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3 years ago

A skydiver falls from rest through air and reaches teminal velocity

Physics

1 answer:

Arte-miy333 [17]3 years ago

3 0

Answer:

B constant at 10m/s²

Explanation:

Newton's second law. An object moving only because of gravity is said to be free falling. All objects accelerate at the same rate of 9.8m/s²

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Does air resistance affect the motion of a falling object differently when the mass is greater?

mixas84 [53]

<span>NO.

Air resistance does not affect the motion of a falling object differently when the mass is greater because the mass of an object does not in any way affect the speed of falling due to gravity, and air resistance depends only on the speed of the object and its surface area.</span>

8 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t

stepan [7]

We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
$F=ma$
where $F=-50 N$ is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
$a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2$

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
$a= \frac{v_f^2-v_i^2}{2d}$
where
$v_f=0$ is the final speed of the sled
$v_i=5 m/s$ is the initial speed
$d$ is the distance covered

By rearranging the equation, we find d:
$d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m$

3 0

3 years ago

Two wires A and B of the same material, having radii in the ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of drif

Brums [2.3K]

uhhhh idk cheif...

thats a big oof right there.

ged is always an option

3 0

4 years ago

Is it possible for one ray to be shorter in length than another?

8090 [49]

Yes, even light rays can vary in wavelength and frequency, if the length of the ray is sorter, it becomes more energetic and has a higher frequency. If you're talking about a ray tracing diagram for lenses or mirrors, the length of the ray doesn't really matter unless you're finding the path length but there are some procedures for that too. Let me know if I missed what you were asking.

8 0

3 years ago