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WINSTONCH [101]
3 years ago
13

A 5kg rock is lifted 2m. Find the amount of work done. ​

Physics
1 answer:
lianna [129]3 years ago
5 0

Answer:

98J

Explanation:

Given parameters:

Mass of rock  = 5kg

Height = 2m

Unknown:

Work done  = ?

Solution:

The amount of work done is given as:

 Work done  = Force x distance

 Work done = Weight x height

  Work done  = mgH

Now insert the parameters and solve;

 Work done  = 5 x 9.8 x 2 = 98J

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Schach [20]

Answer:

The  radius is  R =  0.22 5 \  m

Explanation:

From the question we are told that

    The current is  I  =  61 \ A

     The  magnetic field is  B  =  1.70 *10^{-4} \  T

Generally the magnetic field produced by a current carrying conductor  is mathematically represented as

        B  =  \frac{\mu_o  *  I}{2 *  R }

=>     R  =  \frac{\mu_o  *  I  }{ 2 *  B }

Here  \mu_o is the permeability of free space with value  \mu_o  =  4\pi * 10^{-7} N/A^2

=>    R  =  \frac{  4\pi * 10^{-7}  *   61  }{ 2 *   1.70 *10^{-4} }

=>  R =  0.22 5 \  m

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3 years ago
Which waves have the highest frenquencies?
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I believe the answer is B: UV
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A bird experiences an acceleration of -1.80m/s^2 for 3.81s, and ends up 26.5m to the left of its starting point. What is it's fi
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3 years ago
Three students are having a conversation. It goes like this. Allison says: "This problem says, can you add a resistor to the cir
notsponge [240]

Answer:

Nima and Natasha are absolutely correct.

Explanation:

When connecting two resistors in series, their resistances add:

R_{eq}=R_{1}+R_{2}

which means that whenever we add a resistance in series, their magnitudes will add, giving us a resistance that is greater than the original resistance, which will demand less current from the battery because of ohm's law:

I=\frac{V}{R}

So, the greater the resistance, the smaller the current.

When connecting two resistors in parallel, the reciprocal of ther resistances add:

\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}

or

R_{eq}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}

The equivalent resistance will always be less than the smallest resistor in the circuit, so the equivalent resistance will always decrease as more resistors are added. A decrease in the resistance means that the current will increase.

8 0
3 years ago
Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a dista
luda_lava [24]

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

V = \frac{kq}{r}

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}

Replacing the values we have that

V_1 =  \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}

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Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

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Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

\frac{1}{2} mv^2 = q(V_2-V_1)

Here

m = mass

v = Velocity

q = Charge

V = Voltage

Rearranging to find the velocity

v = \sqrt{ \frac{2q(V_2-V_1)}{m}}

Replacing,

v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}

v = 6.3*10^6m/s

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is 6.3*10^6m/s

6 0
3 years ago
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