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tensa zangetsu [6.8K]
3 years ago
6

At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta

r it is orbiting is 3.0 × 10^23 N. The angle between the planet's momentum and the gravitational force exerted by the star is 63°.
What is the parallel component of the force on the planet by the star?
Physics
1 answer:
Evgesh-ka [11]3 years ago
3 0
727.5256266 AWNSERrRrrRr
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Which statement explains what happens to older crust during seafloor spreading?
ikadub [295]

Answer:

the last one, It moves away from a mid-ocean ridge.

6 0
3 years ago
A bicyclist accelerates from a stop to a speed of 12m/s in 3 seconds.what is her acceleration
AVprozaik [17]

Answer:

4m/s2

Explanation:

acceleration=v-u/t...v=12m/s,u=0m/s,t=3sec

a=12-0/3

=4m/s2

8 0
2 years ago
An airplane travels from st louis to portland, oregon in 4.33 hours. if the distance traveled is 2,742 kilometers, what is the a
Makovka662 [10]

Average speed = (total distance covered) / (total time to cover the distance)

                       =  (2,742 km)  /  (4.33 hours)

                       =  (2,742 / 4.33)    km/hr

                       =      633 km/hr        (rounded)
4 0
3 years ago
Read 2 more answers
alculate the kinetic energies of (a) a 2.00×103-kg automobile moving at 100.0 km/h; (b) an 80.0-kg runner sprinting at 10.0 m/s;
zzz [600]

Answer:

(a) 7.72×10⁵ J

(b) 4000 J

(c) 1.82×10⁻¹⁶ J

Explanation:

Kinetic Energy: This can be defined energy of a body due to its motion. The expression for kinetic energy is given as,

Ek = 1/2mv²................... Equation 1

Where Ek = Kinetic energy, m = mass, v = velocity

(a)

For a moving automobile,

Ek = 1/2mv².

Given: m = 2.0×10³ kg, v = 100 km/h = 100(1000/3600) m/s = 27.78 m/s

Substitute into equation 1

Ek = 1/2(2.0×10³)(27.78²)

Ek = 7.72×10⁵ J

(b)

For a sprinting runner,

Given: m = 80 kg, v = 10 m/s

Substitute into equation 1 above,

Ek = 1/2(80)(10²)

Ek = 40(100)

Ek = 4000 J

(c)

For a moving electron,

Given: m = 9.10×10⁻³¹ kg, v = 2.0×10⁷ m/s

Substitute into equation 1 above,

Ek = 1/2(9.10×10⁻³¹)(2.0×10⁷)²

Ek = 1.82×10⁻¹⁶ J

8 0
3 years ago
Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of
goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north


<u>Explanation</u>


Since displacement is a vector quantity, we take into account the direction.


Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.


We now add to obtain,

4+-2+5+-5

This will simplify to

=4-2+5-5=2

Therefore the combined displacement is 2km north

5 0
3 years ago
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