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oksano4ka [1.4K]
3 years ago
10

A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the

pipe to eject the marshmallow from the right end. The average net force exerted on the marshmallow while it is in the pipe is 0.7 N. The speed of the marshmallow as it leaves the pipe is most nearly
Physics
2 answers:
alexdok [17]3 years ago
5 0

Answer:

15 m/s

Explanation:

Given:

F = 0.7N   m = 0.25g = 0.0025kg   x = 40cm = 0.4m v_{0} = 0

Since F = ma, a = F/m = 0.7N/0.0025kg = 280m/s^{2}

Using the Big Five, v ^{2} = v_0^{2} + 2ax, v = \sqrt{0 + 2*280m/s^2*0.4m} = \sqrt{216} m/s=  15m/s

frosja888 [35]3 years ago
3 0

Answer:

Your answer would be

A person 40 cm- blows into the left end of the pipe to eject the marshmallow from the right end. ... A strain of sound waves is propagated along an organ pipe and gets reflected from an. play · like-icon ... The velocity of sound in air is 340ms^(-1). ... The two pipes are submerged in sea water, arranged as shown in figure. Pipe.Explanation:

I belive this is the answer sorry if im wrong!

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A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of
docker41 [41]

Answer:

The frog takes 8 jumps to reach top of well

Explanation:

Given data

Frog at bottom=17 foot

Each time frog leaps 3 feet

Frog has not reached the top of the well, then the frog slides back 1 foot

To Find

Total number of leaps the frog needed to escape from well

Solution

in 1 jump distance jumped=3+(-1)

                                           =2 feet

                                           =2×1 feet

The "-1" is because the frog goes back

Now After 2 jumps the distance jumped as:

                     Distance Jumped=2+2

                     Distance Jumped=2*2

                                                   =4 feet

Similarly after 7 jumps

                    Distance Jumped=2+2+......+2

                    Distance Jumped=2*7

                                                 =14 feet

Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.

So

              Distance Jumped=(Distance Jumped after 7 jumps)+3

                                           =14+3

                                           =17 feet

The frog takes 8 jumps to reach top of well                

7 0
3 years ago
Yes this nees helppppppppppppppp
kozerog [31]

Answer:

lol

Explanation:

it was funny

6 0
4 years ago
A rock is thrown downward from an unknown height above the ground with an initial speed of 6.1 m/s. It strikes the ground 1.7 s
insens350 [35]

Answer:

24.531 m

Explanation:

t = Time taken = 1.7 s

u = Initial velocity = 6.1 m/s

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

Equation of motion

s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=6.1\times 1.7+\dfrac{1}{2}\times 9.8\times 1.7^2\\\Rightarrow s=24.531\ m

The initial height of the rock above the ground is 24.531 m

7 0
3 years ago
Which objects will likely have the smallest gravitational force between them?
DENIUS [597]

Answer:

A. Two tennis balls that are near each other  

Explanation:

The formula for gravitational force (F) between two objects is

F = \dfrac{Gm_{1}m_{2}}{d^{2} }

where m₁ and m₂ are the masses of the two objects, d is the distance between their centres, and G is the gravitational constant.

Thus, two objects that are far from each other will have a smaller gravitational force. We can eliminate Options C and D.

If the objects are at the same distance, those with the smaller mass will have a smaller force.

The mass of a tennis ball is 57 g.

The mass of a soccer ball is 430 g.

Two tennis balls that are near each other will have a smaller gravitational attraction.

3 0
4 years ago
How far will a freely falling object fall from rest in 5 seconds?​
Travka [436]
<h2><em>how far will a freely falling object fall from rest in 5 seconds?</em></h2>

  • <em>If an object free falls from rest for 5 seconds, its speed will be <u>about 50 m/s.</u></em>

<em><u>hope </u></em><em><u>it</u></em><em><u> helps</u></em>

<em><u>#</u></em><em><u>c</u></em><em><u>a</u></em><em><u>r</u></em><em><u>r</u></em><em><u>y</u></em><em><u> </u></em><em><u>on</u></em><em><u> learning</u></em>

3 0
3 years ago
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