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cricket20 [7]
3 years ago
9

Emilio tries to jump to a nearby dock from a canoe that is floating in the water.Instead of a canoe that is floating in the wate

r.Instead of landing on the dock he falls into the water beside the canoe. Help plz
Physics
1 answer:
elixir [45]3 years ago
7 0
Oooo that ones hard. ummm... idk i think we should just leave it to the experts ya know.
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A bullet with a mass of 0.1 kg is travelling at
mel-nik [20]

The block with the bullet lodged in the block is now travelling at 2.133 m/s.

<h3>What is momentum conservation principle?</h3>

When there is no external force acting on the system, the momentum remains conserved.

For inelastic collision, after collision both objects travel with common speed.

m1u1 + m2u2 =(m1 +m2)v

Substitute initial velocity of bullet u1 =320 m/s , initial velocity of block u2 =0, mass of bullet m1 = 0.1 kg and mass of block m2 = 14.9 kg.

Solve for the final velocity of bullet,

0.1 x 320 + 14.9 x 0 = (0.1 +14.9) x v

v = 2.133 m/s

Thus, the  block with the bullet lodged in block now travelling at 2.133 m/s.

Learn more about momentum conservation principle.

brainly.com/question/14033058

#SPJ1

6 0
2 years ago
If you travel from Yakima to Ellensburg (Yakima to Ellensburg is 50 miles) with a speed of 60 miles/hour for half of the
koban [17]

Answer:

\displaystyle \frac{480}{7}\approx 68.6\; \rm mph.

Explanation:

The average speed of an object is equal to total distance over total time.

  • Distance traveled: \rm 50 \; mi.

How much time is taken? This trip is divided into two halves, each of distance \displaystyle \frac{50}{2} = 25\;\rm mi.

Time spent on the first half of the trip:

\displaystyle t_1 = \frac{s_1}{v_1} = \frac{25}{60} = \frac{5}{12}\; \rm hours.

Similarly, time spent on the second half of the trip:

\displaystyle t_2 = \frac{s_2}{v_2} = \frac{25}{80} = \frac{5}{16}\; \rm hours.

In total:

\displaystyle \frac{5}{12} + \frac{5}{16} = \frac{35}{48} \; \rm hours.

Average speed:

\begin{aligned} \text{Average speed} &= \frac{\text{Total Distance}}{\text{Total Time}}\\ &= 50 \left/\frac{35}{48}\right.\\ &= 50 \cdot \frac{48}{35} \\&= \frac{480}{7}\approx 68.6\; \rm mph \end{aligned}.

This value turned out to be slightly different from the average of the speed during the two halves of the journey. The reason is that the object traveled at each speed for a different amount of time. It spent more time at the slower speed, which gives that speed a greater weight in the average. That explains why the average speed is closer to \rm 60\; mph rather than \rm 80\; mph.

3 0
3 years ago
3 An empty hot tub has a mass of 320 kg. When filled, the tub holds 600 gallons of water (rho = 62.4 lbm/ft3). The local acceler
ehidna [41]

Answer:

Total weight of the hot tub and water is 5676.6 pounds-force

Explanation:

rho = 62.4lbm/ft^3 × 1ft^3/7.481gal = 8.34lbm/gal

Mass of water = rho × volume = 8.34lbm/gal × 600 gallons = 5004lbm = 5004×0.45359kg = 2269.8kg

Total mass of hot tub and water = 320kg + 2269.8kg = 2589.8kg

Local acceleration due to gravity = 32ft/s^2 = 32ft/s^2 × 1m/3.2808ft = 9.75m/s^2

Total weight of hot tub and water = 2589.8kg × 9.75m/s^2 = 25250.55N = 25250.55/4.4482 lbf = 5676.6 lbf

6 0
3 years ago
Devin decides to start a summer job mowing lawns. In his first job, he pushes his lawn mower a total of 1.5 km. The lawn mower w
Xelga [282]

Answer:

1.13\cdot 10^5 J

Explanation:

The amount of work done by Devin when moving the lawn is given by:

W=Fd cos \theta

where

F is the average force applied

d is the displacement of the lawn

\theta is the angle between the direction of the force and the displacement

Assuming that Devin moves the lawn by applying a force parallel to the displacement, we have:

F = 75 N

d = 1.5 km = 1500 m

\theta=0^{\circ}

Therefore, the work done is

W=(75 N)(1500 m)(cos 0^{\circ})=1.13\cdot 10^5 J

5 0
3 years ago
A car moves with 3m/s and travels for 12s. Calculate the distance moved by the car.
DanielleElmas [232]

Answer:

36 m

Explanation:

Distance = rate × time

d = (3 m/s) × (12 s)

d = 36 m

6 0
3 years ago
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