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shusha [124]
3 years ago
10

Which is the 2nd closest star to earth??

Physics
2 answers:
IgorLugansk [536]3 years ago
8 0

Answer:

Proxima Centauri

Explanation:

U 2 can help me by marking as brainliest........

kykrilka [37]3 years ago
5 0
Proxima Centauri is the second closest star to Earth
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9. Steradian is the angel which lies in:<br> a) One dimension<br> b) Two dimensions
Marizza181 [45]

Answer:

B

Explanation:

B

8 0
3 years ago
Read 2 more answers
A bullet is fired straight up from a gun with a
melamori03 [73]

Answer: 815.51 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the initial velocity of the bullet has only y-component, since it was fired straight up. In addition, we are dealing with constant acceleration (due gravity), therefore the following equations will be useful to solve this problem:

V=V_{o}+gt (1)

V^{2}=V_{o}^{2}+2gy (2)

Where:

V is the final velocity of the bullet

V_{o}=152 m/s is the initial velocity of the bullet

g=-9.8 m/s^{2} is the acceleration due gravity, always directed downwards

t=6.9 s is the time

y is the vertical position of the bullet at t=6.9 s

Let's begin by finding V from (1):

V=152 m/s-9.8 m/s^{2}(6.9 s) (3)

V=84.38 m/s (4)

Now we have to substitute (4) in (2):

(84.38 m/s)^{2}=(152 m/s)^{2}-2(9.8 m/s^{2})y (5)

Isolating y:

y=815.511 m This is the displacement  of the bullet after 6.9 s

8 0
3 years ago
What is the resultant force of the free body diagram?
GaryK [48]

1) Magnitude of the net force: 20.8 N

2) Angle: 35.2^{\circ} clockwise from positive x-direction

Explanation:

1)

First of all, we need to find the components of the resultant force along the horizontal and vertical direction.

We observe that there are two forces acting along the horizontal direction:

F_1=33 N\\F_2 = 16 N

We notice that they act in opposite directions, so the net force in the horizontal direction is the difference between these two forces:

F_x =F_1-F_2=33-16=17 N (to the right)

Instead, there is only one force acting in the vertical direction, so the net force in the vertical direction is:

F_y=F_3=12 N (downward)

Therefore, the magnitude of the net force can be found by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}=\sqrt{17^2+12^2}=20.8 N

2)

Now we have to find the angle of the resultant net force. We can do it by using the following equation:

tan \theta = \frac{F_y}{F_x}

where

\theta is the angle, measured as clockwise from the positive x-direction (because the vertical net force is downward)

F_y = 12 N is the vertical net force

F_x=17 N is the horizontal net force

Solving, we find:

tan \theta=\frac{12}{17}=0.706

And so the angle is

\theta=tan^{-1}(0.706)=35.2^{\circ}

In a direction down with respect to the right.

Learn more about forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

About addition of vectors:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

4 0
4 years ago
Review for qtr 3 exam
Naddik [55]

Answer:

Uhm where are the questions and answers?

3 0
2 years ago
A person uses a constant force to push a 14.0 kg crate 1.80 m up a frictionless 10⁰ incline and to also increase its speed from
zubka84 [21]

a) Making h as the shorter cathetus, we have,

h=1.8*sin10=0.312

We proceed to calculate the work done by gravity

W_g=-mgh

W_g=-140*0.312=-43.76J

b) It's necessary to calculate the Kinetic Energy, we use the equation of KE,

\Delta KE = \frac{1}{2}m(v_2^2-v_1^2)

\Delta KE = \frac{1}{2}14(1.5^2-0.5^2)=14J

c) By work energy theorem

W_{net}=\Delta KE \rightarrow W_net = 14J

d) we calculate net work through the law of conservation

W_{net}=W_f+W_g=14

\rightarrow W_f = 14-W_g=14-(-43.76)

W_f=57.76J

6 0
4 years ago
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