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kotegsom [21]
3 years ago
9

A ball thrown vertically upward is caught by the thrower after 2.93 s. Find the initial velocity of the ball. The acceleration o

f gravity is 9.8 m/s 2 . Answer in units of m/s.
Physics
1 answer:
almond37 [142]3 years ago
7 0

Answer:

The initial velocity of the ball is 28.714 m/s

Explanation:

Given;

time of flight of the ball, t = 2.93 s

acceleration due to gravity, g = 9.8 m/s²

initial velocity of the ball, u = ?

The initial velocity of the ball is given by;

v = u + (-g)t

where;

v is the final speed of the ball at the given time, = 0

g is negative because of upward motion

0 = u -gt

u = gt

u = (9.8 x 2.93)

u = 28.714 m/s

Therefore, the initial velocity of the ball is 28.714 m/s

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Answer:

ion is an electrically charged particle

Explanation:

there are 2 types of ions. anion an cation anion is negatively charged

cation is positively charged

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How does the matter and (solar) energy move through the sphere?
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3 years ago
How much force was applied to an object if was moved 2 meters and the work done on the object was 8 joules?
serious [3.7K]

Answer:

The answer is 4N or B

Explanation:

Just the equation W = F x D.

We have W = 8 J and D = 2 m using algebra ....

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3 years ago
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Signment
jeka94

Answer:

v = 5.15 m/s

Explanation:

At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N

As the cable tension is less than this value, the car must be accelerating downward.

7730 = 984(9.81 - a)

a = 1.95 m/s²

kinematic equations s = ut + ½at² and v = u + at

-5.00 = u(4.00) + ½(-1.95)4.00²

u = 2.65 m/s    the car's initial velocity was upward at 2.65 m/s

v = 2.65 + (-1.95)(4.00)

v = -5.15 m/s

3 0
3 years ago
A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo
WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

acceleration=\frac{30}{5}=6m/s^{2}

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0
4 years ago
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