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ANTONII [103]
4 years ago
15

An astronomer sees two stars in the sky. Both stars are equally far from Earth, but the first star is brighter than the second s

tar. Which of the following is a valid conclusion about the first star?
It is larger than the second star. It is smaller than the second star. It is older than the second star. It is younger than the second star.
Physics
1 answer:
Margaret [11]4 years ago
3 0
It is larger than the second star.
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A lever in which the load is between the fulcrum and the applied force is a ________.
MrMuchimi
Hello there.

<span>A lever in which the load is between the fulcrum and the applied force is a ________.

Answer: It is a second class lever.

Hope This Helps You!
Good Luck Studying ^-^</span>
4 0
3 years ago
If Sally is standing on a 200m tall cliff and throws a ball at 40m/s at a 30° angle to the horizontal: a. What is the ball's ini
Irina-Kira [14]

Answer:

(a) 20 m/s (j) m/s

(b) 20√3 m/s (i) m/s

(c) 2.04 s

(d) 20.4 m

Explanation:

In order to solve the problem, you have to apply the <em>Projectil Motion</em> equations.

For part (a) and (b) you have to obtain the components of the initial velocity vector. The direction forms a 30° angle to the horizontal and the modulus (speed) was given. Therefore:

Applying trigonometric identities (Because the initial velocity is the hypotenuse of a right triangle with angle 30° to the horizontal)

Vx: 40Cos(30°)=20√3 m/s

Vy:40Sin(30°)= 20 m/s

The initial velocity in the y direction is: 20 m/s (j) m/s

The initial velocity in the x direction is: 20√3 m/s (i) m/s

Where i and j are the unit vectors.

For part (c) you have to apply the following vertical motion equation:

Vy=Voy-gt

where Voy is the initial velocity, g is gravity and t is the time

The ball reaches its max height when Vy=0 therefore:

0=Voy-gt

Solving for t:

t=Voy/g=20/9.8= 2.04 seconds

For part (d) you have to apply the other vertical motion equation which is:

y=yo+Voyt-0.5gt²

Where yo is the initial position.

Replacing t=2.04 s, yo=0 m, Voy=20 m/s and solving for y:

y=0+(20)(2.04)-(0.5)(9.8)(2.04)²

y=20.4 m

3 0
3 years ago
A ball is thrown vertically downwards with a speed 7.3 m/s from the top of a 51 m tall building. With what speed will it hit the
ddd [48]

Answer:

32.46m/s

Explanation:

Hello,

To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are the follow

\frac {Vf^{2}-Vo^2}{2.a} =X

Where

Vf = final speed

Vo = Initial speed =7.3m/S

A = g=acceleration =9.81m/s^2

X = displacement =51m}

solving for Vf

Vf=\sqrt{Vo^2+2gX}\\Vf=\sqrt{7.3^2+2(9.81)(51)}=32.46m/s

the speed with the ball hits the ground is 32.46m/s

8 0
3 years ago
Read 2 more answers
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the pro
Novay_Z [31]

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) ×  (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

6 0
3 years ago
Newton’s third law says that every time there is an ___________force, there is also a__________ force that is __________ in size
lesantik [10]

Newton’s third law of motion says that every time there is an ____action_______ force, there is also a ______reaction________ force that is ___equal________ in size and acts in the ___opposite____________ direction. This means that forces ALWAYS occur in __pairs________.

Explanation:

3 0
3 years ago
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