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3 years ago

A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?

Physics

1 answer:

lesya692 [45]3 years ago

3 0

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee = 5kg

Final speed = 12m/s

Unknown:

Impulse of the frisbee = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

Impulse = mass (Final velocity - Initial velocity)

Impulse = 5(12 - 0) = 60kgm/s

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At an accident scene on a level road, investigators measure a car's skid mark to be 93 m long. It was a rainy day and the coeffi

USPshnik [31]

Answer:

u = 25 m/s

Explanation:

given,

length of skid = 93 m

coefficient of friction = 0.35

final velocity = 0 m/s

initial velocity = ?

force here is friction f = μ mg

F = ma

now com paring

-μ mg = m a

a = - μ g

a = - 0.35 x 9.8

a = -3.43 m/s²

we know,

v² = u² + 2 a s

0 = u² - 2 x 3.43 x 93

u² = 637.98

u = 25.26 m/s

u = 25 m/s (two significant figure)

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3 years ago

A battery is used to power a flashlight. When the flashlight is in use, what type of energy is lost during energy transformation

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Answer:

The answer is chemical energy

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3 years ago

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A positive test charge of 5.00 E-5 C is places in an electric field. The force on it is 0.751 N. The magnitude of the electric f

zalisa [80]

Explanation:

Charge=5.00 E-5

Force=0.751N

F=qE

0.751=5.00 E-5*E

E=1.502*10⁴

option a

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3 years ago

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Find the potential energy of a 40kg cart that is on a hill that is 10 meters high

Vilka [71]

Explanation:

P.E. = mgh

PE = 40 × 10 × 10

PE = 4000 Joule

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2 years ago

A power plant produces 1000 MW to supply a city 40 km away. Current flows from the power plant on a single wire of resistance 0.

nikitadnepr [17]

Answer:

Current = 8696 A

Fraction of power lost = $\dfrac{80}{529}$ = 0.151

Explanation:

Electric power is given by

$P=IV$

where I is the current and V is the voltage.

$I=\dfrac{P}{V}$

Using values from the question,

$I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}$

The power loss is given by

$P_\text{loss} = I^2R$

where R is the resistance of the wire. From the question, the wire has a resistance of $0.050\Omega$ per km. Since resistance is proportional to length, the resistance of the wire is

$R = 0.050\times40 = 2\Omega$

Hence,

$P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2$

The fraction lost = $\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151$

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3 years ago