h2>The answer got is reasonable.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 300 m/s

Acceleration, a = ?

Final velocity, v = 400 m/s

Displacement,s = 4 km = 4000 m

Substituting

v² = u² + 2as

400² = 300² + 2 x a x 4000

a = 8.75 m/s² = 8.8 m/s²

The acceleration is 8.8 m/s²

The answer got is reasonable.

7 0

3 years ago

A man does 4,780 J of work in the process of pushing his 2.70 103 kg truck from rest to a speed of v, over a distance of 25.5 m.

wolverine [178]

Answer:

(A) Velocity will be 1.88 m/sec

(b) Force will be 187.45 N

Explanation:

We have given work done = 4780 j

Distance d = 25.5 m

(A) Mass of the truck m = $m=2.70\times 10^3kg$

We know that kinetic energy is given by

$KE=\frac{1}{2}mv^2$

So $v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2\times 4780}{2.7\times 10^3}}=1.88m/sec$

(B) We know that work done is given by

W = Fd

So $F=\frac{W}{d}=\frac{4780}{25.5}=187.45N$

4 0

3 years ago