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Art [367]
1 year ago
9

How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?

Physics
1 answer:
Zarrin [17]1 year ago
6 0

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

d =  \frac{ \:Δv}{e}

=  \frac{1.5v}{100v/m}

= 1.5 \times 10 {}^{ - 2} m

learn more about potential difference from here: brainly.com/question/28166044

#SPJ4

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A 28 kg student stand on the surface of the Earth. What is the magnitude of the
Contact [7]

Answer:

F = 274.68[N]

Explanation:

The gravitational force is equal to the weight of a body, or this case that of a person. Weight can be calculated by means of the product of mass by gravitational acceleration. In this way we have the following equation:

F=m*g

where:

F = force or weight [N]

m = mass = 28 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

F=28*9.81\\F=274.68[N]

4 0
3 years ago
On a sky coaster (human pendulum) that reaches 10 meters from it's equilibrium position, a man of 120 kg is able to reach a maxi
dimaraw [331]

Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

v=√(2*9.81*10)=14.0071410359145 m/s

Rounding off, v is approximately 14 m/s

6 0
3 years ago
You have just landed on Planet X. You take out a ball of mass 101 g , release it from rest from a height of 16.0 m and measure t
IRINA_888 [86]

Answer:

0.3817 N

Explanation:

Remark

One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.

Givens

m = 101 g

vi = 0       That's what at rest means.

t = 2.91 s

d = 16 m

F= ?

Formulas

d = vi*t + 1/2*a * t^2

Force = m * a

Solution

16 = 0 + 1/2 a * 2.91^2

16 = 4.234 a                       Divide by 4.234

16/4.234 = a

a = 3.779

F = m * a

a = 3.779

m = 101 g = 1 kg / 1000 grams

m = 0.101 kg

F = 0.101 * 3.779

F = 0.3817N

8 0
3 years ago
The logarithm of x, written log(x), tells you the power to which you would raise 10 to get x. So, if y=log(x), then x=10^y. It i
fomenos

To solve this problem it is necessary to apply the rules and concepts related to logarithmic operations.

From the definition of logarithm we know that,

Log_{10}(10) = 1

In this way for the given example we have that a logarithm with base 10 expressed in the problem can be represented as,

log_{10}(1,000,000)

We can express this also as,

log_{10}(10^6)

By properties of the logarithms we know that the logarithm of a power of a number is equal to the product between the exponent of the power and the logarithm of the number.

So this can be expressed as

6*log_{10}(10)

Since the definition of the base logarithm 10 of 10 is equal to 1 then

6*1=6

The value of the given logarithm is equal to 6

8 0
4 years ago
When you put a pot of water on the stove, the stove transfers thermal energy to the water. As the water gains large
STALIN [3.7K]

Answer:

It releases some of the energy into the atmosphere as hot steam.

Explanation:

8 0
3 years ago
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