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SashulF [63]
3 years ago
14

What is the average velocity of the time interval 0 to 4 seconds

Physics
1 answer:
Anvisha [2.4K]3 years ago
7 0
Average velocity over a given time interval is the distance traveled divided by the time:
v= \frac{60m-20m}{4s-0s} =40/4=10 \frac{m}{s} 
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Please help!!30 points and will give the brainliest answer for a clear explanation.I need it really fast. Thank you!
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I would rather be hit by the deflated ball because it wouldn't hurt as bad because it wouldn't have a lot of weight to hurt me in anyway
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3 years ago
Kissa and Lily ran a race in gym class Lily was off balance so she did not start until 2 seconds had passed
Paul [167]
Average speed = (distance covered) / (time to cover the distance)

Tissa covered 60 meters in 10 seconds. Her average speed was

(60 m) / (10 sec) = 6 m/s.

That's the slope of the dotted line.

Lilly covered 60 meters in 8 seconds. Her average speed was

(60 m) / (8 sec) = 7.5 m/s .

That's the slope of the solid line.

Lilly covered the same distance in less time, and both girls
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3 0
2 years ago
How much current is in a circuit that includes a 9-volt battery with a resistance of 3
Darya [45]
Just remember
Voltage = current times resistance

current = voltage over resistance

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7 0
3 years ago
During nuclear decay, a new isotope is created. How is the nucleus of the new isotope different from the parent if the parent is
STALIN [3.7K]

Answer:

Alpha decay will produce a daughter nucleus with more protons and beta decay will produce a daughter nucleus with fewer protons than the parent nucleus has.

3 0
3 years ago
Read 2 more answers
A boy throws a ball with an initial velocity of 19.6 m/s. What maximum height does the ball reach?
Wewaii [24]
<h2>Hello!</h2>

The answer is: 19.59 m

<h2>Why?</h2>

Since there is no information about the launch type, we can assume that the ball is thrown vertically upward.

When the ball reaches the maximum height, just at that moment, the velocity turns to 0, and after that moment, the ball starts falling, so:

We will use the following formula:

Vf^2=Vi^2+2*g*s

Where:

Vf= Final velocity = 0

Vi= Initial velocity = \frac{19.6m}{s}

g = Gravity Acceleration = \frac{9.81m}{s^{2} }

s = Traveled distance

0=19.6^2+2*-9.81*s\\s=\frac{19.6^2}{2*9.81}=\frac{384.16}{19.62}=19.59m

Have a nice day!

8 0
3 years ago
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