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2 years ago

I will mark brainliest! :D SO please help me M8

Physics

2 answers:

Kobotan [32]2 years ago

7 0

Answer: 100 J

Explanation: 1/2 5 x 2^2 = 100

Hope this made any sense.

steposvetlana [31]2 years ago

5 0

A 50-kg object should have 100J kinetic energy if it is moving at a velocity of 2 m/s.

Explanation: Just use the formula! Remember to:

Plug in the weight/mass to find m

Plug in the speed/velocity to find v

So the equation changes from 1/2 x M x V^2

to...

1/2 x 50 x 2^2

this is equal to 100.

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Will give correct answer brainliest

sineoko [7]

$200 \times 80\%$

<h2>calculate</h2>

$200 \times \frac{80}{100}$

<h2>reduce the numbers</h2>

$2 \times 80$

<h2>multiply</h2>

$= 160$

<h2>there for we have a solution to the equation</h2>

hope it helps

#carry on learning

mark me as brainlist plss

6 0

3 years ago

Read 2 more answers

This is the phase of matter with no fixed shape but fixed volume.

Nataly_w [17]

A liquid is a matter that neither has a fixed shape but fixed volume...in college really need to know this.

GOOD LUCK

6 0

3 years ago

Read 2 more answers

An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?

Norma-Jean [14]

Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s

6 0

2 years ago

14. When incoming light rays strike a flat, plane mirror at an

Mariulka [41]

Answer:

close to the mirrors surface

Explanation:

this is because angle of incidence equals angle of reflection

3 0

3 years ago

2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he

Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, $v_s$ = d/t₂

∴ $v_s$ = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, $v_s$ ≈ 5.79 m/s.

5 0

3 years ago