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3 years ago

How can you remove petrol stains from your clothes? Explain.

1 answer:

5 0

Allow it to air dry and then brush the baking soda off of the clothing. This step can be repeated until the gasoline is fully removed. Alternatively, you can soak the stained clothes overnight submerged in water with one cup of baking soda added. Wash, rinse, and check for odor. Hope this helps!!

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What words can you spell out of the periodic table

loris [4]

GeNiUS is a word written with the element symbols for germanium, nickel, uranium, and sulfur. Anne Marie Helmenstine, Ph.

7 0

4 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

4 years ago

Every substance has a specific value of heat required to change the temperature of 1 gram of the substance by 1 degree Celsius.

zimovet [89]

I believe the correct answer from the choices listed above is option C. The units for the specific heat capacity or C would be J/g °C. It is <span>the heat required to raise the temperature of the unit mass of a given substance by a given amount .</span>

8 0

3 years ago

Read 2 more answers

What two quantities affect the force of attraction between two charges

miss Akunina [59]

1. the amount of electric charge that they carried
2. the distance between them

5 0

4 years ago

Read 2 more answers

g (15 points) Picture taken from problem 13-35. A 2000 kg truck is traveling at 25 meters/second. The driver suddenly hits the b

otez555 [7]

Answer:

Δx = 3.99 m

Explanation:

To determine distance, use kinetic energy

will make it short and easy.

KE=1/2mv2 and KE=Δxmgμ

Set the equations equal to each other

1/2mv2=Δxmgμ (Note: The masses cancel )

1/2v2=Δxgμ Solve for Δx

where g=9.8

Δx=v2/(2gμ) Δx = 25 / (2 * 9.8 * 0.32) Δx = 3.99 m

Please let me know if its correct, if not report it so we can correct it.

6 0

4 years ago