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dalvyx [7]
3 years ago
9

3. During a race, a sprinter increases from 5.0 m/s to 7.5 m/s over a period of 1.25s. What is the sprinter’s average accelerati

on during this period?
Also please try and explain
Please and Thank you
Physics
1 answer:
Rasek [7]3 years ago
5 0
(7.5)-(5)
----------- = 2 m/s^2
(1.25-0)

Average acceleration is calculated using the equation V(f)-V(i) / t(f)-t(i). (Final velocity minus initial velocity divided by final time minus initial time)
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Three ropes A, B and C are tied together in one single knot K. (See figure.)
noname [10]

The tension in the rope B is determined as 10.9 N.

<h3>Vertical angle of cable B</h3>

tanθ = (6 - 4)/(5 - 0)

tan θ = (2)/(5)

tan θ = 0.4

θ = arc tan(0.4) = 21.8 ⁰

<h3>Angle between B and C</h3>

θ = 21.8 ⁰ + 21.8 ⁰ = 43.6⁰

Apply cosine rule to determine the tension in rope B;

A² = B² + C² - 2BC(cos A)

B = C

A² = B² + B² - (2B²)(cos A)

A² = 2B² - 2B²(cos 43.6)

A² = 0.55B²

B² = A²/0.55

B² = 65.3/0.55

B² = 118.73

B = √(118.73)

B = 10.9 N

Thus, the tension in the rope B is determined as 10.9 N.

Learn more about tension here: brainly.com/question/24994188

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3 0
2 years ago
Due to the tilt of the Earth's axis,
Lynna [10]
6 months with no sunrise and the other 6 without a sunset, at some places on Earth, are the result of the orientation of Earth's axis.
3 0
3 years ago
What are stepdown transformers used for
lesya692 [45]

Answer:

Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.

Explanation:

pls mark me as brainlist

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7 0
3 years ago
A 2-kg toy car accelerates from 0 to 5 m/s2. It
yan [13]
10 joules of work is done by the object
8 0
3 years ago
The output voltage of a voltage amplifier has been found to decrease by 20% when a load resistance of 1 k is connected. What is
yKpoI14uk [10]

We use the voltage division problem  between the load resistance, amplifier output resistance as

V_{out} = V_{amlifire} \times \frac{R_{load} }{R_{load} + R_{out} }.

Here, V_{out} is the output voltage, V_{amlifire} is the amplifier voltage, R_{load} is the load resistance and R_{out} is the amplifier output resistance.

Therefore,

1-\frac{20}{100} = \frac{1 \ k\Omega }{1 \ k\Omega +R_{out} } \\\\ R_{out} = \frac{1 \ k\Omega }{0.8} -1 \ k\Omega =1250 \Omega -1000 \Omega =250 \Omega.

Thus, the amplifier output resistance is 250 \ \Omega.

4 0
3 years ago
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