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ASHA 777 [7]
3 years ago
12

If we want to know the velocity that an object is traveling, we must know the (2 points)

Physics
2 answers:
Lady_Fox [76]3 years ago
7 0
I’m gonna say B but I’m not sure
Ber [7]3 years ago
7 0

Answer:

it's b

Explanation:

yes its b...................

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Write newtone first law
gtnhenbr [62]

Newtons first law is that an Object that is in motion will stay in motion until an unbalanced force acts upon it, and part 2 is that an object at rest will stay at rest until and unbalanced force acts upon it

4 0
3 years ago
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An electron is accelerated uniformly at a rate of 25,000 km/s from rest in an accelerator, over a distance of 125 km. What is th
Black_prince [1.1K]

Answer: 2500km/s

Explanation:

use equation wich conects distance and  acceleration:

S=Vst+at²/2

Vs=0km/s(from rest)

S=at²/2

S2=at²

t=√2S/a

t=√2*125km/25000 km/s²

t=0.1s

Vf --final velocity

Vs-- start velocity

a=Vf-Vs/t

at=Vf-Vs

Vs=0Km/s

at=Vf

25000Km/s*0.1s=Vf

Vf=2500Km/s

 

3 0
4 years ago
An eagle is flying horizontally 16.4 meters above a lake at a speed of 9.3 m/s, carrying a small pumpkin in its talons. The pump
Dima020 [189]

Answer:

The horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

Explanation:

Given;

height above the ground, h = 16.4 m

speed of the eagle, v = 9.3 m/s

The time it will take the pumpkin to fall at the given height is calculated as;

t = \sqrt{\frac{2h}{g} }\\\\t =  \sqrt{\frac{2*16.4}{9.8} }\\\\t = 1.83 \ s

The horizontal distance traveled at this time is given by;

x = vt

x = (9.3)(1.83)

x = 17.02 m

Therefore, the horizontal distance the pumpkin will travel after it slips from the eagle is 17.02 m

4 0
3 years ago
Democritus thought that matter was made of tiny particles. true or false
Firlakuza [10]

Answer:true

Explanation:because yes

4 0
4 years ago
Read 2 more answers
You throw a tennis ball straight upward. the initial speed is about 12 m/s. part a how long will it take for the ball to reach i
nataly862011 [7]

Answer: 1.2 s

It is given that a ball is thrown vertically upwards with initial velocity, u= 12 m/s.

At the maximum height, instantaneous velocity would be, v= 0 m/s

the ball would decelerate due to gravity, g=-9.8m/s^{2}

using equation of motion,

v-u=gt\\ \Rightarrow t=\frac{v-u}{g} \\ \Rightarrow t= \frac {0-12 m/s}{9.8 m/s^{2}}=1.2s

Hence, the time taken by ball to reach its maximum height is 1.2 s when thrown vertically upward with initial velocity of 12 m/s.

7 0
3 years ago
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