WILL MARK BRAINLIEST!!!!!!!!!!!!!!!!!!

1 answer:

3 0

*FRICTIONAL FORCE* in the opposite direction of the way Bobby is pushing.
Friction is a force which varies but it is always opposing the direction of motion.

*APPLIED FORCE* is the force that Bobby is pushing with.
An applied force is literally the force that is applied to an object.

*WEIGHT FORCE* is also called the force of gravity. It is straight downward.
It is the weight of the object multiplied by the force of gravity. If the TV weighed 100kg, acceleration is always 9.81 m/s^2, so the weight force would be 981 N.

*NORMAL FORCE* is the force which is holding the TV above ground. The ground supplies a force upward against the TV.
Normal force is just the force that prevents the TV from falling through the ground. We don't normally realize it in our everyday life, but the floor must hold everything up because gravity is always "pushing" against it.

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You would convert from grams to moles by using the ___________ of the substance.

Firdavs [7]

You would convert from grams to moles by using the molar mass of the substance. The answer is letter B. for example, the molar mass Carbon dioxide is 44.01 g/mol. It means that for 1 mole of carbon dioxide, it contains 44.01 grams of Carbon dioxide.

7 0

3 years ago

Mrs. Waid drives her car up Route 8 at 80mph. If her car weighs 2500 lbs,

Anvisha [2.4K]

1.358 J is the kinetic energy of the car driven by Mrs. Waid.

<u>Explanation:</u>

Given data:

Velocity at which Mrs. Waid drives her car = 80 mph

In order to convert mph (meter per hour) into mps (meter per second),

$\frac{80}{3600}=0.0222 \mathrm{m} / \mathrm{s}$

Car weighs 2500 lbs, means mass of the car, m = 2500 lbs

I kilo gram = 2.20462 pound

Therefore, 1 pound (lb)= 0.45359237 kilograms (kg).

To converting pounds into kilogram,

$\frac{2500}{0.45359237}=5511.55 \mathrm{kg}$

As we know, the kinetic energy can be defined as directly proportionate to the object’s mass (m) and square of its velocity (v). The expression can be given as below,

$\text { kinetic energy }(K . E)=\frac{1}{2} \times m \times v^{2}$