A student catches a can of soda dropped from some unknown height by Mr. Fineman. If

the can was dropped from rest, and it is traveling at 8.9 m/s before it arrives in the

student's hand...

Answers

Answer 1

Answer:

v² = u² + 2gh

8.9² = 0² + 2(9.8)h

h = 4.0 m

v = at

t = 8.9/9.8

t = 0.91 s


Related Questions

i just want an answer please

Answers

Answer: An answer on what? I’ll never ignore you!

Explanation:

Answer:

an answer on what?

Explanation: Im here to help!!

Use the sentence to answer the question.

Light is affected by gravity.

Which inference can be made based on this fact?

(1 point)

Light behaves differently in space than on Earth.
Light behaves differently in space than on Earth.

Gravity causes light to refract.
Gravity causes light to refract.

Light moves faster in space than on Earth.
Light moves faster in space than on Earth.

Stronger gravity causes an increase in light.

Answers

Answer:

Light behaves differently in space than on Earth.

Explanation:

Because the gravity field is greater near earth than in most of space. Not the areas near stars, black holes, pulsars, and such but in the vast emptyness between the clumpy spots.

I need help. please look at the image below and let me know I need this by 7:20 am pst. ​

Answers

Answer:

3(1.5) = 4.5 V

Explanation:

What causes the difference in the angle of the sun on the Earth's surface throughout the year?

Answers

Answer:

The axis is tilted and points to the North Star no matter where Earth is in its orbit. Because of this, the distribution of the Sun's rays changes. ... It also means that the angle at which sunlight strikes different parts of Earth's surface changes through the year.

Explanation:

Pls sub to bdoggaming if this helped

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!pls

Answers

I think this is the solution:

1: U-1, F,-4
2: Na-6, Mo-1, O-4
3: Bi-1, O-1, C-1, I-1
4: In-9, N-1
5: N-2, H-4, S-1, C-1
6: Ge- 15, N-4
7: N-1, H-4, C-1, I-1, O-3
8: H-7, F-1
9: N-1, O-5, H-1, S-1
10: H-8
11: Nb-1, O-1, C-1, I-3
12: C-3, F-3, S-1, O-3, H-1
13: Ag-1, C-1, N-1, O-1
14: Pb-6, H-1, As-1, O-4

I need a short answer ?

Answers

Answer:

Explanation:

7a) t = d/v = 100/45cos14.5 = 2.29533...= 2.30 s

7b) h = ½(9.81)(2.29533/2)² = 6.46056... = 6.45 m

  or

  h = (45sin14.5)² / (2(9.81)) = 6.47 m

which rounds to the same 6.5 m when limiting to the two significant digits of the initial velocity.

A mars surface exploration vehicle drops a rock off a 1.00 I'm high vertical Cliff. The sound of the rock landing at the base of the cliff is recorded by instruments on the vehicle 27.1 seconds later. Calculate the acceleration due to gravity on Mars given that the speed of sound on Mars is 320 m/s

Answers

The acceleration due to gravity on Mars is 11.81 m/s².

The given parameters:

Height of the cliff, h = 1 mTime of motion of the sound wave, t = 27.1 sSpeed of sound in mass, v = 320 s

The equation of motion to determine the acceleration due to gravity on the moon is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2[/tex]

where;

s is the distance traveledt is the time of motion

Since the time measured is two way time, the new equation for the total distance traveled is calculated as;

[tex]v = \frac{2d}{t} \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{320 \times 27.1}{2} \\\\d = 4,336 \ m[/tex]

The acceleration due to gravity is calculated as follows;

[tex]s = vt + \frac{1}{2} gt^2\\\\4,336 = 0 \ + \ \frac{1}{2} \times g \times (27.1)^2\\\\4,336 = 367.21g\\\\g = \frac{4,336}{367.21} \\\\g = 11.8 1 \ m/s^2[/tex]

Thus, the acceleration due to gravity on Mars is 11.81 m/s².

Learn more about acceleration due to gravity here: https://brainly.com/question/88039

The diagram below shows a 5.00-kilogram block
at rest on a horizontal, frictionless table.
5.00-kg
block
Table
Which of the following is the correct name and strength of the force holding the block up?

Answers

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

Mass of the block, m = 5 kg

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

g is acceleration due to gravity = 10 m/s²

W = 5 x 10

W = 50 N (downwards)

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N (upwards)

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: https://brainly.com/question/14486416

The block will remain on the table because the normal force balances with the weight of the block. The correct answer is  50 N upward normal force

From the diagram shown a 5.00-kilogram block at rest on a horizontal, frictionless table. The weight of the block will act downward which will be

Weight W = mg

let g = 10 m/[tex]s^{2}[/tex]

W = 5 x 10

W = 50 N

The block will also produce an equal but in opposite direction of a normal force which is equal to the weight of the block. That is,

Normal force N = 50 N

The block will remain on the table because the normal force balances with the weight of the block.    

