Ask question

Ask question

All categories

3 years ago

7

URGENT PLEASE ANSWER

1 answer:

olga_2 [115]3 years ago

7 0

Answer:

metal bolt

Explanation:

sorry if my explanation isn't correct but an object like feathers fall slower because of air resistance.

You might be interested in

The diagram shows changes of state between solid, liquid, and gas. The atoms of a substance gain energy during a change of state

Whitepunk [10]

I think it's liquid to gas

3 0

3 years ago

Read 2 more answers

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

How is the mass and speed of a particle related to its kinetic energy and temperature?

Brut [27]

Higher the mass will cause more friction which will make its temperature higher and it would also need more kinetic energy to move at a higher speed<span />

6 0

3 years ago

Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.

eimsori [14]

Answer:

(a). The net gravitational force is $4.20\times10^{-6}\ N$

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

$F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}$

$F_{net}=\dfrac{Gm(M'-M)}{r^2}$

Put the value into the formula

$F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}$

$F_{net}=4.20\times10^{-6}\ N$

The net gravitational force is $4.20\times10^{-6}\ N$

(b). We need to calculate the position

Force from 555 mass = Force from 255 mass

$\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}$

$555(0.390-x)^2=255x^2$

$300x^2-0.78x+84.4155=0$

$x=0.232\ m$

The position is at 0.232 m.

Hence, (a). The net gravitational force is $4.20\times10^{-6}\ N$

(b). The position is at 0.232 m.

5 0

3 years ago

Read 2 more answers

How does a pie chart help form conclusion

daser333 [38]

Its easy to create a visual and see major differences between data. If something takes up a large portion of the pie, its easy for you to tell just by looking at it

3 0

3 years ago