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2 years ago

10

A spring of spring constant 25 N/m is hung vertically and a 0.300 kg mass is attached to one end, causing a displacement of the

end of the spring of _____ m.

1 answer:

Mamont248 [21]2 years ago

3 0

Answer:k=1175

Explanation:thank you for asking

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How can being scientifically literate help you in your own life?

stiv31 [10]

Answer: You can betterly understand what's around you and how it works

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1 year ago

Q12. How big is a Moon? How big is a Mars? What is therefore the weight of the person from Q11 on the Moon? What is the person's

shutvik [7]

Answer:

The moon is 1,079.4 mi.

Mars is 2,106.1 mi

Multiply your weight by the moon's gravity relative to earth's, which is 0.165. Solve the equation. In the example, you would obtain the product 22.28 lbs. So a person weighing 135 pounds on Earth would weigh just over 22 pounds on the moon

Being that Mars has a gravitational force of 3.711m/s2, we multiply the object's mass by this quanitity to calculate an object's weight on mars. So an object or person on Mars would weigh 37.83% its weight on earth.

Explanation:

~Hope this helps

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2 years ago

A load of bricks is being lifted by a crane at a steady velocity of 5.0m/s when one bricks falls off 6.0 m above the ground

Ganezh [65]

LOOK OUT BELOW ! ! !

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2 years ago

A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric fiel

AveGali [126]

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

$j=\frac {I}{A}$ where I is current and A is area and $A=\pi r^{2}$

Substituting this into the first equation then $E=P\times \frac {I}{\pi r^{2}}$

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

$E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c$

4 0

2 years ago

A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

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2 years ago