The force that causes an object to follow a

When a container decreases in size, what will happen to an amount of gas in the container??

alexgriva [62]

The gas would also decrease in size since the container lost gas to decrease the size of the container.

7 0

2 years ago

A balloon will stick to a wooden wall if the balloon is charged

grigory [225]

<h2>Answer: either way</h2>

The balloon contains neutral charge atoms, that is, it has the same number of electrons (negative charge), protons (positive charge) and neutrons (no charge).

Then, when two objects come into contact, the electrons of one of them can become part of the other.

Thus, by bringing the balloon closer to the wall, the wall, which is also made up of atoms, will reorder its charges in such a way that its electrons or protons become part of the balloon, charging it.

7 0

2 years ago

How long will it take for a boat traveling at 36 k/h to go 126 km?

brilliants [131]

3 hours 30 min i believe because 126/36 is 3.5 and 3.5 hours is 3 hours and 30 min. Brainliest pls

5 0

2 years ago

a flashlight has a resistance of 2.4 ohms. what voltage is applied by the batteries if the current in the circuit is 2.5A?

Mashcka [7]

E = I R

That means

Voltage = (current) x (resistance)

= (2.5 A) x (2.4 ohms)

= 6 volts .

8 0

3 years ago

Read 2 more answers

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

3 years ago