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3 years ago

Help please. I will give brainliest.

Physics

1 answer:

ser-zykov [4K]3 years ago

7 0

Answer:

D, E, F, G are correct.

Explanation:

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A stone falls from rest from the top of a cliff. A second stone is thrown downward from the same height 2.7 s later with an init

Darina [25.2K]

Answer:4.05 s

Explanation:

Given

First stone is drop from cliff and second stone is thrown with a speed of 52.92 m/s after 2.7 s

Both hit the ground at the same time

Let h be the height of cliff and it reaches after time t

$h=\frac{gt^2}{2}$

For second stone

$h=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$ ---2

Equating 1 &2 we get

$\frac{gt^2}{2}=52.92\times \left ( t-2.7\right )+\frac{g\left ( t-2.7\right )^2}{2}$

$\frac{g}{2}\left ( t-t+2.7\right )\left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$13.23\times \left ( 2t-2.7\right )-\left ( t-2.7\right )52.92=0$

$26.46t-35.721-52.92t+142.884=0$

$t=4.05 s$

4 0

3 years ago

A football punter accelerates a football from rest to a speed of 15 m/s during the time in which his toe is in contact with the

KengaRu [80]

Use F=ma formula
F=m v/t. a=v/t
put values u will get answer.
answer should be around 30N .

7 0

4 years ago

What is the function of dermal tissue

VladimirAG [237]

It covers the younger part of the planet.

3 0

3 years ago

Calculate the electric field strength where a test charge of 2 coulombs is repelled by a force of 2 newtons. ________ N/C

Nitella [24]

Answer:

1 N/C

Explanation:

Please Mark Me Brainly!!

Hope This Help??

7 0

3 years ago

What amount of heat is required to increase the temperature of 75.0 grams of gold from 150°C to 250°C? The specific heat of gold

aleksley [76]

Answer:

<h2>B. 980 joules</h2>

Explanation:

Given the following data

initial temperature T1= 150 °C

final temperature T2= 250 °C

specific heat of gold c= 0.13 J/g°C

mass of gold m= 75.0 grams

we can use the expression stated below to solve for the quantity of heat

$Q= mc(T2-T1)---------1$

Substituting our known data into the expression we can solve for the value of Q

$Q= 75*0.13(250-150)---------1\\\\Q= 75*0.13(100)\\\\Q= 975 Joules$

<em><u>The quantity of heat need to raise the temperature from 150°C to 250°C is 975 J</u></em>

4 0

3 years ago