All categories

3 years ago

What conditions must be met if a man-made satellite is to attain a circular orbit around the earth?(choose as many as apply)

Physics

1 answer:

Hunter-Best [27]3 years ago

5 0

<span>C) It must be launched with a tangential speed greater than 8 km/s.</span>

You might be interested in

Which type of graph would be best for showing the percentage of people in a family with different jobs? A. Circle graph B. Scatt

musickatia [10]

Answer:

Explanation:

A circle graph typically represents numbers in percentages, used to visualize a part to whole relationship or a composition

6 0

4 years ago

a 20-kg object travelling at 20 m/s collides head on with an 18-kg object travelling at 17 m/s.If they were locked together afte

djverab [1.8K]

Answer:

2.47 m/s

Explanation:

Momentum = Mass X Velocity

If they were locked together, it means its a perfectly inelastic collision. Therefore,

Total momentum before = Total momentum after

Total momentum before = (20 X 20) - (18 X 17)

= 94

Total momentum after = 94

Y = Object speed after collision

94 = (20+18)Y

Y = 2.47368421 m/s

3 0

2 years ago

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

= $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

= $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

ma = qE

a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

= $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0

3 years ago

If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t

Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

μ₀ =permeability of free space 4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0

3 years ago

A policeman is chasing a criminal across a rooftop at 10 m/s. He decides to jump to the next building which is 2 meters across f

valina [46]

Answer:

They will meet at a distance of 7.57 m

Given:

Initial velocity of policeman in the x- direction, $u_{x} = 10 m/s$

The distance between the buildings, $d_{x} = 2.0 m$

The building is lower by a height, h = 2.5 m

Solution:

Now,

When the policeman jumps from a height of 2.5 m, then his initial velocity, u was 0.

Thus

From the second eqn of motion, we can write:

$h = ut + \frac{1}{2}gt^{2}$

$h = \frac{1}{2}gt^{2}$

$2.5 = \frac{1}{2}\times 10\times t^{2}$

t = 0.707 s

Now,

When the policeman was chasing across:

$d_{x} = u_{x}t + \frac{1}{2}gt^{2}$

$d_{x} = 10\times 0.707 + \frac{1}{2}\times 10\times 0.5 = 9.57 m$

The distance they will meet at:

9.57 - 2.0 = 7.57 m

8 0

3 years ago