3 years ago

PLEASE HURRY Which of the following measurements is a measurement of velocity?

Physics

1 answer:

madreJ [45]3 years ago

6 0

Answer:

C) 350 m/s N

Explanation:

Velocity is measured in miles per hour or metres per second.

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The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r

pogonyaev

Given:

L = 1 mH = $1\times 10^{-3}$ H

total Resistance, R = 11 $\Omega$

current at t = 0 s,

$I_{o}$ = 2.8 A

Formula used:

$I = I_{o}\times e^-{\frac{R}{L}t}$

Solution:

Using the given formula:

current after t = 0.5 ms = $0.5\times 10^{-3} s$

for the inductive circuit:

$I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}$

$I = 2.8\times e^-5.5$

I =0.011 A

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2 years ago

A vase falls from a windowsill 50.0 m above the sidewalk. How fast is the base moving when it hits the ground?

scZoUnD [109]

Answer:

The answer is A sorry if i'm wrong

Explanation:

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3 years ago

Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna

son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0

2 years ago

Ninais ba ni floranteng mapalapit ang kanyang loob kay adolfo

luda_lava [24]

Answer:

Explanation:

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2 years ago

Part complete How long must a simple pendulum be if it is to make exactly one swing per five seconds?

shtirl [24]

Answer:

$L=6.21m$

Explanation:

For the simple pendulum problem we need to remember that:

$\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0$ ,

where $\theta$ is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:

$\omega^{2}=\frac{g}{L}$ ,

where $\omega$ is the angular frequency.

There is also an equation that relates the oscillation period and the angular frequeny:

$\omega=\frac{2\pi}{T}$ ,

where T is the oscillation period. Now, we can easily solve for L:

$(\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m$

3 0

3 years ago