A young man heaves a 6.5 kg rock at a velocity of 6.9 m/s. What is the kinetic energy of the rock?

If we double the mass of the sun, what would happen to the gravitational force between the sun and earth?

Sphinxa [80]

It would increase
because the force is directly proportional to the Value of masses given

4 0

3 years ago

An input for of 80 N is used to lift an object weighing 240 N with a system of pulleys. How far must the rope around the pulleys

sammy [17]

Answer:

4.2 m

Explanation:

Note: If energy is conserved, i.e no work is done against friction

Work input = work output.

Work output = Force output × distance,

Work input = force input × distance moved moved.

Therefore,

input force×distance moved = output force × distance moved........................Equation 1

Given: input force = 80 N, output force = 240 N, output distance = 1.4 m

Let input distance = d

Substitute into equation 1

80×d = 240×1.4

80d = 336

d = 336/80

d = 4.2 m.

Thus the rope around the pulley must be pulled 4.2 m

7 0

3 years ago

The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

polet [3.4K]

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

$Power = \frac{Energy}{time}$

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, $s = ut + 0.5at^{2}$

The drill head starts from rest, u = 0 m/s

$5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s$

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

6 0

3 years ago

Determine

djyliett [7]

(i) The total capacitance for the circuit is 5 μF.

(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.

(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.

<h3>Total capacitance of the circuit</h3>

The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.

<h3>C1 and C2 are in series </h3>

$\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F$

<h3>C1 and C2 are parallel to C3</h3>

$C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F$

<h3>C(123) is series to C5 and C6</h3>

$\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F$

<h3>C7 and C8 are in series</h3>

$\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F$

<h3>Total capaciatnce of the circuit</h3>

Ct + C(78) = 2 μF + 3 μF = 5 μF

<h3 /><h3>Total charge stored in the circuit</h3>

The total charge stored in the capacitor is calculated as follows;

Q = CV

Q = (5 x 10⁻⁶) x (20)

Q = 1 x 10⁻⁴ C

<h3>Charge stored in 3μF capacitor</h3>

Q = (3 x 10⁻⁶) x (20)

Q = 6 x 10⁻⁶ C

Learn more about capacitance of capacitor here: brainly.com/question/13578522

8 0

2 years ago

A ball is dropped from a cliff. determine how far the ball fell after 7.5 seconds

grin007 [14]

Answer:

The ball fell 275.625 meters after 7.5 seconds

Explanation:

If an object is left on free air (no friction), it describes an accelerated motion in the vertical direction, powered exclusively by the acceleration of gravity. The formulas needed to compute the different magnitudes involved are

$V_f=gt$

$\displaystyle y=\frac{gt^2}{2}$

Where $V_f$ is the final speed of the object in free fall, assumed positive downwards, t is the time elapsed since the release and y is the vertical distance traveled by the object

The ball was dropped from a cliff. We need to calculate the vertical distance the ball went down in t=7.5 seconds. We'll use the formula

$\displaystyle y=\frac{gt^2}{2}$

$\displaystyle y=\frac{(9.8)(7.5)^2}{2}$

$Y=275.625\ m$

5 0

3 years ago