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A hovering mosquito is hit by a raindrop that is 50 times as massive and falling at 8.5 m/sm/s , a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic

Answers

Answer 1

Answer:

Answer:

$v = 8.333\,(m)/(s)$

Explanation:

The system mosquito-raindrop is described by the Principle of Momentum Conservation:

$(50\cdot m)\cdot (8.5\,(m)/(s))+m\cdot (0\,(m)/(s) ) = 51\cdot m \cdot v$

$425 = 51\cdot v$

The final speed is:

$v = 8.333\,(m)/(s)$

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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

Answers

Answer:

The speed of the arrow immediately after it leaves the bow is 38.73 m/s

Explanation:

given information:

force, F = 150 N

x = 50 cm = 0.5 m

mass of arrow, m = 50 g = 0.05 kg

We start from the force of the spring

F = kx

k = $(F)/(x)$

= $(150)/(0.5)$

= 300 N/m

The potential energy, EP of the spring is

EP = $(1)/(2) kx^(2)$

the kinetic energy, EK of the spring

EK = $(1)/(2) mv^(2)$

According to conservative energy,

EP = EK

$(1)/(2) kx^(2)$ = $(1)/(2) mv^(2)$

$kx^(2)$ = $mv^(2)$

$v^(2)$ = $(kx^(2) )/(m)$

v = $x\sqrt{(k)/(m) }$

= $0.5\sqrt{(300)/(0.05) }$

= 38.73 m/s

Final answer:

Using Hooke's Law, we can determine the speed of the arrow. The speed of the arrow immediately after it leaves the bow is approximately 38.7 m/s.

Explanation:

In this problem, we can use Hooke's Law to determine the speed of the arrow. Hooke's Law states that the force exerted by a spring is proportional to the displacement of the spring:

F = -kx

Where F is the force, x is the displacement, and k is the spring constant.

In this case, the force exerted by the bow on the arrow is acting like a spring force. The force of the bow is 150N, and the displacement is 50cm (which is equivalent to 0.5m). So we can set up the equation as:

150N = -k * 0.5m

Now we can solve for k:

k = -150N / 0.5m = -300 N/m

Now that we have the spring constant, we can use it to find the potential energy stored in the bow:

PE = 0.5kx^2 = 0.5*(-300N/m)*(0.5m)^2 = 37.5 J

Next, we can use the conservation of energy to find the kinetic energy of the arrow right after it leaves the bow. The potential energy stored in the bow is converted into kinetic energy:

KE = PE = 37.5 J

The kinetic energy is given by the equation:

KE = 0.5mv^2

Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for v:

v = sqrt(2KE/m) = sqrt(2*37.5 J / 0.05 kg) = sqrt(1500) m/s ≈ 38.7 m/s.

Learn more about speed of the arrow here:

brainly.com/question/32696507

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A 0.157kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.850m/s . It has a head-on collision with a 0.306kg glider that is moving to the left with a speed of 2.26m/s . Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.157kg glider.Find the magnitude of the final velocity of the 0.306kg glider.

Answers

Answer:

V1 = -3.260 m/s, V2 = 1.303 m/s

Explanation:

Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s

mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)

To Find: Final Velocity of Left Glider is V1=? m/s and Velocity of right Glider is V2 =? m/s (After Collision)

from law of conservation of momentum and energy we deduce a formula:

V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)

V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)

V1 = -0.273 -2.987

V1 = -3.260 m/s

and V2 Formula

V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)

V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)

V2 = 0.727 + 0.576

V2 = 1.303 m/s

-0.149, 0.463

A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the bottom. It then slides on the horizontal surface. The coefficient of friction between the block and the horizontal surface is 0.20. How far does the block slide on the horizontal surface before it comes to rest?

Answers

The block slide on the horizontal surface is "24.99 m" far.

According to the question,

Vertical height = 5.0 m
Coefficient of friction = 0.20

Let,

The time taken be "t".

Now,

→ $s = ut+ (1)/(2) at^2$

By substituting the values, we get

$5 = (1)/(2)* 9.8* t^2$

$t = 1.01 \ sec$

The final velocity will be:

→ $v_1 = gt$

$= 9.8* 1.01$

$= 9.899 \ m/s$

Now,

→ $t = (u)/(a)$

$= (9.899)/(0.2* 9.8)$

$= 5.05 \ seconds$

hence,

The distance will be:

→ $s = ut+0.5* at^2$

$= 9.899(5.05)-0.5* (0.2* 9.8* 5.05^2)$

$= 24.99 \ m$

Thus the above approach is right.

Learn more about friction here:

brainly.com/question/18851133

Answer:

The block slides on the horizontal surface 25 m before coming to rest.

Explanation:

Hi there!

For this problem, we have to use the energy-conservation theorem. Initially, the block has only gravitational potential energy (PE) that can be calculated as follows:

PE = m · g · h

Where:

m = mass of the block.

g = acceleration due to gravity.

h = height at which the block is located.

As the block starts to slide down the track, its height diminishes as well as its potential energy. Due to the conservation of energy, energy can´t disappear, so the loss of potential energy is compensated by an increase of kinetic energy (KE). In other words, as the block slides, the potential energy is converted into kinetic energy. The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

m = mass of the block.

v = speed of the block.

Then, at the bottom of the ramp, the kinetic energy of the block will be equal to the potential energy that the block had at the top of the ramp.

Initial PE = KE at the bottom

When the block starts sliding horizontally, friction force does work to stop the block. According to the energy-work theorem, the change in the kinetic energy of an object is equal to the net work done on that object. In other words, the amount of work needed to stop the block is equal to its kinetic energy. Then, the work done by friction will be equal to the kinetic energy of the block at the bottom, that is equal to the potential energy of the block at the top of the track:

initial PE = KE at the bottom = work done by friction

The work done by friction is calculated as follows:

W = Fr · Δx

Where:

W = work

Fr = friction force.

Δx = traveled distance.

And the friction force is calculated as follows:

Fr = μ · N

Where:

μ = coefficient of friction.

N = normal force.

Since the block is not accelerated in the vertical direction, in this case, the normal force is equal to the weight (w) of the block:

Sum of vertical forces = ∑Fy = N - w = 0 ⇒N = w

And the weight is calculated as follows:

w = m · g

Where m is the mass of the block and g the acceleration due to gravity.

Then, the work done by friction can be expressed as follows:

W = μ · m · g · Δx

Using the equation:

intial PE = work done by friction

m · g · h = μ · m · g · Δx

Solving for Δx

h/μ = Δx

5.0 m / 0.20 = Δx

Δx = 25 m

The block slides on the horizontal surface 25 m before coming to rest.

The data listed below are for mechanical waves. Which wave has the greatest energy?

Answers

theres no data listed below

Answer:

amplitude = 14 cm; wavelength = 7 cm; period = 12 seconds

Explanation:

A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed by 11 cm from its uncompressed length. The compressed spring and block rests at the bottom of an incline of 28◦ with the spring lying along the surface of the ramp.After all the external forces are removed (so the compressed spring releases the mass) how far D along the plane will the block move before coming to a stop? Answer in units of m.

Answers

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy= $(1)/(2) kx^(2) +PE1$