What is the force required to accelerate 1 kilogram of mass at 1 meter per second per second.1 Newton

1 pound

1 kilometer

1 gram

Answers

Answer 1

Answer:

Answer:

it's answer is 1 newton

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After a package is dropped from the plane, how long will it take for it to reach sea level from the time it is dropped? assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. express your answer numerically in seconds. neglect air resistance.

Answers

It will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

We can assume that the package, like the plane, has an initial velocity of 342 km/hour in the horizontal direction. We also assume that air resistance can be neglected.

Assuming that the package was dropped from rest at a height of h, the time it takes for the package to reach sea level can be calculated using the equation:

h = (1/2) * g * t²

where g is the acceleration due to gravity (9.8 m/s²) and t is the time it takes for the package to reach sea level.

Solving for t, we get:

t = sqrt(2h/g)

To convert the initial velocity of the package from km/hour to m/s, we can use the conversion factor:

1 km/hour = 0.2778 m/s

Therefore, the initial velocity of the package is:

v0 = 342 km/hour * 0.2778 m/s/km/hour = 95.0 m/s

if the package was dropped from a height of 5000 meters, the time it takes for the package to reach sea level is:

h = 5000 m

t = sqrt(2h/g) = sqrt(2*5000/9.8) = 32.0 seconds

Therefore, it will take approximately 32.0 seconds for the package to reach sea level from the time it is dropped, assuming that air resistance can be neglected.

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Final answer:

The time a dropped package takes to reach sea level from a plane is determined by its vertical motion. If the package retains only horizontal velocity when released, the time taken would be calculated using the height from which the object is dropped. However, to give a numerical value of time, we need to know the exact height.

Explanation:

The time it takes for the package dropped from the plane to reach sea level is determined exclusively by the package's vertical motion, assuming the package does not face air resistance. Specifically, the time of flight for a projectile launched and landing at the same elevation is governed by the equation: t = 2*v/g, where v represents the initial vertical velocity and g is the acceleration due to gravity. From the scenario, it seems the package retains only horizontal velocity when released since it's dropped down directly rather than being thrown downward, hence rendering initial vertical velocity as zero. Simply put, the package only begins to accelerate in the vertical direction once it's dropped, meaning the time taken would be calculated using the equation: t = √(2h/g), h being the height from which the object is dropped.

In the provided context, unfortunately, we need the height from which the package is dropped to give a specific numerical value of the time in seconds. If we knew the height of the plane at the time the package was dropped, we'd recalculate the time in seconds more precisely.

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A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam

Answers

Answer:

the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

Explanation:

Given the data in the question;

initial velocity; u = 0 m/s

height; h = 2.5 m

we find the velocity of the ball just before it touches the foam.

using the equation of motion;

v² = u² + 2gh

we know that acceleration due gravity g = 9.81 m/s²

so we substitute

v² = ( 0 )² + ( 2 × 9.81 × 2.5 )

v² = 49.05

v = √49.05

v = 7.00357 m/s

Now as the ball touches the foam

final velocity v₀ = 0 m/s

compresses S = 3 cm = 0.03 m

v₀² = v² + 2as

we substitute

( 0 )² = 49.05 + 0.06a

0.06a = -49.05

a = -49.05 / 0.06

a = -817.5 m/s²

Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²

A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.

Answers

Answer:

42.67N

Explanation:

Step one:

Given

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

Required

Force F

the expression for the force is

F=mΔv/t

F=0.32*(22-14)/0.06

F=(0.32*8)/0.06

F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?

Answers

Answer:

ratio of the piccolo's length to the flute's length is 0.4916

Explanation:

given data

frequency of piccolo = 522.5 Hz

frequency of flute = 256.9 Hz

to find out

ratio of the piccolo's length to the flute's length

solution

we get here length of tube that is express as

length of tube = velocity of sound ÷ fundamental frequency .......................1

so here ratio of Piccolo length to flute that is

$(L\ picco)/(L\ flute) = (f\ flute)/(f \ piccolo)$

$(l \ piccolo)/(L\ flute) = (256.9)/(522.5)$ = 0.4916

so ratio of the piccolo's length to the flute's length is 0.4916

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$

s = 136.89 m

Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m

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Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=(v-u)/(t)$

$a=(13.41-31.29)/(5)=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576* t$

$t=(31.29)/(3.576)=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2* (-3.576)\cdot s$

s=136.89 m

A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?