Ask question

Ask question

All categories

vladimir2022 [97]

3 years ago

13

Please help on this one?

1 answer:

Schach [20]3 years ago

6 0

Chemical energy. Elimination is useful to answer this question.

You might be interested in

A 200 kg crate is pulled along a level surface by an engine. the coefficient of friction between crate and surface is 0.4.

siniylev [52]

Normal force = 200(9.81) = 1962 N
Friction force = 1962 * .4 = 784.8
power = force * distance / time = 784.8 * 5 = 3924 Watts

Watts = Joules / sec
3924 x 180 = 706,320 Joules

4 0

3 years ago

Read 2 more answers

5. average A body sets off from rest with a constant acceleration of 8.0 m/s? What distance will it have covered after 3.0 s? 6.

chubhunter [2.5K]

Answer:

$\boxed {\boxed {\sf 36 \ meters}}$

Explanation:

We are asked to find the distance a body covers. We know the initial velocity, acceleration, and time, so we will use the following kinematic equation.

$d= v_i t+ \frac {1}{2} \ at^2$

The body starts at rest with an initial velocity of 0 meters per second. The acceleration is 8 meters per second squared. The time is 3.0 seconds.

$v_i$ = 0 m/s
a= 8 m/s²
t= 3 s

Substitute the values into the formula.

$d= (0 \ m/s)(3 \ s) + \frac{1}{2} (8 \ m/s^2)(3 \ s)^2$

Multiply the first set of parentheses.

$d= ( 0 \ m/s * 3 \ s) + \frac{1}{2} ( 8 \ m/s^2)(3 \ s)^2$

$d=0 \ m + \frac{1}{2} ( 8 \ m/s^2)(3 \ s)^2$

Solve the exponent.

(3 s)²= 3 s* 3 s= 9 s²

$d= 0 \ m + \frac{1}{2}( 8 \ m/s^2)(9 \ s^2)$

Multiply again.

$d= 0 \ m + \frac{1}{2} ( 72 \ m)$

$d= 36 \ m$

The body will cover a distance of <u>36 meters</u>.

5 0

2 years ago

A point charge of +3 C is located at the origin of a coordinate system and a second point charge of -6 C is at x = 1.0 m. At w

Butoxors [25]

Answer:

The point at which the electrical potential is zero is x = +0.33 m.

Explanation:

By definition the electrical potential is:

$V_{E} = \frac{K*q}{r}$

Where:

K: is Coulomb's constant = 9x10⁹ N*m²/C²

q: is the charge

r: is the distance

The point at which the electrical potential is zero can be calculated as follows:

$V_{1} + V_{2} = 0$

$K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0$ (1)

q₁ is the first charge = +3 mC

r₁ is the distance from the point to the first charge

q₂ is the first charge = -6 mC

r₂ is the distance from the point to the second charge

By replacing r₁ = 1 - r₂ into equation (1) we have:

$K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0$ (2)

By solving equation (2) for r₂:

$r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m$

Therefore, the point at which the electrical potential is zero is x = +0.33 m.

I hope it helps you!

8 0

3 years ago

Find the volume of a box with a length of 5 cm, a width pf 5cm, and a height of 10cm.

NNADVOKAT [17]

Answer:

250 cm³

Explanation:

given,

Length of the Box, L = 5 cm

Width of Box, W = 5 cm

height of the box, H = 10 cm

Volume of the Box = L W H

V = 5 x 5 x 10

V = 250 cm³

Volume of the box is equal to 250 cm³

7 0

3 years ago

What do you need to calculate the mechanical advantage of a block and tackle? A.resistance force

Degger [83]

I think D is the right answer

God Bless

5 0

3 years ago

Read 2 more answers