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2 years ago

An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?

Physics

2 answers:

anyanavicka [17]2 years ago

8 0

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

Papessa [141]2 years ago

3 0

Answer:

10 seconds

Explanation:

It takes 10 seconds for an object to fall from an elevation of 250 ft.

250 = 0 + (0) t + ½ (5) t²

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A hot air balloon is hovering in the air when it drops a 40 Kg food package to some lost golfers. If the package is dropped from

UNO [17]

We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:

Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.

(1/2)*m*v²=m*g*h, masses cancel out and we get:

(1/2)*v²=g*h, and we multiply by 2 both sides of the equation

v²=2*g*h, and we take the square root to get v:

v=√(2*g*h)

v=99.04 m/s

So the package is moving with the speed of v= 99.04 m/s when it hits the ground.

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2 years ago

Two planets have masses 2 x 10^23 kg and 5 x 10^22 kg, and the distance between them is 3 x 10^16 m. Calculate the gravitational

xxTIMURxx [149]

Explanation:

given,

mass of one planet (m1)=2*10^23 kg

mass of another planet (m2)=5*10^22kg

distance between them(d)=3*10^16m

gravitational constant(G)=6.67*10^-11Nm^2kg^-2

gravitational force between them(F)=?

we know,

F=Gm1m2/d^2

or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2

or, F=6.67*2*5*10^-11+23+22/3*3*10^32

or, F=66.7*10^34/9*10^32

or, F=7.41*10^34-32

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A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

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