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3 years ago

Which projectiles will be visibly affected by air resistance when they fall?

1 answer:

4 0

Any object that is launched as a projectile will lose speed and, as a result, altitude, as it travels through the air. The rate at which the object loses speed and altitude depends on the amount of force that way applied to it when it was launched. It is also dependent on the size and shape of the item. This is why something like, say, a football is much faster to fall to the ground than a bullet.

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If we mechanically expand a gas into a larger volume, while keeping the temperature constant, what happens to the pressure?

bezimeni [28]

C. The pressure decreases, and we can precisely predict by how much.

4 0

3 years ago

What is the momentum of a 800 kg car traveling at 20 m/s

BabaBlast [244]

Answer: The momentum of the car is 16000 kg.m/s

Explanation:

Momentum is defined as the force that keeps the object moving. It is also defined as the product of mass and velocity of an object.

Mathematically,

$p=m\times v$

where,

p = momentum of the car = ?

m = mass of the car = 800 kg

v = velocity of the car = 20 m/s

Putting values in above equation, we get:

$p=800kg\times 20m/s\\\\p=16000kg.m/s$

Hence, the momentum of the car is 16000 kg.m/s

5 0

3 years ago

Read 2 more answers

Please help!!!!!!!!!!!!!!!!

Serjik [45]

Answer:

jeez i dont get it hold up

Explanation:

where de numbers at

4 0

3 years ago

PLS HELP MEE ill literally suck you off JUST PLS and mark you brainliest

Svetlanka [38]

Do i have to answer the question and still have the offer on the table

7 0

3 years ago

A car is traveling with an initial velocity of 15 m/s when the driver decides to go faster by accelerating at 5 m/s squared for

tiny-mole [99]

Answer:

x = 400 [m]

Explanation:

To solve this problem we must use the following kinematics equations, first, we find the final speed, and then we proceed to find the distance traveled.

$v_{f}=v_{i} +(a*t)\\$

where:

Vf = final velocity [m/s]

Vi = initial velocity = 15 [m/s]

a = acceleration = 5 [m/s^2]

t = time = 10 [s]

Note: the positive sign in the Equation indicates that the car is accelerating, i.e. its speed is increasing.

Now replacing

Vf = 15 + (5*10)

Vf = 65 [m/s]

Now using the second equation:

$v_{f} ^{2}= v_{i} ^{2}+(2*a*x)$

where:

x = distance traveled [m]

x = (65^2 - 15^2)/ (2*5)

x = 400 [m]

8 0

4 years ago