Therefore, the correct name and strength of the force holding the block up is 50 N upward normal force.

Learn more about stability here: https://brainly.com/question/517289

Explain how the linear rate spring operates?

Answers

A linear spring has the same diameter along its entire length, and this uniform diameter gives it a constant spring rate.

A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. The total mechanical energy is 25 J. The maximum speed of the block is:

Answers

Answer:

Explanation:

easy way

when system is all kinetic energy, velocity is at a maximum

E = ½mv²

v = √(2E/m) = √(2(25)/0.5) = √100 = 10 m/s

harder way

ω = √(k/m) = √(80/0.5) = √160 rad/s

When the system is entirely spring potential, the amplitude A is

E = ½kA²

A = √(2E/k) = √(2(25)/80) = 0.790569... = 0.79 m

maximum velocity is ωΑ = 0.79√160 = 10 m/s

Cody hits up food king and uses a scale to weigh the mass of an apple. if the spring potential energy in the scale is .09 j and is spring is stretched 0.6 meters, calculate the spring constant

Answers

Answer:

oK so  here's  what you should do is  add .09 and 0.6

Explanation:

2 examples of non fossil fuels ?

Answers

Answer:

-> Hydropower

-> Solar power

Explanation:

-> Hydropower

[] The power of water! It is the use of falling or fast-running water to produce electricity for power. Impoundments or da*ms are mainly used in this type of power source.

-> Solar power

[] The power of the sun! It is the use of sunlight, or solar energy, to produce electricity for power. You have probably heard of solar panels, and this is the main way to collect it.

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

Answer:

Wind energy and solar power

Explanation:

they do not use fossil fuels

If you apply a net force of 100 N to the hoverboard, and it accelerates
2m/s/s, how much mass does it have?

Answers

Answer:

50 kg

Explanation:

The mass of an object given only the force acting on it and it's acceleration can be found by using the formula

[tex]m = \frac{f}{a} \\ [/tex]

f is the force in N

a is the acceleration in m/s²

From the question

f = 100 N

a = 2 m/s²

We have

[tex]m = \frac{100}{2} = 50 \\ [/tex]

We have the final answer as

50 kg

Hope this helps you

A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m when it begins to roll down the incline. If rolling and sliding friction are neglected, what is the linear velocity, in m/s, of the center-of-mass of the wheel when it reached the bottom of the incline?

Answers

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

why is it so important that you take care of your nervous system?

Answers

Answer:

The nervous system handles the stress response, which, if overworked, can eventually lead to diseases ranging from high blood pressure to diabetes.

Explanation:

hope I helped

12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]

Answers

Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

[tex]KE = \frac{1}{2}mv^2[/tex]

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s

A 300 cm rope under a tension of 120 N is set into oscillation. The mass density of the rope is 120 g/cm. What is the frequency of the first harmonic mode (m

Answers

Answer:

Explanation:

f = [tex]\sqrt{T/(m/L)} / 2L[/tex]

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = [tex]\sqrt{120/12} /(2(3))[/tex])

f = [tex]\sqrt{10\\}[/tex]/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

A wagon of dog treats (combined mass 55 kg) is rolling at 2.1 m/s. A dog with mass 21 kg dives into the wagon, colliding with just enough momentum to make both stop. If the collision between the dog and the wagon lasts 0.1 s, what is the magnitude of the average force that will be exerted on the dog by the collision with the wagon

Answers

Answer:

Explanation:

An impulse results in a change of momentum

If the wagon and dog both stop, they must have had equal and opposite momentums

FΔt = mΔv

F = mΔv/Δt = m(v₁ - v₀)/(t₁ - t₀)

v₁ = t₀ = 0

F = m(v₀)/t₁

F = 55(2.1)/0.1 = 1155 N

We could have also figured the dog's initial velocity and used the dog's mass in the equation as well. Result would be identical.

8 N to the left , and 4 N to the right. Find the net force. Is this balanced?

Answers

Explanation:

12N by first law of newton is net force after colloision

A 200-kg, 2.0-m-radius, merry-go-round in the shape of a flat, uniform, circular disk parallel to level ground is rotating at 1.2 cycles/second about an axis through its center of mass and perpendicular to the ground. A 50-kg boy jumps onto the edge of the merry-go round and lands at a fixed point. What is the angular velocity of the merry-go-round after the boy lands on it

Answers

Answer:

Explanation:

Conservation of angular momentum.

