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2 years ago

The black hole is___ times smaller that the star. I need answers please

Physics

1 answer:

mojhsa [17]2 years ago

3 0

Answer:

The answer is 24 (for the first question).

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>

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Consider the two-body situation at the right. A 300kg crate rests on an inclined plane and is connected by a cable to a 100 kg m

trasher [3.6K]

Answer:

a= 0.578 m/s

T = 1037.8 N

Explanation:

Data

m₁= 300 kg

m₂= 100 kg

inclined plane, θ = 30°

μk = 0.120

Newton's second law to m₁:

We define the x-axis in the direction parallel to the movement of the 300kg (m₁) crate on the ramp and the y-axis in the direction perpendicular to it.

∑F = m₁*a Formula (1)

Forces acting on m₁

W₁: m₁ weight : In vertical direction

N : Normal force : perpendicular to the inclined plane

f : Friction force: parallel to the inclined plane

T: cable tension : parallel to inclined plane

Calculated of the W₁

W₁=m₁*g

W₁= 300kg* 9.8 m/s² = 2940 N

x-y weight components

W₁x= W₁sin θ =2940 N*sin(30)° =1470 N

W₁y= W₁cos θ =2940 N *cos(30)° =2156.4 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - W₁y = 0

N = W₁y

N = 2156.4 N

Calculated of the f

f = μk* N= (0.120)*(2156.4 N)

f = 258.77 N

Newton's second law to m₁ in direction x-axis :

∑Fx = m₁*ax ,ax =a

We assume that m₁ descends on the inclined plane and we positively take the direction of movement:

wx-f-T = m*a

wx - f - m*a =T

1470 -258.77 -300*a =T

T= 1211.23-300*a Equation (1)

Newton's second law to m₂

∑Fy = m₂*ay ,ay =a

Forces acting on m₂

W₂: m₂ weight : In vertical direction

T: cable tension:In vertical direction

Calculated of the W₂

W₂=m₂*g

W₂= 100kg* 9.8 m/s² = 980 N

∑Fy = m₂*a

Because we assume that m₁ descends on the inclined plane, then, m₂ ascends vertically, we take positive the direction of movement:

T-W₂ = m₂*a

T-980 = 100*a

T = 980 + 100*a Equation (2)

Problem development

Equation (1) = Equation (2) = T

1211.23-300*a= 980 + 100*a

1211.23- 980 = 100*a + 300*a

231.23 = 400*a

a= 231.23 / 400

a= 0.578 m/s

Because the acceleration tested positive then effectively m₁ descends on the inclined plane and m₂ ascends vertically.

We replace a= 0.578 m/s in the equatión (2)

T = 980 + 100* (0.578 )

T = 1037.8 N

5 0

3 years ago

if the Horse and Rider have a combined mass of 572 kg what force would be required to accelerate them 5 kph per second

Vlad [161]

Answer:

Force required to accelerate = 794.44 N

Explanation:

Force required = Mass of horse x Acceleration of horse

Mass of horse and rider, m= 572 kg

Acceleration of horse and rider, a = 5 kph per second

$=\frac{5*1000}{60*60} =1.39 m/s^2$

Force required = ma

= 572 x 1.39 = 794.44 N

Force required to accelerate = 794.44 N

8 0

3 years ago

An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16

Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg- $m^{2}$

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7× $10^{6}$ J

Angular velocity of the flywheel is 16000 $\frac{rev}{min}$ = 1675.51 $\frac{rad}{sec}$

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E = $\frac{1}{2}$ $Iw^{2}$

By rearranging the equation:

I = $\frac{2E}{w_{2} }$

I = 2.63 kg- $m^{2}$

Thus the moment of inertia of the flywheel is 2.63 kg- $m^{2}$ .

Learn more about moment of inertia here;

brainly.com/question/13449336

#SPJ4

7 0

1 year ago

I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?

OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

7 0

3 years ago

A subatomic particle that has a positive charge and that is located in the nucleus of an atom

wolverine [178]

Answer:

Explanation:

3 0

2 years ago

Read 2 more answers