B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

Earth's polar ice caps, sea ice, glaciers, snow cover, and permafrost make up which sphere?

julsineya [31]

The Cryosphere "A subsystem of hydrosphere that consists of all the frozen water(snow and ice) on Earth.

3 0

3 years ago

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A car has a momentum of 20,000 kg • m/s. What would the car’s momentum be if its velocity doubles?

AnnyKZ [126]

To answer this question, it helps enormously if you know
the formula for momentum:

Momentum = (mass) x (speed) .

Looking at the formula, you can see that momentum is directly
proportional to speed. So if speed doubles, so does momentum.

If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.

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3 years ago

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a magnet has a 20 cm magnetic field if a piece of metal is 18 cm from the magnet will it be attracted or not

mina [271]

Answer:

yes

Explanation:

The metal is closer than 20 cm to the magnet which is in the magnetic field.

4 0

3 years ago

A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand

motikmotik

Given:

V1 = 4m3

T1 = 290k

P1 = 475 kpa = 475000 Pa

V2 = 6.5m3

T2 = 277K

Required:

P

Solution:

n = PV/RT

n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)

n = 788 moles

P = nRT/V

P = (788 moles)(8.314 Pa-m3/mol-K)(277K)/(6.5m3)

P = 279,204 Pa or 279 kPa

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3 years ago

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A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its eq

timama [110]

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

<h3>Total energy of the mass</h3>

The total energy possessed by the mass under the simple harmonic motion is calculated as follows;

U = ¹/₂kA²

where;

k is the spring constant
A is the amplitude of the oscillation

<h3>Potential energy of the mass at 5 cm from equilibrium point</h3>

P.E = ¹/₂k(Δx)²

<h3>Kinetic energy of mass</h3>

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

<h3>Percentage of its energy that is kinetic</h3>

$K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%$

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: brainly.com/question/25959744

3 0

2 years ago