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2 years ago

If you throw a ball down ward then acceleration immeditely after leaving your hand is

Physics

1 answer:

bezimeni [28]2 years ago

7 0

Answer:

9.8m/s²

Explanation:

The acceleration of the ball thrown after leaving my hand is 9.8m/s². This will be the acceleration due to gravity on the body.

Acceleration due to gravity is caused by the pull of the earth on a massive object.
The value of this acceleration is 9.8m/s².
As the ball nears the surface, it comes near zero.

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What is the wavelength of an earthquake wave if it has a speed of 12 km/s and a frequency of 15 Hz

professor190 [17]

First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency

3 0

3 years ago

67 points plus brainlest if done correctly.I will report you if you answer 3 or less of the questions, also must post all the an

Annette [7]

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3 0

3 years ago

The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ahrayia [7]

<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> $D$ :

$D={C}_{d}\frac{\rho V^{2} }{2}A$ (1)

Where:

$C_ {d}$ is the drag coefficient

$\rho$ is the density of the fluid (air for example)

$V$ is the velocity

$A$ is the transversal area of the object

So, this force is proportional to the transversal area of the falling element and to the square of the velocity.

2. Its <u>weight </u>due to the gravity force $W$ :

$W=m.g$

(2)

Where:

$m$ is the mass of the object

$g$ is the acceleration due gravity

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

$D=W$ (3)

${C}_{d}\frac{\rho V^{2} }{2}A=m.g$ (4)

$V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}$ (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0

2 years ago

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent

tensa zangetsu [6.8K]

Answer:

(a): The car's relative position to the base of the cliff is x= 32.52m.

(b): The lenght of the car in the ir is tfall= 1.78 sec.

Explanation:

Vo= 0

V= ?

d= 50m

h= 30m

a= 4 m/s²

t= √(2*d/a)

t= 5 sec

V= a*t

V= 20 m/s

Vx= V * cos(24º)

Vx= 18.27 m/s

Vy= V* sin(24º)

Vy= 8.13 m/s

h= Vy*t + g*t²/2

clearing t:

tfall= 1.78 sec (b)

x= Vx * tfall

x= 32.52 m (a)

4 0

3 years ago

The pressure inside a sealed container of methane gas (CH4) is 35.0 kPa. If this 80.0 L sample

STatiana [176]

Answer:

<h3>The answer is 30.43 L</h3>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{35000 \times 80}{92000} = \frac{2800000}{92000} = \frac{700}{23} \\ = 30.434782...$

We have the final answer as

Hope this helps you

4 0

2 years ago