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3 years ago

If the gravitational field strength at the top of Mount Everest is 9.772 N/kg, approximately how tall (in feet) is

Physics

1 answer:

Delicious77 [7]3 years ago

6 0

Answer:

The height of the Everest mountain is, x = 8514.087 m

Explanation:

Given data,

The gravitational field strength at the top of mount Everest, gₓ = 9.772 m/s²

The formula for gravitational field strength is,

Where, x is the height from the surface of the Earth

Therefore,

(R+x)² = GM/gₓ

x = √(GM/gₓ) - R

Substituting the values,

x = √(6.67408 x 10⁻¹¹ X 5.972 x 10²⁴ / 9.772) - 6.378 x 10⁶

x = 8514.087 m

Therefore, the height of the Everest mountain is, x = 8514.087 m

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What property describes the ability of a material to be flattened into thin sheets by hammering? options:

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10. A worker uses a pulley system to raise a 24.0 kg carton 16.5 m. A force of 129 N is exerted and the rope is pulled 33.0 m. a

lozanna [386]

Answer: Machanical advantage of the machine is 1.86

Explanation: Machanical advantage of a machine is the ratio of the Force to overcome which is the load in this case 24kg * 10= 240N to the force exerted(Effort) to overcome the load in this case 129N.

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3 years ago

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3 years ago

A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with

Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

$V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)$

Charge density of the shell is,

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}$

Here, $R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q$

$\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}$

The electric field is $E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}$

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

For R <r <2R According to gauss law

$E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)$

substitute $\rho=\frac{-3Q}{28\piR^3}$

$E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}$

The net charge enclosed for each r in this range is positive and the electric field is outward

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3 years ago

A 150kg person stands on a compression spring with spring constant 10, 000 N/m and norminal.Length of 0.50.What is the togal len

Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity $g=9.8m/sec^2$

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

$mg=kx$

$150\times 9.8=10000\times x$

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

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3 years ago