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Butoxors [25]
2 years ago
8

A stationary 15 kg object is located in a table near the surface of the earth. The coefficient of static friction between the su

rfaces is 0.40 and the coefficient of kinetic friction is 0.25.
A horizontal force of 40 N is applied to the object.
Determine the acceleration of the object.
Physics
1 answer:
Marta_Voda [28]2 years ago
8 0

The net force acting on the object perpendicular to the table is

∑ F[perp] = F[normal] - mg = 0

where mg is the weight of the object. Then

F[normal] = mg = (15 kg) (9.8 m/s²) = 147 N

The maximum magnitude of static friction is then

0.40 F[normal] = 58.8 N

which means the applied 40 N force is not enough to make the object start to move. So the object has zero acceleration and does not move.

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8 0
3 years ago
According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
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Answer:

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7 0
2 years ago
Read 2 more answers
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

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