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VLD [36.1K]
3 years ago
12

Question 8

Physics
1 answer:
Viefleur [7K]3 years ago
3 0

Answer:

96 Joules

Explanation:

The formula for work is Fnet times displacement (F x d = w) which, in this case, 48N is the Fnet and 2m as the displacement. Then all we need to do is multiply these two and we get 96 Joules.

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How much work is done by the force lifting a
Ann [662]

The work done in lifting the hamburger is equal to the increase in gravitational potential energy of the hamburger, given by

W=\Delta U=mg \Delta h

where

m=0.1 kg is the mass of the hamburger

g=9.81 m/s^2 is the gravitational acceleration

\Delta h=0.3 m is the increase in height of the hamburger


Substituting numbers into the equation, we find

W=(0.1 kg)(9.81 m/s^2)(0.3 m)=0.3 J


So, the correct answer is

(3) 0.3 J

3 0
3 years ago
Is eureka a better conductor than copper
cricket20 [7]
Nope Copper is a better conductor
5 0
2 years ago
A 120 resistor a 60 ohm resistor and a 40 ohm resistor are connected in parallel to a 120 volt power source. what is the current
larisa [96]

Answer:

6 A

Explanation:

First of all, we need to calculate the equivalent resistance of the circuit. The three resistors are connected in parallel, so their equivalent resistance is given by:

\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{120 \Omega}+\frac{1}{60 \Omega}+\frac{1}{40 \Omega}=\frac{3+2+1}{120 \Omega}=\frac{6}{120 \Omega}\\R_T = \frac{120}{6} \Omega

And now we can use Ohm's law to find the current in the circuit:

V=R_T II=\frac{V}{R_T}=\frac{120 V}{\frac{120}{6}\Omega}=6 A

6 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
3 years ago
A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tunin
JulijaS [17]

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

f_s = f(\dfrac{v}{v-(-v_s)})

f_s = f(\dfrac{v}{v+v_s})

v_s = v[\dfrac{f_s}{f_o}-1]

      = 343[\dfrac{508}{490}-1]

      v_s=12.6 m/s

distance the tunning fork has fallen

y_1=\dfrac{v^2}{2a_y}

     =\dfrac{12.6^2}{2\times 9.8}

     =8.1 m

now, time required for the observed will be

t = \dfrac{8.1}{343} = 0.023 s

now, for the distance calculation

y_2 = u\ t + \dfrac{1}{2}at^2

  = 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m

3 0
2 years ago
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