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Virty [35]
3 years ago
5

15) A 1720 kg car accelerates at a rate of 3.0 m/s2. How much force is the car's engine producing? (use the formula F=ma

Physics
2 answers:
AleksAgata [21]3 years ago
5 0

Answer:

5160 N

Explanation:

Anastasy [175]3 years ago
4 0

Answer:

<h3>The answer is 5160 N</h3>

Explanation:

To find the force acting on an object given it's mass and acceleration we use the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 1720 kg

acceleration = 3.0 m/s²

We have

Force = 1720 × 3

We have the final answer as

<h3>5160 N</h3>

Hope this helps you

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No, the number of particles does not change as the substance changes its state.

<h3><u>Explanation: </u></h3>

Change of state from one phase to another is achieved by providing or absorbing heat or pressure. For instance, liquid water if heated becomes vapour steam and if cooled becomes solid ice. Vapour can be compressed to form liquid water again and thus change of state is a reversible action.

The "chemical composition of the matter remains the same" irrespective of its state. Unless a chemical change is carried out, no change occurs with the number of particles. Phase change only affects the "arrangement of molecules", its structure and its motion.

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Answer:

B and D is the answer

Explanation:

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Although the Earth has a lesser mass than the Sun, it is far closer to you, allowing for a stronger pull.

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3 years ago
A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases he
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The skater's final angular speed is equal to 12 rad/s.

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The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.

In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.

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7 0
2 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
3 years ago
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