A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds. a 1247 m
b 0.013 m
c 130 m
d 76.92 m

Answers

Answer 1

Answer:

Option C is the correct answer

Explanation:

We have equation of motion , $s= ut+(1)/(2) at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = horizontal velocity = 100 m/s, acceleration = 0 $m/s^2$ , we need to calculate displacement when time = 1.3 seconds.

Substituting

$s= 100*1.3+(1)/(2) *0*1.3^2 = 130 meter$

So, bullet will travel a distance of 130 meter in 1.3 seconds.

Option C is the correct answer

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Answers

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What causes resistance in an electrical current

Answers

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A sled with rider having a combined mass of 110 kg travels over the perfectly smooth icy hill shown in the accompanying figure. How far does the sled land from the foot of the cliff?

Answers

The sled land is "25.531 m" far from the foot of the cliff.

According to the question,

The velocity of the top = $V_t$

then,

→ $0.5 \ Vt^2-0.5 \ Vi^2 = -mgh$

→ $0.5 \ Vt^2 = 0.5 \ Vi^2- 9.8* 11$

→ $0.5 \ Vt^2 = 0.5 \ Vi^2- 107.8$

→ $=17.04 \ m/sec$

Now,

The projectile motion in horizontal direction,

→ $S_y = 11 = ut-0.5 \ gt^2$

$= 0-0.5* 9.8 \ t^2$

or,

→ $S_x = 17.04* 1.498$

$= 25.53 \ m$

Thus the above answer is the correct one.

Learn more:

brainly.com/question/13027204

Find the speed the sled has at the top of the hill from the law of conservation of mechanical energy. Equate the kinetic energy at the bottom of the hill to the kinetic plus potential energy at the top :

0.5mv₀² = 0.5mv² + mgh
v = √[v₀² - 2gh]
= √[(22.5m/s)² - (2 x 9.80m/s² x 11.0m)]
= 17.0m/s

From the time independent kinematics equation, find the vertical component of the sleds final velocity (note that the vertical component of the sleds initial velocity is zero) :

v² = v₀² + 2gΔy
= 0 + 2(-9.80m/s²)(-11.0m)
= -14.7m/s (select the neg root, because motion is downward)

With this you can find the time vertically which is the same horizontally :

v = v₀ + gt
t = (v - v₀) / g
= (-14.7m/s - 0) / -9.80m/s²
= 1.50s

Now, the horizontal distance is :

Δx = (v₀ + v)t / 2
= (17.0m/s + 17.0m/s)1.50s / 2
= 25.5m

tell me if you need more help :)

How do geologist locate the epicenter of an earthquake?

Answers

by taking measurements from 3 seismograms.

Ball X is dropped from a height of 1.25 meters above the ground. At the same time and from exactly the same height, Ball Y is shot outhorizontally at an initial velocity of 8.0 m/s. Air resistance is negligible and assume g =-10 m/s?
How far away from Ball X will Ball Y hit the ground?

Answers

Answer: The distance between Ball X and Ball Y is 4.0 meters

Explanation:

First we have to use the information we have from Ball Y to find time as well as it’s distance.

Ball Y’s Given:

Vx=8.0m/s

ax=0m/s^2

x=?

y=1.25m

ay=-10m/s^2

To find the time, use the equation t=(2y/g)^1/2.

t=(2(1.25)/10)^1/2

t=(2.5/10)^1/2

t=(0.25)^1/2

t=0.5s

Now we have to calculate how far away Ball Y fell from the base of the cliff. To do this use the equation x=(Vx)(t) and plug in the values.

x=(8.0m/s)(0.5s)

x=4.0m

Because Ball X is dropped from the cliff, we can conclude that the ball is 0meters away from the cliff. With this in mind, we can assume that the distance between Ball X and Ball Y is 4.0m.

Total Distance = Ball Y - Ball X

Total Distance = 4.0 m - 0 m

Total Distance = 4.0 m

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