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4 years ago

8

Why does Gravity have a lesser than grater effect on higher altitudes?

1 answer:

IrinaVladis [17]4 years ago

4 0

The strength of the gravitational force between two objects depends on the distance between their centers.

At higher altitude, the center of the Earth is farther from the center of YOU, so the strength of the gravitational force between those two objects is less.

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How long does it take for mars to rotate on its axis

Svetllana [295]

Hello There!

It takes the planet Mars around 24 hours, 37 minutes, 23 seconds to rotate on its axis. This is around the same amount of time that it takes our planet to rotate once on its axis.

8 0

3 years ago

Read 2 more answers

A thin hoop is supported in a vertical plane by a nail. What should the radius of the hoop be in order for it to have a period o

maxonik [38]

Answer:

0.124 m

Explanation:

the period of a simple pendulum with a small amplitude is given as

T = 2π [√(I/mgd)]

From the above stated formula,

I = moment of inertia

m = mass of the pendulum

g = acceleration due to gravity, 9.8 m/s²

d = distance from rotation axis due to center of gravity

Also, moment of Inertia, I = 2mr², if we substitute this in the above formula, we have

T = 2π [√(2mr²/mgd)]

If we assume that our r = d, then we have

T = 2π [√(2r/g)]

If we make r the subject of the formula in the above equation, we get

r = gT² / 8π²

r = (9.8 * 1²) / 8 * π²

r = 9.8 / 78.98

r = 0.124 m

Thus, the radius of the hoop is 0.124 m

4 0

3 years ago

If it takes 35J to lift up an object to some height, then how much gravitational potential energy will that object have?

uysha [10]

35j because if it takes so much to lift it’s that much pullin down

8 0

3 years ago

Two point charges of 60.0 C and -12.0 C are separated by a distance of 20.0 cm. A 7.00 C charge is placed midway between these t

iogann1982 [59]

Answer:

$4.5\times 10^{14} N$

Explanation:

$q_1=60 C$

$q_2=-12 C$

$q_3=7 C$

Distance between q1 and q2,d=20 cm

$r=\frac{d}{2}=\frac{20}{2}=10 cm=10\times 10^{-2} m$

1m =100 cm

Electric force on charge q3 due to charge q1

$F_1=\frac{kq_1q_3}{r^2}=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}$ (away from q1)

Electric force on charge q3 due to charge q2

$F_2=\frac{kq_2q_3}{r^2}=\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}$ (attract toward q2)

Net force= $F=F_1+F_2$

$F=\frac{9\times 10^9\times 60\times 7}{(10\times 10^{-2})^2}+\frac{9\times 10^9\times 12\times 7}{(10\times 10^{-2})^2}$

$F=\frac{9\times 10^9\times 7}{(10\times 10^{-2})^2}(60+12)$

$F=4.5\times 10^{14} N$

5 0

4 years ago

What is the minimum amount of energy required to completely melt a 7.25-kg lead brick which has a starting temperature of 18.0 °

Delvig [45]

Answer: c. 4.56 × 105 J

Explanation:

Given that

mass of lead brick, m= 7.25kg

Temperature T1 = 18.0 °C

Temperature T2 = 328 °C

specific heat capacity of lead, c = 128 J/(kg∙C°)

latent heat of fusion Lfusion =23,200 J/kg

Amount of energy Q =?

Using the formulae

Amount of energy ,Q =mc ( T2-T1)+ mLfusion

7.25kg x 128 J/(kg∙C°) x (328-18°C) + 7.25kg x 23200 J/kg

=455880J

=4.56 x 10^5 J

5 0

3 years ago