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attashe74 [19]
2 years ago
10

HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

Physics
1 answer:
bagirrra123 [75]2 years ago
6 0

Answer:

C

Explanation:

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cupoosta [38]
Will this one work?...................

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The player that possess the best ball handling skills and usually handles the ball often
velikii [3]

The player that possess the best ball handling skills and usually handles

the ball often is referred to as the point guard.

Point guards in basketball are referred to as the players who have certain

characteristics such as speed ,an excellent ball handling skills and

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4 0
3 years ago
A 30 kg student drops down from the monkey bars. The acceleration due to gravity is -9.8 m/s^2. Neglecting air drag, what is the
Sholpan [36]

The net force on the student is A) -294 N

Explanation:

Neglecting air resistance, there is only one force acting on the student: the force of gravity, which is given by

F=mg

where

m is the mass of the student

g is the acceleration of gravity

In this problem, we have:

m = 30 kg is the mass of the student

g=-9.8 m/s^2 is the acceleration of gravity, where the negative sign means the direction is downward

Substituting, we find the force of gravity on the student:

F=(30)(-9.8)=-294 N

And since this is the only force acting on the student, it is also the net force on him.

Learn more about gravitational force:

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8 0
3 years ago
A 244.0 N block is at rest on a flat, frictionless table. A hooked cable applies an upward force of 24.0 N on the block. What is
blagie [28]

Answer:

268N

Explanation:

The upward force acting on the block are the reaction and the hooked table..

The total normal force acting = normal reaction + 24N

Note that the normal reaction is always equal the weight of the table

Hence the normal force acting in the block is 244.0+24 = 268.0N

4 0
3 years ago
A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.
cupoosta [38]

Answer:

a) The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

Explanation:

a) The vectorial equation of momentum is represented by the following expression:

\vec p = m\cdot \vec v (1)

Where:

\vec p - Vector momentum, measured in kilogram-meters per second.

m - Mass of the particle, measured in kilograms.

\vec v - Vector velocity, measured in meters per second.

If we know that m = 2.93\,kg and \vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right], then the momentum is:

\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]

\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]

The x and y components of the momentum are 8.731\,\frac{kg\cdot m}{s} and -11.661\,\frac{kg\cdot m}{s}, respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}} (2)

\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right) (3)

Where:

\|\vec p\| - Magnitude of momentum, measured in kilogram-meters per second.

\theta - Direction of momentum, measured in sexagesimal degrees.

If we know that p_{x} = 8.731\,\frac{kg\cdot m}{s} and p_{y} = -11.661\,\frac{kg\cdot m}{s}, then the magnitude and direction of momentum are, respectively:

\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}

\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}

\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)

\theta \approx 306.823^{\circ}

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

6 0
3 years ago
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