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3 years ago

Is my answer to part A correct ? And can someone please answer part 2 and if I am wrong correct part A ? !!!!!!

Physics

1 answer:

levacccp [35]3 years ago

5 0

In A, the left side is exactly what you should write, to lead you to the answer ... correct units and everything. But when you tried to walk across the 'equals' sign, you slipped, fell, and got demolished on the rocks below. "19.6" is not the product of the things on the left side. Do the arithmetic again please.

B. When the height doubles, so does the potential energy.

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7. A car is travelling along a road at 30 ms when a pedestrian steps into the road 55 m ahead. The

trasher [3.6K]

Answer: Car collide with man

Explanation:

Given

Speed of car is $u=30\ m/s$

Distance of the man from the car is $s=55\ m$

Reaction time $t_r=0.5\ s$

Rate of deceleration $a_d=-10\ m/s^2$

Distance traveled in the reaction time $d_o=30\times 0.5=15\ m$

Net effective distance to cover $d=55-15=40\ m$

Distance required to stop the car

$\Rightarrow v^2-30^2=2(-10)(s)\\\Rightarrow 0-900=-20s\\\Rightarrow s=45\ m$

Require distance is more than that of net effective distance. Hence, car collides with the man.

6 0

3 years ago

Q7. What caused the early materials to clump together?

tekilochka [14]

Answer:

C. pressure

because heat expands and so does explosions and gravity is just gravity.

4 0

3 years ago

What happens when white light shines through a translucent, red, glass window? a) All colors of light except red are transmitted

GREYUIT [131]

Answer:

b. Red light is transmitted through and reflected by the glass

Explanation:

Give me brainliest plz!

3 0

3 years ago

A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?

olchik [2.2K]

Answer:

$m=5.78\times 10^{-3}\ g$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Explanation:

Given:

quantity of charge transferred, $Q=31\ C$

No. of electrons in the given amount of charge:

As we have charge on one electron $1.602\times 10^{-19}\ C$

so,

$n_e=\frac{Q}{e}$

$n_e=\frac{31}{1.602\times 10^{-19}}$

$n_e=1.935\times 10^{20}$ is the no. of electrons

Now if each water molecules donates one electron:

Then we require $n=1.935\times 10^{20}$ molecules.

Now the no. of moles in this many molecules:

$n_m=\frac{n}{N_A}$

where

$N_A=6.022\times 10^{23}$ Avogadro No.

$n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}$

$n_m=3.213\times 10^{-4}\ moles$

We have molecular mass of water as M=18 g/mol.

So, the mass of water in the obtained moles:

$n_m=\frac{m}{M}$

where:

m = mass in gram

$3.213\times 10^{-4}=\frac{m}{18}$

$m=5.78\times 10^{-3}\ g$

7 0

3 years ago

a father and his son want to play on a seesaw. where on the seesaw should each of them sit to balance the torque

joja [24]

On the opposite sides, the father would out weigh the sons weight, simply because he is bigger than the son

The definite answer is “the opposites side of the seesaw”

Hope this helps you ☁︎☀︎☁︎

3 0

3 years ago