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nexus9112 [7]
3 years ago
14

A 0.250 kg block on a vertical spring with spring constant of 4.45 ✕ 103 N/m is pushed downward, compressing the spring 0.080 m.

When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
Physics
1 answer:
Elza [17]3 years ago
6 0

Answer:h=5.81 m

Explanation:

Given

Mass of block(m)=0.250 kg

Spring Constant k=4.45\times 10^3 N/m

Initial elongation =0.080 m=8 cm

Thus Initial Potential Energy stored =Final Potential Energy stored in Block

P_i=\frac{kx^2}{2}

P_i=\frac{4.45\times 10^3\times 64\times 10^{-4}}{2}=14.24 J

P_f=mgh=0.25\times 9.8\times h

P_i=P_f

14.24 =0.25\times 9.8\times h

h=\frac{14.24}{0.25\times 9.8}=5.81 m

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densk [106]

Explanation:

V=IR; I=V/R

V=120V, R=10

I= 120/10=12

The current in the circuit is 12A

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A dvd drive has a maximum speed of 72000 revolutions per minute. if a dvd has a diamter of 12 what is the linear speed
Brilliant_brown [7]
The question isn't clear enough, I think it ask us to calculate the linear speed of a point at the edge of the DVD.
Now let's imagine we're a point at the edge of the DVD, we're undergoing a circular motion. Each minute we will complete a circular track 7200 times, now we need to know the distance we travel each turn. The perimeter of the DVD, a circular object is:
P=2\pi.R
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v=\frac{d}{t}
We now need to know how much distance is traveled during a minute or 60 seconds:
D=7200\times 2\pi\times R
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v=\frac{7200\times2\pi\times R}{60}
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Where the distance units were named units as the length unit is not specified in this exercise.<span />
7 0
4 years ago
A force of 2 newtons is required to stretch a spring 4 cm. The amount of force required to stretch the same spring 8 cm is _____
laiz [17]

Answer will be c for this one!

6 0
4 years ago
One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large
svet-max [94.6K]

Given:

• Mass, m = 0.200 mg

,

• Speed, v = 3.00 x 10³ m/s

,

• Time, t = 6.00 x 10^⁻⁸ s.

Let's calculate the force exerted.

Using the inpulse-momentum theroerm, we have:

impulse = change in momemntum

Where:

Impulse = force x time

change in momentum = mass x velocity.

Thus, we have:

F\times6.00\times10^{-8}=(0.200\times10^{-6})\times(3.00\times10^3)

Let's solve for the force F:

\begin{gathered} F=\frac{(0.200\times10^{-6})(3.00\times10^3)}{6.00\times10^{-8}} \\  \\ F=10000N \end{gathered}

Therefore, the force exterted is 10000 N.

ANSWER:'

10000 N

4 0
1 year ago
At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig
Anettt [7]

Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

Ball 1: m₁ = 0.5kg, u (initial velocity) =0, t = 2seconds

Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

<u>Second step:</u> we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;  

y = ut - \frac{gt^2}{2}

y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}

y_{1} = -19.6m

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1}  = 5.4m[/tex]

<u>Third step</u>: we determine the y-coordinate of ball 2 after 2 seconds

y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}

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<u>Fourth step: </u>we determine the y-component of the center mass of the two balls

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y = \frac{(0.5)X(5.4) +(0.25) X (10.4)}{(0.5 +0.25) }

y = 7.1m

<u>Fifth step:</u> we solve B part of the question; velocity of the center mass of the two balls

Velocity = \frac{distance of center mass of the two balls (y)}{time}

velocity = \frac{7.1 m}{2 s}

velocity = 3.55m/s

<u>Sixth step:</u> we solve C part of the question; acceleration of the center mass of the two balls

acceleration = \frac{velocity}{time}

acceleration = \frac{3.55}{2}

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