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Alex777 [14]
2 years ago
11

A block slides down an inclined plane from rest. Initially the block is at 4.5m above the ground. Find the speed of the block wh

en it is 1.5m above the ground. 1) 7.7m/s 2) 9.4m/s 3) 5.4m/s 4) 3.2m/s
Physics
1 answer:
WINSTONCH [101]2 years ago
7 0

Since, no external force is acting , so the system is in equilibrium .

Initial total energy = Final total energy

mg(4.5) = mg(1.5) + \dfrac{mv^2}{2}\\\\\dfrac{v^2}{2}=3\times g \\\\v^2=3\times 9.8\times 2\\\\v = \sqrt{58.8}\ m/s\\\\v = 7.67 \ m/s ( Here , g = acceleration due to gravity = 9.8 m/s² )

Therefore, option 1) is correct.

Hence, this is the required solution.

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A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d
prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance  

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

V_{t}= V_{o}e^{\frac{-t}{R*C} }

Voltage in this case is the energy dissipated so

E_{t}= E_{o}e^{\frac{-t}{R*C} }

\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }

\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }

Using the equation to find capacitance

ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }

C= 5.32x10^{-6} F

C= 5.32 uF because u is the symbol for micro that is equal to 10^{-6}

8 0
3 years ago
sonic is sliding down a frictionless 15m tall hill. He starts at the top with a velocity of 10m/s. At the bottom of the hill he
podryga [215]

Answer:

The maximum speed of sonic at the bottom of the hill is equal to 19.85m/s and the spring constant of the spring is equal to (497.4xmass of sonic) N/m

Energy approach has been used to sole the problem.

The points of interest for the analysis of the problem are point 1 the top of the hill and point 2 the bottom of the hill just before hitting the spring

The maximum velocity of sonic is independent of the his mass or the geometry. It is only depends on the vertical distance involved

Explanation:

The step by step solution to the problem can be found in the attachment below. The principle of energy conservation has been applied to solve the problem. This means that if energy disappears in one form it will appear in another.

As in this problem, the potential and kinetic energy at the top of the hill were converted to only kinetic energy at the bottom of the hill. This kinetic energy too got converted into elastic potential energy .

x = compression of the spring = 0.89

5 0
3 years ago
Which form of the law of conservation of energy describes the motion of the block as it slides on the floor from the bottom of t
aivan3 [116]

Answer:

Explanation:

Let assume begins movement at zero point, that is, height is equal to zero. The block has an initial linear kinetic energy and no gravitational potential energy and end with no linear kinetic energy, some gravitational potential energy and work losses due to slide friction. In mathematical terms, this system can be model as follows:

K_{1} = U_{2} + W_{loss, 1 \longrightarrow 2}

Where K, U, W are linear kinetic energy, gravitational potential energy and work, respectively.

8 0
3 years ago
19. State any 3 applications of capillary action (3mk)​
zlopas [31]

Explanation:

1. Movement of water, food and mineral salts in plants

2. Absorption of water by towels when wiping our bodies

3. It is used to absorb ink using a blotting paper or tissue

7 0
3 years ago
A 0.245 kg ball is thrown straight up from 2.07 m above the ground. Its initial vertical speed is 8.00 m/s. A short time later,
iris [78.8K]

Answer:

The work done by gravity is 4.975 \: Joules

Explanation:

The data given in the question is :

Mass is 0.245 kg

Height from ground is 2.07 m

As we know , the work done is state function , it depends on initial and final position not on the path followed.

So, work done by gravity = change in potential energy

Work done = Initial potential energy - final potential energy

Insert values from question

Work done = mass \times gravity \times (change \: in \: height)

Work done = 0.245 kg \times 9.81 m/s^{2} \times 2.07 m

So, work done = 4.975 Joules

Hence the work done by gravity is 4.975 \: Joules

3 0
3 years ago
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