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3 years ago

What is the net force on a 10-kg solid steel sphere falling in air at terminal speed?

1 answer:

5 0

When we refer to the term terminal velocity, we will therefore understand that the velocity is constant at that point. For this, if the speed is constant, the acceleration is zero.

Taking Newton's second law as a reference, we have that the force is equivalent is the product between mass and acceleration, therefore

$F= ma$

Since acceleration is the change of speed in an instant of time, we have to

$F =m \frac{\Delta V}{t}$

However, the change in speed is zero since this is constant, so we will have to

$F= (10)(0)$

$F = 0$

Therefore the net force will be zero.

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On a day the air is still a windmill still posses what type of energy is that ?

Montano1993 [528]

Well, it's up on top of a pole or pedestal of some sort,
so it has some gravitational potential energy relative to
the ground. In other words, if it somehow became detached
from its structure and fell to the ground, it would make quite
an energetic splat when it got there.

Also, the windmill is at the temperature of the air around it,
which is far from Absolute Zero, so the windmill holds a lot of
thermal (heat) energy.

Then I guess there's the matter of the chemical energy in the
molecules of the material that the windmill is made of, and the
nuclear energy in its atoms.

6 0

3 years ago

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then rel

pshichka [43]

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

5 0

3 years ago

An offshore rig begins to spill oil in a circular patch centered on the rig. If the radius ofthe oil spill increases at a rate o

Tresset [83]

Answer:

Explanation:

The area of a circle is given b

A=πr²

Then,

dA/dr= 2πr

Also give that,

dr/dt=30m/hr, r=100m

We want to find dA/dt

dA/dt=dA/dr×dr/dt

dA/dt= 2πr × 30

dA/dt= 2 ×π ×100×30

dA/dt=18849.56m²/s.

The rate of the area is 18849.56m²/s

7 0

3 years ago

Read 2 more answers

- A cannon of 2000 kg fires a shell of 10 kg at

fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

$p=0$

After the shot, the momentum is:

$p=MV+mv$

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V, we find the velocity of the cannon:

$V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s$

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

$a=-0.25 m/s^2$ is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

$s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m$

7 0

4 years ago

An object is 10 cm from the mirror, its height is 1 cm and the focal length is 5 cm. What is the image height? (Indicate the obj

boyakko [2]

The image height is -10 cm (the image is upside down)

Explanation:

We can solve the problem by using the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 5 cm is the focal length of the mirror

p = 10 cm is the distance of the object from the mirror

q is the distance of the image from the mirror

Solving for q, we find:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}\\\rightarrow q= 10 cm$

So, the distance of the image from the mirror is 10 cm.

Now we can find the image height by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where

y' is the height of the image

y = 1 cm is the height of the object

and using

p = 10 cm

q = 10 cm

We find the size of the image:

$y' = -\frac{qy}{p}=-\frac{(10)(1)}{10}=-10 cm$

where the negative sign indicates that the image is upside down.

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8 0

3 years ago