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maria [59]
3 years ago
15

What is the average velocity of Runner C for the first 15 seconds? Question 5 options: 1.3 m/s -2.7 m/s -1.3 m/s 2.7 m/s

Physics
2 answers:
kogti [31]3 years ago
7 0

Answer:

2.7 m/s

Explanation:

Took the test and got 100%

kipiarov [429]3 years ago
3 0

Answer:

Average velocity (v) of an object is equal to its final velocity (v) plus initial velocity (u), divided by two.

v¯¯¯=(v+u)2

Where:

v¯¯¯ = average velocity

v = final velocity

u = initial velocity

The average velocity calculator solves for the average velocity using the same method as finding the average of any two numbers. The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.

Explanation:

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What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?
AnnZ [28]
V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
F = 0.08810409956 Hz
4 0
3 years ago
The average speed of an object which moves 100 m in 5 seconds is?
CaHeK987 [17]

Answer:

A car

Explanation:

A car can travel 100 m in 5 seconds

Hope this helps!

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Answer:

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Explanation:

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5. Psychologists begin their studies by framing ____.
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3 years ago
A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge
Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

\tau=RC

where

R=50,000 \Omega is the resistance

C=2.0\mu F=2.0\cdot 10^{-6}F is the capacitance

Substituting,

\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s

The charge on a charging capacitor is given by

Q(t)=Q_0 (1-e^{-t/\tau} ) (1)

where

Q_0 is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Q(t)=0.90 Q_0

Substituting this into eq.(1) we find

0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s

4 0
3 years ago
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