Help me I got this photo please

"The International Space Station (ISS) orbits at a distance of 350 km above the surface of the Earth. (a) Determine the gravitat

vagabundo [1.1K]

Answer:

(a) g = 8.82158145 $m/s^2$ .

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

$g = \frac{M_{(earth)} }{r^2} G$ , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145 $m/s^2$ .

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

$a_{centripetal}=\frac{V^2}{r} =g$ here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c) S = vT, here T is time period or time required to complete one full revolution.

S = earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

3 0

3 years ago

The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about

nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s$

a) The vertical speed when the player leaves the ground is 4.45 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s$

Time taken to reach the maximum height is 0.45 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s$

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0

3 years ago

A student performs a lab measuring the velocities of toy cars of different masses

stiv31 [10]

The answer is Car 1 and Car 2.

5 0

3 years ago

Read 2 more answers

ماهي قوة التركيز من العين عندما ينضر الكاءن من ٢٠ الى ٥٠٠ من عينه

Debora [2.8K]

Answer:

What is the power of focus from the eye when a subject looks from 20 to 500 from its eye?

Explanation:

Is that your question?

4 0

2 years ago

Solve for the unknown in each of these circuits

Tamiku [17]

Ohms law: This law relates voltage difference between two points. Mathematically, the law states that V=IR;

Where

V = voltage difference ; in volts

I = Current ; in Amperes

R = Resistance ; in ohms

1. Answer : given that R = 10 ; V= 12 V ; I = ?

From ohms law, I = V/R

= 12/10

= 1.2 Amp.

2. Answer: given that R = 10 ; V= ? ; I = 5

From ohms law, V = IR

= 10×5 = 50 V

3 . Answer: given that R = ? ; V= 120 ; I = 5

From ohms law, R = V/I

= 120/5

= 24 Ω

4 . Answer: given that R = ? ; V= 10 ; I = 20

From ohms law, R = V/I

= 10/20

= 0.5 Ω

5 . Answer: given that R = 480 ; V= 24 ; I = ?

From ohms law, I = V/R

= 24/480

= 0.05 A

6. Answer: given that R = 150 ; V= ? ; I = 1

From ohms law, V = IR

= 1 × 150

= 150 V

6 0

3 years ago