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Alekssandra [29.7K]
3 years ago
12

How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J

c.4.09 x 10^5 J d.4.77 x 10^5 J
Physics
1 answer:
algol133 years ago
6 0

Answer:

Work done by the frictional force is 3.41\times 10^5\ J

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W=k_f-k_i

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)

W = −340605 J

or

W=3.41\times 10^5\ J

Hence, the correct option is (a).

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The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum
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Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is 1.20×10^{-2}\ \mu F.

We need to calculate the resonance frequency

Using formula of frequency

f=\dfrac{1}{2\pi\sqrt{LC}}

Put the value into the formula

f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}

f=2356.88\ Hz

We need to calculate the maximum current

Using formula of current

I=\dfrac{V_{c}}{X_{c}}

I=2\pi\times f\times C\times V_{c}

I=2\pi\times2356.88\times1.20\times10^{-8}\times590

I=0.1048\ A

Impedance of the circuit is

z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}

At resonance frequency X_{L}=X_{C}

Z=R

We need to calculate the maximum voltage

Using ohm's law

V=I\times R

V=0.1048\times400

V=41.92\ V

Hence, The maximum voltage is 41.92 V.

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3 years ago
What is accurate about the planet’s climate system?
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The climate system is a highly complex global system consisting of five major components: the atmosphere, the ocean, the cryosphere (cryosphere), the land surface, the biosphere, and the interactions between them.

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1. The following can be inferred from Newton’s second law of motion except:
lora16 [44]

Answer: 1.d) The acceleration of an object is always less than the acceleration due to gravity, g (9.81m/s^-2)

2.a)acceleration decreases

Explanation:

Newton's second law:

Newton's second law states that the acceleration of an object is defined by two variables which is the total force acting on the object and the mass of that object. The acceleration is directly proportional to the net force that is applied on an object and inversely proportional to the mass of that object.

When the force applied on an object is increased so does the acceleration of an object however if the mass increase the acceleration decreases.

This can be felt when you look at the truck which usually carry heavy loads they seem to drive slow due to the load hence their acceleration is decreased by the mass that these truck carry .

7 0
3 years ago
A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c
natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

mg = \rho V_{displaced} g

V_{displaced} = \frac{m}{\rho}

V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

V\rho_w = m_{ice}

V = \frac{0.200}{1000} = 2\times 10^{-4} m^3

So here extra volume that rise in the level will be given as

\Dleta V = V - V_{displaced}

\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}

(\pi (0.039^2) h = 0.41 \times 10^{-4}

h = 0.86 cm

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

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