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Valentin [98]
3 years ago
8

How is velocity determined from a distance over time graph

Physics
1 answer:
Elden [556K]3 years ago
7 0

Answer:

The slope of a distance vs time graph represents velocity.

Explanation:

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A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s
lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0
3 years ago
What is the best estimate of the frequency of the wave shown below
GrogVix [38]

Answer:

D. 2.5 Hz

Explanation:

Frequency = speed of wave / wavelength

= 335 /140 ( from graph)

= 2.4

5 0
3 years ago
Which memory disorder is a common illness on television and films
Alenkinab [10]
amnesia is the most common illness used in tv an films
5 0
3 years ago
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Someone plzzz helpppppp with this last question
vredina [299]

Answer:

I dont know someone deleted answers. But they were wrong. INERTIA IS CORRECT I DID THIS IN MY SCHOOL

C IS CORRECT

7 0
3 years ago
A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around
Alex_Xolod [135]

Answer:

\alpha =\frac{m*g*R}{I-m*R^2}

a = \frac{m*g*R^2}{I-m*R^2}

T=\frac{I*m*g}{I-m*R^2}

Explanation:

By analyzing the torque on the wheel we get:

T*R=I*\alpha    Solving for T:   T=I/R*\alpha

On the object:

T-m*g = -m*a    Replacing our previous value for T:

I/R*\alpha-m*g = -m*a

The relation between angular and linear acceleration is:

a=\alpha*R

So,

I/R*\alpha-m*g = -m*\alpha*R

Solving for α:

\alpha =\frac{R*m*g}{I+m*R^2}

The linear acceleration will be:

a =\frac{R^2*m*g}{I+m*R^2}

And finally, the tension will be:

T =\frac{I*m*g}{I+m*R^2}

These are the values of all the variables: α, a, T

8 0
3 years ago
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