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2 years ago

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Darryl throws a basketball at the gym floor.The ball bounces once on the floor comes to rest in his coach's hands.At which point

are all the forces on the basketball balanced?
a.the moment the ball leaves Darryl's handB.the moment the ball touches the floorC.when the ball is in the airD.when the ball starts falling downward before coming to restE.the moment the ball comes to complete rest

2 answers:

Arada [10]2 years ago

6 0

B. The Moment the ball touches the floor.

pshichka [43]2 years ago

6 0

Answer:

E.the moment the ball comes to complete rest

Explanation:

The position of basket ball is said to be at balanced position when net force on the basket ball is zero and its velocity is not changing with time.

So here when ball starts to fall then the velocity will start increasing so this is not balanced situation.

When ball touches the floor then after that ball rebound in opposite direction so this is also unbalanced situation as velocity is changing here.

When ball is moving in air then due to gravity the speed is continuously changing so it is also unbalanced situation.

So here the correct balanced position is when ball comes to rest in the hand of the man because after that its velocity is not changing.

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Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of ma

tiny-mole [99]

Explanation:

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2 years ago

A TV satellite broadcasts at a frequency of 5000 MHz, (1 MHz = 1 million Hertz). What is the wavelength of this radiation?

Nitella [24]

Answer:

$\lambda=0.06\ m$

Explanation:

Given:

frequency of the broadcast, $f=5000\ MHz=5\times 10^9\ Hz$

we have the speed of the radiation equal to the speed of light, $c=3\times 10^8\ m.s^{-1}$

The broadcast waves are the electromagnetic waves but it can travel only upto a hundred kilometers without any loss of information carried by it.

<u>The relation between the frequency and the wavelength:</u>

$\lambda=\frac{c}{f}$

$\lambda=\frac{3\times 10^8}{5\times 10^9}$

$\lambda=0.06\ m$

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2 years ago

Read 2 more answers

a rocket has a mass 250(10^3) slugs on earth. Specify its mass in si units and its weight in si unites. if the rocket is on the

Katena32 [7]

Answer:

$W_{earth} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 5.91 * 10^6 N$ : Rocket weight on moon

Explanation:

Conceptual analysis

Weight is the force with which a body is attracted due to the action of gravity and is calculated using the following formula:

W = m × g Formula (1)

W: weight

m: mass

g: acceleration due to gravity

The mass of a body on the moon is equal to the mass of a body on the earth

The acceleration due to gravity on a body is different on the moon and on the earth

Equivalences

1 slug = 14.59 kg

Known data

$m_{earth} = m_{moon} = 250 * 10^3 slug = 250* 10^3slug * \frac{14.59kg}{1slug} = 3647.5* 10^3 kg$

$g_{moon}= 1.62 \frac{m}{s^2}$

$g_{earth}= 9.8 \frac{m}{s^2}$

Problem development

To calculate the weight of the rocket on the moon and on earth we replace the data in formula (1):

$W_{earth} = 3647.5* 10^3 kg * 9.8 \frac{m}{s^2} = 35.74 * 10^6 N$ : Rocket weight on earth

$W_{moon} = 3647.5* 10^3 kg * 1.62 \frac{m}{s^2} = 5.91 * 10^6 N$ : Rocket weight on moon

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2 years ago

Burning coal, which is how many power plants generate electricity, releases a number of harmful byproducts. Particulate pollutio

kiruha [24]

Answer:

$7.25\times 10^{-16}\ C$

Explanation:

F = Force = $7.25\times 10^{-11}\ N$

E = Electric field = $1\times 10^5\ N/C$

Force is given by

$F=qE$

$\Rightarrow q=\dfrac{F}{E}$

$\Rightarrow q=\dfrac{7.25\times 10^{-11}}{1\times 10^5}$

$\Rightarrow q=7.25\times 10^{-16}\ C$

The charge on the particle is $7.25\times 10^{-16}\ C$

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2 years ago

At what angle should the roadway on a curve with a 50m radius be banked to allow cars to negotiate the curve at 12 m/s even if t

just olya [345]

Answer:

The banking angle of the road is 16.38 degrees.

Explanation:

Given:

Radius of the roadway on curve, R = 50 m

velocity of the moving car, V = 12 m/s

The banking angle can be calculated by using the formula below:

If there's no frictional force, then net force is equal to centripetal force.

MgTanΦ = (MV^2)/ R

TanΦ = V^2 / gR

Where;

Φ is the banking angle

g is acceleration due to gravity

TanΦ = (12 x 12) / (9.8 x 50)

TanΦ = 0.2939

Φ = arctan (0.2939)

Φ = 16.38 degrees

Therefore, the banking angle of the road is 16.38 degrees.

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