Disk            I = ½MR²

Point mass I = mR²  (boy)

Initial angular momentum

L₀ = Iω = ½MR²ω₀

Final angular momentum

L₁ = Iω = (½MR² + mR²)ω₁

as momentum is conserved, these are equal

(½MR² + mR²)ω₁ = ½MR²ω₀

                       ω₁ = ω₀(½MR²/ (½MR² + mR²))

                       ω₁ = ω₀(½M/ (½M + m))

                       ω₁ = 1.2(½(200)/ (½(200) + 50))

                       ω₁ = 1.2(⅔)

                      ω₁ = 0.8 cycles/second or 0.8(2π) = 1.6π rad/s

5.000 km =
3.125
mi
8.000 fl oz =
mL

Answers

Answer:

236.588 mL

Explanation:

The formula for an approximate result is to multiply the volume value by 29.574

[tex]8.000 \times 29.574 = 236.588[/tex]

Is that what you were asking for?

During take-off a 8kg model rocket is burning fuel causing its speed to increase
at a rate of 4m/s2 despite experiencing a 90N drag.

What’s is the strength of the thrust?
(Answer unit is in N)( and the answer isn’t 212)

Answers

The strength of the thrust is 122 newtons.

The motion of the rocket is described by the second Newton's law, whose model is shown below:

[tex]\Sigma F = F - D = m\cdot a[/tex] (1)

Where:

[tex]F[/tex] - Thrust, in newtons[tex]D[/tex] - Drag, in newtons[tex]m[/tex] - Mass of the rocket, in kilograms[tex]a[/tex] - Net acceleration of the rocket, in meters per square second

If we know that [tex]D = 90\,N[/tex], [tex]m = 8\,kg[/tex] and [tex]a = 4\,\frac{m}{s^{2}}[/tex], then the strength of the thrust is:

[tex]F = D + m\cdot a[/tex]

[tex]F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 122\,N[/tex]

The strength of the thrust is 122 newtons.

To learn more on Newton's laws, we kindly invite to check this verified question: https://brainly.com/question/13678295

explain the process of convergence and divergence ! HELPPP

Answers

Answer:

Divergence generally means two things are moving apart while convergence implies that two forces are moving together. ... Divergence indicates that two trends move further away from each other while convergence indicates how they move closer together.

Explanation:

A car travels a certain distance from A to B with a speed of 60km/hr and then returns along the same path to the starting point with a speed of 40km/hr. Find the average speed and average velocity.
a) Km/hr
b) m/s

wrong answers will be reported!

Answers

Answer:

Explanation:

Speed is total distance traveled over time taken to do so.

If AB is measured in kilometers, time (t) for the whole trip is

t = AB/60 + AB/40  

t = 2AB/120 + 3AB/120

t = 5AB/120 hrs

Average speed is distance over time

s = 2AB / (5AB/120)

s = 2(120)/5

s = 48 km/hr

s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s

Velocity is displacement over time.

As displacement is zero, velocity is zero

v = 0 km/hr = 0 m/s

Pretty harsh reporting answers just because they are wrong.

An object, initially traveling at a velocity of 73 m/s, experiences an acceleration of -9.8 m/s^2. How much time will it take it to come to rest?

Answers

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

yayy here you are f, r, e, e, p, o, i, n, t, s

Answers

Answer:

Albert Einstein Albert Einstein was a German-born theoretical physicist, widely acknowledged to be one of the greatest physicists of all time. Einstein is best known for developing the theory of relativity, but he also made important contributions to the development of the theory of quantum mechanics.

Explanation:

Thank you so much buddy !!

An object accelerates from rest to 93 m/s over a distance of 49 m. What acceleration did it experience?

Answers

Answer:

Explanation:

acceleration=  change in velocity/time taken

acceleration=  93/49

=2.02

What are sound detectors?

Answers

Answer:

A sound detector comes in the shape of a rectangular board which comprises a microphone as well as a processing circuitry.

Answer:

Sound detection sensor works similarly to our Ears, having diaphragm which converts vibration into signals.

What initial speed v is required if the blocks m1 =2.5 kg and m2=1.5 kg are to travel a distance d =7.0cm before coming to rest? Assume the coefficient of kinetic friction between m1 and the tabletop is ųk=0.21

Answers

Answer:

OPTRIMUM PRIDE URGH URGH URGH

Explanation:

AHHAAHAHAHAHA

HELPPP!! Thanks!

If you only wanted to increase the particle motion of a gas without increasing any of its other properties, which would the most correct situation?


a. Keep the gas at a constant pressure and keep the temperature constant, but increase the volume of the gas

b. Keep the gas in a fixed container at constant pressure and increase the temperature

c. Keep the gas in a fixed container at constant pressure and decrease the temperature

d. Keep the gas at a constant volume and keep the temperature constant, but decrease the pressure of the gas

Answers

Answer:c i think

Explanation: not sure

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