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2 years ago

PLEASE I NEED HELP CLICK ON THIS IMAGE

Physics

1 answer:

3241004551 [841]2 years ago

6 0

I believe you are incorrect. A weathered mountain would appear more jagged.

I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.

<em>If this is incorrect, please, don't refrain to tell me.</em>

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Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S

Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0

2 years ago

A World War II bomber flies horizontally over level terrain, with a speed of 287 m/s relative to the ground and at an altitude o

Scorpion4ik [409]

Answer: 7.38 km

Explanation: The attachment shows the illustration diagram for the question.

The range of the bomb's motion as obtained from the equations of motion,

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2

R = 287 √(2×3240/9.8) = 7380 m = 7.38 km

6 0

2 years ago

A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on

lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0

2 years ago

A 32-kg child decides to make a raft out of empty 1.0-L soda bottles and duct tape. Neglecting the mass of the duct tape and pla

Vedmedyk [2.9K]

Answer:

32 bottles

Explanation:

If we create a free body diagram on the child we have his weight and the bouyant force

W-B=0

They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced

W=B=Ww

Mg=mg

32 kg=mass of water displaced

1 kilogram per liter (kg/L) is the density of water, this means that 32 Liters of water are displaced and since the bottles can retain 1 liter, the child needs 32 bottles

6 0

3 years ago

Which bulbs would be on and or off if :

insens350 [35]

Bulbs c and b would still be screwed in if they were in to begin with and bulbs A, D, and E. would be unscrewed

6 0

2 years ago

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