All categories

3 years ago

A book is sitting on a table until I decide I needed more room sol pushed it 2 points

Physics

1 answer:

EleoNora [17]3 years ago

5 0

Answer:

is their a picture or something

Explanation:

You might be interested in

Define 1 pascal pressure

Rasek [7]

Answer:For example, standard atmospheric pressure (or 1 atm) is defined as 101.325 kPa. The millibar, a unit of air pressure often used in meteorology, is equal to 100 Pa. (For comparison, one pound per square inch equals 6.895 kPa.)

Explanation:A pascal is a pressure of one newton per square metre, or, in SI base units, one kilogram per metre per second squared.

I hope this helps.... I'm sorry if it doesn't

5 0

2 years ago

Which of the following is a subsurface event takes place during the rock cycle

alukav5142 [94]

Answer:

The answer is A. Cementing...

Explanation:

hope this helps

4 0

3 years ago

Read 2 more answers

You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$

6 0

3 years ago

What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL

pogonyaev

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of object = $6.0g/ml$

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

$6.0g/ml=\frac{3.0g}{\text{Volume of substance}}$

${\text{Volume of substance}}=0.50ml$

Thus the volume of an irregularly shaped object is 0.50 ml

4 0

3 years ago

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by $F=kx$

So $63.5=k\times 0.0531$

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

$W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J$

8 0

3 years ago