A boxer throws a punch with a force of 1,400 N that lasts 0.02 s. What is the impulse of this punch? (1 point) 28 kg⋅m/s 28 kilograms times meters per second 280 kg⋅m/s 280 kilograms times meters per second 70,000 kg⋅m/s 70,000 kilograms times meters per second 7,000 kg⋅m/s

Answers

Answer 1

The impulse of the boxers punch is 28 kgm/s.

The given parameters;

applied force by the boxer, F = 1400 Ntime of force action, t = 0.02 s

The impulse of the boxers punch is calculated as follows;

[tex]J = Ft[/tex]

where;

F is the applied force (N)t is the time of force action (s)

The magnitude of the impulse is calculated as follows;

[tex]J = 1400 \times 0.02 \\\\J = 28 \ kg m/s[/tex]

Thus, the impulse of the boxers punch is 28 kgm/s.

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Explanation:

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Answer:

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Without friction, the car cannot stop...

What is the sum of 130+125+191? A. 335. B. 456. C. 446. D. 426.
If we minus 712 from 1500, how much do we get? A. 788. B. 778. C. 768. D. 758.
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1. C- 446
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1. Frequency

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When a log of wood is put over the freshwater, the log start to float. Option B is correct.

Density:

It can be defined as the amount of substance per unit volume. It is usually denoted by D. It is measured in the [tex]\bold { kg/m^3}[/tex].

The less dense object always floats over the more dense substance.

For example- Here, The density of the log is 0.8 [tex]\bold{ cm^3}[/tex] and the density of the of water is 1.0 [tex]\bold{ cm^3}[/tex] .

Since, the density of the log is lesser than the density of the water.

Therefore, when a log of wood is put over the freshwater, the log start to float.

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Objects want to continue doing what they’re doing because they are “lazy.” This is called law of inertia.

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is being acted upon by an external force. This law is also called the law of inertia because it depends on mass.

From the given question, we can fill gaps as follows;

Objects want to continue doing what they’re doing because they are “lazy.” This is called law of inertia.

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When mechanical waves interact with material, it transfer its energy to the material.

What are mechanical waves?

Mechanical waves are the type of waves that require a medium for transporting their energy from one region to another. Mechanical waves depend on particle interaction in order to transport their energy.

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Answer:

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Answers

The statement  " Brand-W motor does 7,640 N of your toughest work." can be truthfully use.

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During chemical reactions, bonds between atoms break and form Electrical fields of charged particles interact, bonding those with opposite charges.

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Answer:
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Answer:

Explanation:

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Explanation:

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Answer:

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Explanation:

Scientific notation is a way of expressing numbers that are too large or too small to be conveniently written in decimal form. It may be referred to as scientific form or standard index form, or standard form in the UK.

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A(n) 1700 kg car is moving along a level road at 21 m/s. The driver accelerates, and in the next 10 s the engine provides 22000 J of additional energy. If 3666.67 J of this energy must be used to overcome friction, what is the final speed of the car

Answers

The final speed of the car at the given conditions is 30.1 m/s.

The given parameters:

Mass of the car, m = 1700 kgVelocity of the car, v = 21 m/sTime of motion, t = 10 sAdditional energy provided by the engine, E₁ = 22,000 JEnergy used in overcoming friction, E₂ = 3,666.67 J

The change in the energy applied to the car is calculated as;

[tex]\Delta E = E_1 - E_2\\\\\Delta E = 22,000 \ J \ - \ 3,666.67 \ J\\\\\Delta E = 18,333.33 \ J[/tex]

The final speed of the car is calculated as follows;

[tex]\Delta E = \frac{1}{2} m(v_f^2 - v_0^2)\\\\v_f^2 - v_0^2 = \frac{2\Delta E}{m} \\\\v_f^2 = \frac{2\Delta E}{m} + v_0^2\\\\v_f = \sqrt{ \frac{2\Delta E}{m} + v_0^2} \\\\v_f = \sqrt{ \frac{2\times 18,333.4}{1700} + (21)^2} \\\\v_f = 30.1 \ m/s[/tex]

Thus, the final speed of the car at the given conditions is 30.1 m/s.

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In a car engine, fuel is ignited within a cylinder to cause an explosion that moves a piston. If the explosion exerts 300 Pa of pressure on a piston with a surface area of 80 cm^2?

Answers

The force exerted by the explosion on the piston is 2.4 N.

The given parameters:

Pressure of the explosion on piston, P = 300 Pa (N/m²)Surface area of the piston, A = 80 cm²

The force exerted on the piston by the explosion is calculated as follows;

F = PA

where;

A is the area of the piston in m²

The surface area of the piston in square meter is calculated as follows;

[tex]A = 80 \ cm^2 \times \frac{(1m)^2}{(100 \ cm)^2} \\\\A = 0.008 \ m^2[/tex]

The force exerted by the explosion on the piston is calculated as follows;

[tex]F = 300 \ (N/m^2)\ \times \ 0.008 \ (m^2)\\\\F = 2.4 \ N[/tex]

The complete question is here:

In a car engine, fuel is ignited within a cylinder to cause an explosion that moves a piston. If the explosion exerts 300 Pa of pressure on a piston with a surface area of 80 cm^2, what is the force exerted on the piston?

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desventajas de la fricción cinética

Answers

Answer:

Please translate this into Spanish! Sorry I can't :(

Explanation:

Disadvantages of Friction:

Friction produces unnecessary heat leading to the wastage of energy.

The force of friction acts in the opposite direction of motion, so friction slows down the motion of moving objects.

Forest fires are caused due to the friction between tree branches.

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The answer would be 20000

The answer in standard form/scientific notation would be 2 x 10^4

(The exponent is 4 because that's how many digits after the 2 there is)

Find the distance traveled in one back-and-forth swing by the weight of a 12 in. Pendulum that swings through a 75 degree angle.

Answers

The distance traveled by pendulum, in one back-and-forth swing is 75.75 inches.

The period of pendulum can be calculated by

[tex]T = 2\pi \sqrt {\dfrac Lg}[/tex]

Where,

[tex]T[/tex] - period

[tex]L[/tex] - length = 12 inches

[tex]g[/tex]- gravitational acceleration  = [tex]\bold {9.8\rm \ m/s^2}[/tex]

Put the values,

[tex]T = 2\pi \sqrt {\dfrac {12}{9.8}}\\\\T = 2 \times 3.14 \times \sqrt {0.122}\\\\T = 2.191[/tex]

Now, the angular displacement of the pendulum can be calculated by,

[tex]\theta = A\times\rm \ cos(\omega T)[/tex]

Where,

[tex]A[/tex]- amplitude

[tex]\theta[/tex] - angle  = [tex]75^o[/tex]

[tex]\omega[/tex] - angular displacement = [tex]2\pi/T[/tex] = 2.866 m

Put the values and calculate for [tex]\omega[/tex],

[tex]75 = A\times{\rm \ cos}(2.866\times 2.191)\\\\75 =A \times cos\ 6.26\\\\A =\dfrac {75}{0.99}\\\\A = 75.75 \rm \ inches[/tex]

Therefore, the distance traveled by pendulum, in one back-and-forth swing is 75.75 inches.

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