How many neutrons does C-13 have?<br> A. 6<br> B. 7<br> C. 13<br> D. 19

Suppose you want to use this human engine to lift a 2.35 kg box from the floor to a tabletop 1.30 m above the floor. How much mu

morpeh [17]

Answer:

The increase in the gravitational potential energy is 29.93 joules.

Explanation:

Given that,

Mass of the box, m = 2.35 kg

It is lifted from the floor to a tabletop 1.30 m above the floor, h = 1.3 m

We need to find the increase the gravitational potential energy. Initial it will placed at ground i.e. its initial gravitational potential is equal to 0. The increase in the gravitational potential energy is given by :

$U=mgh$

$U=2.35\ kg\times 9.8\ m/s^2\times 1.3\ m$

U = 29.93 Joules

So, the increase in the gravitational potential energy is 29.93 joules. Hence, this is the required solution.

6 0

2 years ago

Chen is testing the friction of three surfaces. He pushes the same ball across three different surfaces with the same force and

Law Incorporation [45]

Answer:

im no proffesional

Explanation:

but i tghink you need a proffessional for this one

4 0

2 years ago

NEED HELP

ch4aika [34]

Answer: C.

Explanation:

Just took quiz

8 0

2 years ago

Got an F in Physical Science. HELP ME PLZZZ

Dmitriy789 [7]

i know can you plzz help me with this question im sorry i didt answer your question i just need hel.

7 0

2 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$

Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

5 0

2 years ago

How many neutrons does C-13 have? A. 6 B. 7 C. 13 D. 19