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Vesna [10]
3 years ago
7

How much momentum does a 77 kg football player have if he is running with a speed of 4 m/s

Physics
1 answer:
worty [1.4K]3 years ago
4 0

Answer:

308 N-s

Explanation:

Momentum is given by

P= mv

P= 77 x 4

P= 308

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Punishment is used to B. decrease behavior
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Impulse is (force) x (time).

That's the <em>last choice</em> on the list.  It could be 'D', or '4', or 'Δ', or 'ד' etc.

If you play with it for a while, you discover that impulse has the same units as momentum, which certainly gives you something to think about.

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How would Newton's Three laws apply to a Ferris Wheel?
VladimirAG [237]

Answer:

Explanation:

Although this isn't as overtly involved, Newton's Third Law of Motion is also important because it helps you to better design the Ferris Wheel so that the centripetal force of the wheel's center is opposed by an equal force from the objects being spun around by the wheel.

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8 0
3 years ago
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

T_{f} =  (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

T_{f}  = 466.8 K

so change in temperature ΔT

ΔT =  466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x_{i\\}² - x_{f\\}² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q =  16 J

Therefore, the required heat energy is 16 J

5 0
3 years ago
Each of the gears a and b has a mass of 675 g and has a radius of gyration of 40 mm, while gear c has a mass of 3. 6 kg and a ra
navik [9.2K]

9.87 seconds

The time required for this system to come to rest is equal to 9.87 seconds.

We have the following data:

Mass of gear A = 675 g to kg = 0.675 kg.

Radius of gear A = 40 mm to m = 0.04 m.

Mass of gear C = 3.6 kg.

Radius of gear C = 100 mm to m = 0.1 m.

How can I calculate the time needed?

We would need to figure out the moment of inertia for gears A and C in order to compute the time needed for this system to come to rest.

Mathematically, the following formula can be used to determine the moment of inertia for a gear:

I = mr²

Where:

m is the mass.

r is the radius.

We have, For gear A:

I = mr²

I = 0.675 × 0.04²

I = 0.675 × 0.0016

I = 1.08 × 10⁻³ kg·m².

We have, For gear C:

I = mr²

I = 3.6 × 0.1²

I = 3.6 × 0.01

I = 0.036 kg·m².

The initial angular velocity of gear C would therefore be converted as follows from rotations per minute (rpm) to radians per second (rad/s):

ωc₁ = 2000 × 2π/60

ωc₁ = 4000π/60

ωc₁ = 209.44 rad/s.

Also, the initial angular velocity of gears A and B is given by:

ωA₁ = ωB₁ = rc/rA × (ωc₁)

ωA₁ = ωB₁ = 0.15/0.06 × (209.44)

ωA₁ = ωB₁ = 2.5 × (209.44)

ωA₁ = ωB₁ = 523.60 rad/s.

Taking the moment about A, we have:

I_A·ωA₁ + rA∫F_{AC}dt - M(f)_A·t = 0

On Substituting the given parameters into the formula, we have;

(1.08 × 10⁻³)·(523.60) + 0.06∫F_{AC}dt - 0.15t = 0

0.15t - 0.06∫F_{AC}dt = 0.56549   ----->equation 1.

Similarly, the moment about B is given by:

0.15t - 0.06∫F_{BC}dt = 0.56549    ------>equation 2.

Note: Let x = ∫F_{BC}dt + ∫F_{AC}dt

Adding eqn. 1 & eqn. 2, we have:

0.3t - 0.06x = (0.56549) × 2

0.3t - 0.06x = 1.13098  ------>equation 3.

Taking the moment about A, we have:

Ic·ωc₁ - rC∫F_{AC}dt - rC∫F_{BC}dt - Mc(f)_A·t = 0

0.036(209.44) - 0.3t - 0.15(∫F_{BC}dt + ∫F_{AC}dt) = 0

0.3t + 0.15x = 7.5398    ------->equation 4.

Solving eqn. 3 and eqn. 4 simultaneously, we have:

x = 30.5 Ns.

Time, t = 9.87 seconds.

To learn more about moment of inertia visit:

brainly.com/question/15246709

#SPJ4

6 0
2 years ago
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