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attashe74 [19]
3 years ago
11

g A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has

a radius that is 90 % of Earth's (where Earth's radius is approximately 6370 km ). What is the gravitational acceleration of the meteoroid
Physics
1 answer:
Nikitich [7]3 years ago
3 0

Answer:

Explanation:

gravitational acceleration of meteoroid

=  GM / R²

M is mass of planet , R is radius of orbit of meteoroid from the Centre of the planet .

R = (.9 x 6370 + 600 )x 10³ m

= 6333 x 10³ m

M , mass of the planet = 5.97 x 10²⁴ kg .

gravitational acceleration of meteoroid

=  GM / R²

=  (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ kg  / (6333 x 10³ m)²

9.92m/s²

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12. A car is travelling at 30 m/s when the driver sees a red light in the distance and immediately applies the brakes. The car c
Triss [41]

Answer:

22.5 m

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

4 0
3 years ago
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
slamgirl [31]

Answer:

18.89cm

Explanation:

As we know that the person is standing 5m in front of the camera

d_0=5m=500cm

The focal length of the lens =50cm

f=50 cm

By Lens formula we have:

\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm

By the formula of magnification

\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm

The height of the image formed is 18.89cm.

5 0
3 years ago
A particle on a spring moves in simple harmonic motion along the x axis between turning points at x1 = 95 cm and x2 = 135 cm. (i
uranmaximum [27]

Answer:

(i) x = 115\,cm, (ii) x = 95\,cm, (iii) x = 95\,cm

Explanation:

(i) x_{1} and x_{2} represent the points where particle has a velocity of zero and spring reach maximum deformation, Given the absence of non-conservative force and by the Principle of Energy Conservation, the position where particle is at maximum speed is average of both extreme positions:

x = 115\,cm

(ii) Maximum accelerations is reached at x_{1} and x_{2}.

x = 95\,cm

(iii) Greatest net forces exerted on the particle are reached at  x_{1} and x_{2}.

x = 95\,cm

8 0
3 years ago
Please answer it is urgent <br>I will make u brainlist
adelina 88 [10]
For B, it is because water is a really good conductor of electricity, so the electrician will get shocked
6 0
4 years ago
A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
Blababa [14]

Answer:

The maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

\tau=NIAB\ \sin\theta

For maximum torque, \theta=90^{\circ}

\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m

So, the maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

4 0
3 years ago
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