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2 years ago

14

If I have an object that starts from rest and reaches a velocity of 25m/s over a distance of 11m what is the acceleration of the

object
28.4m/s
28m/s
28.4m/s^2
28m/s^2

1 answer:

Elodia [21]2 years ago

3 0

Answer:

Here is something that may help you!!

Explanation:

I found it in a cite (not that I'm plagiarizing, or anything).

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If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com

Lelu [443]

Answer:

<em> B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.</em>

<em></em>

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = $100*10^{-6} R$

for dry skin, voltage will be

V = IR = $100*10^{-6}*200R = 0.02R$

Comparing both voltages

$0.02R$ ÷ $100*10^{-6} R$ = 200

<em>this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.</em>

4 0

3 years ago

Which environmental change occurs quickly

never [62]

Air and water have a good day

3 0

3 years ago

What will change the velocity of a periodic wave?

Fantom [35]

<span>it will be changed by changing the medium of the wave</span>

8 0

3 years ago

Which choice correctly describes what happens during heating?

Masja [62]

Answer:

Option 4

Explanation:

During heating actually heat transfer takes place from a body at higher temperature to a body at lower temperature and the heat transfer takes place until both attain the same temperature

Therefore heat transfer depends on the temperature of the systems

Now while comparing the thermal energies of the systems, if both the systems have same mass then the system which is at higher temperature has greater thermal energy when compared to the system which is at lower temperature

So in this case assuming that both the systems have same mass then the energy will leave the system with greater thermal energy and go into the system with less thermal energy as the system with greater thermal energy in this case will be at higher temperature and we are considering this assumption because thermal energy not only depends on temperature but also depends on mass of the system

7 0

3 years ago

*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa

Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
This force is not a new force, is just the net force aiming to the center of the circle.
In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
So, we can write the following expression:

$F_{cent} = F_{n} - F_{g} (1)$

It can be showed that the centripetal force is related to the speed by the following expression:
$F_{cent} = m*\frac{v^{2}}{r} (2)$

The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
Replacing (2) in (1), and solving for Fn, we get:

$F_{n} = m*\frac{v^{2} }{r} + m*g (3)$

Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

$F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g} (4)$

Replacing Fg by its value, simplifying, and solving for v, we get:

$v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)$

3 0

2 years ago