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RideAnS [48]
4 years ago
9

If I drop a watermelon from the top of one of the tower dorms at CSU, and it takes 3.34 seconds to hit the ground, calculate how

tall the building is in meters
Physics
1 answer:
WINSTONCH [101]4 years ago
4 0
T= 3.34

Vi= 0

A= 9.81

D= ?

d=Vit+1/2at^2

d= 1/2(9.81)(3.34)2

d= 54.7 or 55 meters tall
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Distance divided by time what is velocity
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Yup displacement covered /time taken. bcz both velocity and displacement are vectors
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Which type of graph is best for analyzing quantitative dependent and independent variables?
34kurt

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line graph

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8 0
3 years ago
Find the magnitude of the average induced emf in the coil when the magnet is turned off and the field decreases to 0 T in 2.8 s
Ostrovityanka [42]

Answer:

<em>2.15 mV</em>

<em></em>

Explanation:

The complete question is

You have a coil of wire with 17 turns each of 1.5 cm radius. You place the plane of the coil perpendicular to a 0.50-TB?  field produced by the poles of an electromagnet.

Find the magnitude of the average induced emf in the coil when the magnet is turned off and the field decreases to 0 T in 1.9 s .

Radius of the coil = 1.5 cm = 0.015 m

number of turns = 17

Initial magnetic field B_{1} = 0.50 T

Final magnetic field B_{2} = 0 T

time taken dt = 1.9 s

Area pf the coil = πr^{2} = 3.142 x 0.015^{2} = 7.1 x 10^-4 m^2

magnetic flux = BA

initial flux = B_{1}A = 0.5 x 7.1 x 10^-4 = 3.55 x 10^-4 Wb

Final flux = B_{2}A = 0 x 7.1 x 10^-4 = 0 Wb

change in flux dФ = 3.55 x 10^-4  -  0 = 3.55 x 10^-4 Wb

Induced EMF E = NdФ/dt = (17 x 3.55 x 10^-4)/2.8 = 2.15 x 10^-3 V

==> <em>2.15 mV</em>

7 0
4 years ago
Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o
disa [49]

Answer:

ΔV = 20.1 V

Explanation:

As the positive plates are connected to each other, the capacitors are connected in parallel, so the total system load is the sum of the charges on each capacitor.

               Q = Q₁ + Q₂

                 

The charge on each capacitor is

            Q₁ = C₁ ΔV₁

            Q₁ = 24 10⁻⁶ 25

            Q₁ = 6.00 10⁻⁴ C

            Q₂ = C₂ ΔV₂

            Q₂ = 13 10⁻⁶ 11

            Q₂ = 1.43 10⁻⁴ C

The total set charge is

            Q = (6 + 1.43) 10⁻⁴

            Q = 7.43 10⁻⁴ C

The equivalent capacitance is

           C_eq = C₁ + C₂

           C_eq = (24 + 13) 10⁻⁶

           C_eq = 37 10⁻⁶ F

Let's use the relationship to find the voltage

          Q = C_eq ΔV

          ΔV = Q / C_eq

          ΔV = 7.43 10⁻⁴ / 37 10⁻⁶

          ΔV = 2.008 10¹

          ΔV = 20.1 V

This voltage is constant in the combination so it is also the voltage in capacitor C1

3 0
4 years ago
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