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Mumz [18]
2 years ago
12

Suppose you are driving in a convertible with the top down at constant speed. You point a rifle straight upward and fire it. In

the absence of air resistance, would the bullet land
(a) behind you,
(b) ahead of you, or
(c) in the barrel of the rifle?
Physics
1 answer:
lyudmila [28]2 years ago
4 0

Answer:

(c) in the barrel of the rifle

Explanation:

When we are moving in a convertible with constant speed so as per inertia all the objects are moving with same speed as that the speed of convertible.

So here we can say that bullet when shot from the rifle then its speed is same as that of convertible in horizontal direction

So here we can say that if air friction is absent then bullet will move same horizontal distance as the distance moved by you in the convertible

so after it it will reach again at the same position then the displacement of you and bullet is same in horizontal direction so we can say that correct answer will be

(c) in the barrel of the rifle

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The proper IUPAC naming convention for the compound symbol 'KI' is
Darya [45]
Proablyly a or c idk 
6 0
2 years ago
A dog walks 12 meters to the east and then 16 meters back to the west for this motion what is the distance moved What is the mag
FinnZ [79.3K]

The distance is 28 meters, and the displacement is -4.

For the distance it would be 12 + 16 = 28.

For the displacement it would be 12 - 16 = -4.

would really appreciate a brainliest! Hope this helped!

6 0
2 years ago
an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Nata [24]

The y-component of the acceleration is 0.28 m/s^2

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

Here we have:

  • Initial velocity is u=2.25 m/s at \theta_1=50.0^{\circ}, so its y-component is u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s
  • Final velocity is v=4.65 m/s at \theta_2=120.0^{\circ}, so its y-component is v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
2 years ago
A ball is kicked from level ground at an angle 60° with initial velocity 10 m/s. The distance the ball travels, in meters, is:
RUDIKE [14]

Answer:

<em>the ball travels a distance of 8.84 m</em>

Explanation:

Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R = (U²sin2∅)/g.............................. Equation 1

Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.

<em>Given: U = 10 m/s, ∅ = 60°</em>

<em>Constant: g = 9.8 m/s²</em>

Substituting these values into equation 1

R = [10²×sin(2×60)]/9.8

R = (100sin120)/9.8

R = 100×0.8660/9.8

R = 86.60/9.8

R = 8.84 m

<em>Therefore the ball travels a distance of 8.84 m</em>

4 0
3 years ago
PHYSICS QUESTION - CAN ANYONE HELP NOW?
GrogVix [38]
It's velocity when it strikes the ground is. D. 232.9 kg.m/s<span>.
</span>
I hope this helps!!!







3 0
2 years ago
Read 2 more answers
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