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umka2103 [35]
3 years ago
15

A 1.4kg ball is swung in a vertical circle with a radius of 0.8m. If the frequency is 1.5Hz, find the tension in the rope at the

bottom position.​
Physics
1 answer:
Artyom0805 [142]3 years ago
4 0

Answer:

43

Explanation:

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Does gravity have an effect on the potential difference of a battery?
Sonja [21]

Answer: The correct answer is "No".

Explanation:

Gravity: It is the force which causes object to fall on the earth. It is the force which attracts bodies towards each other.

Potential difference: It is defined as the potential acting between the two points. The work done in moving the unit positive charge from one location to the another location.

The potential difference in battery is caused by the electrodes. There are two terminals in battery: Negative terminal which is at lower potential and Positive terminal which is at higher potential. It forces the electrons to flow in the circuit which constitutes the current.

The gravity and the potential difference have no relation between them.

Therefore, gravity have no effect on the potential difference of a battery.

5 0
3 years ago
Help please asap due 20 minutes please help me ​
Finger [1]
Did you turn it in yet?
4 0
3 years ago
The electric field between two parallel plates is uniform, with magnitude 628 N/C. A proton is held stationary at the positive p
aliina [53]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

F_{p} = q_{p}E

and, we know that, F = ma

So,

m_{p}a = q_{p}E

a = \frac{q_{p}.E }{m_{p} }      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2at^{2}

Here, u = 0.

S = 1/2at^{2}

Put equation 1 into the above equation:

S = 1/2 x (\frac{q_{p}.E }{m_{p} }  )t^{2}      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x (\frac{q_{e}.E }{m_{e} }  )t^{2}    Equation 3

We know that the charge of electron is equal to the charge of proton so,

q_{p} = q_{e} = q

By dividing the equation 2 by equation 3, we get:

\frac{S}{D-S} = \frac{m_{e} }{m_{p} }

Solve the above equation for S,

Sm_{p} = m_{e}D - m_{e}S

So,

S = \frac{m_{e}.D }{(m_{e} + m_{p})  }

Plugging in the values,

As we know the mass of electron is 9.1 x 10^{-31} and the mass of proton is 1.67 x 10^{-27}

S = \frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27}  }

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = \frac{m_{Cl}.D }{(m_{Cl} + m_{Na})  }

As we know the mass of chlorine is 35.5 and of sodium is 23

S = \frac{35.5 . 4.22}{(35.5 + 23)}

S = 2.56 cm

7 0
2 years ago
What are the benefits of physical exercises?
makvit [3.9K]

Answer:

Manage your weight

Have lower blood pressure

Lower your risk of falls

it reduces your risk of heart attack

7 0
2 years ago
How do you calculate the net force, i need a full explanation PLEASE
Lina20 [59]

Answer:

Once you have drawn the free-body diagram, you can use vector addition to find the net force acting on the object. We will consider three cases as we explore this idea:

Case 1: All forces lie on the same line.

If all of the forces lie on the same line (pointing left and right only, or up and down only, for example), determining the net force is as straightforward as adding the magnitudes of the forces in the positive direction, and subtracting off the magnitudes of the forces in the negative direction. (If two forces are equal and opposite, as is the case with the book resting on the table, the net force = 0)

Example: Consider a 1-kg ball falling due to gravity, experiencing an air resistance force of 5 N. There is a downward force on it due to gravity of 1 kg × 9.8 m/s2 = 9.8 N, and an upward force of 5 N. If we use the convention that up is positive, then the net force is 5 N - 9.8 N = -4.8 N, indicating a net force of 4.8 N in the downward direction.

Case 2: All forces lie on perpendicular axes and add to 0 along one axis.

In this case, due to forces adding to 0 in one direction, we only need to focus on the perpendicular direction when determining the net force. (Though knowledge that the forces in the first direction add to 0 can sometimes give us information about the forces in the perpendicular direction, such as when determining frictional forces in terms of the normal force magnitude.)

Example: A 0.25-kg toy car is pushed across the floor with a 3-N force acting to the right. A 2-N force of friction acts to oppose this motion. Note that gravity also acts downward on this car with a force of 0.25 kg × 9.8 m/s2= 2.45 N, and a normal force acts upward, also with 2.45 N. (How do we know this? Because there is no change in motion in the vertical direction as the car is pushed across the floor, hence the net force in the vertical direction must be 0.) This makes everything simplify to the one-dimensional case because the only forces that don’t cancel out are all along one direction. The net force on the car is then 3 N - 2 N = 1 N to the right.

Case 3: All forces are not confined to a line and do not lie on perpendicular axes.

If we know what direction the acceleration will be in, we will choose a coordinate system where that direction lies on the positive x-axis or the positive y-axis. From there, we break each force vector into x- and y-components. Since motion in one direction is constant, the sum of the forces in that direction must be 0. The forces in the other direction are then the only contributors to the net force and this case has reduced to Case 2.

If we do not know what direction the acceleration will be in, we can choose any Cartesian coordinate system, though it is usually most convenient to choose one in which one or more of the forces lie on an axis. Break each force vector into x- and y-components. Determine the net force in the x direction and the net force in the y direction separately. The result gives the x- and y-coordinates of the net force.

Example: A 0.25-kg car rolls without friction down a 30-degree incline due to gravity.

We will use a coordinate system aligned with the ramp as shown. The free-body diagram consists of gravity acting straight down and the normal force acting perpendicular to the surface.

We must break the gravitational force in to x- and y-components, which gives:

F_{gx} = F_g\sin(\theta)\\ F_{gy} = F_g\cos(\theta)F

gx

​

=F

g

​

sin(θ)

F

gy

​

=F

g

​

cos(θ)

Since motion in the y direction is constant, we know that the net force in the y direction must be 0:

F_N - F_{gy} = 0F

N

​

−F

gy

​

=0

(Note: This equation allows us to determine the magnitude of the normal force.)

In the x direction, the only force is Fgx, hence:

F_{net} = F_{gx} = F_g\sin(\theta) = mg\sin(\theta) = 0.25\times9.8\times\sin(30) = 1.23 \text{ N}F

net

​

=F

gx

​

=F

g

​

sin(θ)=mgsin(θ)=0.25×9.8×sin(30)=1.23 N

7 0
3 years ago
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