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Leni [432]
2 years ago
5

What happens to a falling object when the force of air resistance equals the force of gravity?

Physics
2 answers:
12345 [234]2 years ago
7 0

Answer:

A falling object will continue to accelerate to higher speeds until they encounter an amount of air resistance that is equal to their weight.

Explanation:

borishaifa [10]2 years ago
7 0

Answer:

so its basically termonal velocity

Explanation:

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A student throws a baseball horizontally at 25 meters per second from a cliff 45 meters above the level ground. Approximately ho
Delicious77 [7]
First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed. 
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m 
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s 
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use: 
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s 
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
8 0
3 years ago
Looking that following diagram of bar magnets, determine if the magnets will or will not connect (attract) and why.
vovangra [49]

Answer: They will NOT connect because like poles are facing each other, and like poles repel each other.

3 0
3 years ago
Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an
8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 m/s^2 where as radial the component of acceleration is 8.77 m/s^2

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

r=e^u

So the transverse component of acceleration are given as

a_{\theta}=(ru''+2r'u')\\

Here

r=e^u\\r=e^{\pi/4}\\r=2.1932 m

r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m

So

a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\

The transverse component of acceleration is 26.32 m/s^2

The radial component is given as

a_r=r''-r\theta'^2

Here

r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m

So

a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2

The radial component of acceleration is 8.77 m/s^2

6 0
3 years ago
What type of radioactive decay is shown in this equation?
svp [43]
There is no <span>radioactive decay</span>
6 0
3 years ago
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st
liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

4 0
2 years ago
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