PHYSICS COLLEGE

A ball having a mass of 500 grams is dropped from a height of 9.00 meters what is its kinetic energy when it hits the ground

Answers

Answer 1

Answer:

Answer:

where m = mass, g = acceleration due to gravity (9.8 m/s^2), h = height

Given m = 500g = 0.5 kg, h = 9 meters

0.5*9*9.8 = 44.1 joules

Explanation:

Answer 2

Answer:

Answer:44.1

Explanation:

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A battery lighting a bulb is an example of _ energy converting to _ energy.

A wire with mass 90.0 g is stretched so that its ends are tied down at points 98.0 cm apart. The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude of 0.300 cm at the antinodes. Part A What is the speed of propagation of transverse waves in the wire

Answers

Answer:

118 m/s

Explanation:

Given :

We know that

$f\ =\ (1)/(2l) \sqrt{(T)/(U) }$ ......Eq(1)

Where $\sqrt{(T)/(u) }$ =v

l=length

f=frequency

l= 98.0 cm= 0.98 m

f=60.0 Hz

Now from the Eq(1)

$f\ =\ (v)/(2l)$

This equation can be written as

v=2fl.............Eq(2)

Putting the value f and l in Eq(2)

v=2*60*0.98

v=117.6 m/s ~ 118 m/s

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = (1)/(12)ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= (1)/(12)ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$((1)/(12)ML^2 +2mr_1^2) \omega_i = ((1)/(12)ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s$

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Suppose you left a 100-W light bulb on continuously for one month. If the electricity generation and transmission efficiency is 30%, how much chemical energy (in joules) was wasted at the power plant for this oversight? If the fuel consumption for one meal in Cambodia using a kerosene wick stove is 6 MJ (1 MJ = 1,000,000 joules), how many equivalent meals could be cooked with this wasted energy.

Answers

The wasted chemical energy be "8.64 × 10⁸ J" and the equivalent meals could be cooked be "144".

Chemical energy

According to the question,

Bulb power, P = 100 W

Time, t = 1 month or,

= 1 × 30 × 24

= 720 h

Efficiency, η = 30% or,

= 0.30

Fuel consumption, E = 6 MJ or,

= 6 × 10⁶ J

Energy consumed be:

→ $E_c$ = P × t

By substituting the values,

= 100 × 720

= 72 kWh

Wasted energy be:

→ $E_g$ = $(E_c)/(\eta)$

= $(72000)/(0.3)$

= 240 kWh or,

= 240 × 3.6 × 10⁶

= 8.64 × 10⁸ J

and,

The no. of meals be:

→ N = $(8.64* 10^8)/(6* 10^6)$

= 144 meals

Thus the answers above are correct.

Find out more information about chemical energy here:

brainly.com/question/347340

Answer:

$E_g = 240 \ kWh$

$N = 144 \ meals$

Explanation:

From the question we are told that

The power rating of the bulb is P = 100 W

The duration is t = 1 month = 1 * 30 * 24 = 720 h

The efficiency is $\eta = 30\% = 0.30$

The fuel consumption for one meal is $E = 6 MJ = 6 *10^6 J$

Generally the energy consumed by the bulb is mathematically represented as

$E_c = P * t$

=> $E_c = 100 * 720$

=> $E_c = 72\ k Wh$

Generally the energy generated at the power plant that was wasted by the bulb is mathematically represented as

$E_g = (E_c)/(\eta)$

=> $E_g = (72000)/(0.3)$

=> $E_g = 240 \ kWh$

Converting this value to Joules

$E_g = 240 * 3.6 * 10^(6) = 8.64*10^8$

Generally the number of means that would be cooked is

$N = (8.64*10^8 )/(6 *10^6)$

=> $N = 144 \ meals$

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

Answers

Given :

∅ = 60⁰

u = 4 m/s

g = 10m/s²

to find :

T = ?

Solution :

as per formula,

$t = (2u \: sin \theta)/(g)$

now put the value : $t \: = (2 * 4 * sin \: 60)/(10)$

as we know $sin60 \: = ( √(3) )/(2)$

therefore,

$t \: = (8 * ( √(3) )/(2) )/(10)$

as we solve this we get,

$t \: = ( 2√(3) )/(5)$

that's t = 0.69 sec

$\sf\fbox\red{\:I \:hope \:it's \:helpful \:to \:you}$

0.8 seconds

Explanation:

time of flight = 2u/g

u=4m/s

g=10

= 8/10

= 0.8 sec

just a trial...not sure!!!

You spray your sister with water from a garden hose. The water is supplied to the hose at a rate of 0.111 liters per second and the diameter of the nozzle you hold is 5.79 mm. At what speed does the water exit the nozzle

Answers

Answer:

29.5 m/s

Explanation:

Volumetric flowrate = (average velocity of flow) × (cross sectional area)

Volumetric flowrate = 0.111 liters/s = 0.000111 m³/s

Cross sectional Area of flow = πr²

Diameter = 0.00579 m,

Radius, r = d/2 = 0.002895 m

A = π(0.002895)² = 0.0000037629 m²

Velocity of flow = (volumetric flow rate)/(cross sectional Area of flow)

v = 0.000111/0.0000037629

v = 29.5 m/s

Given Information:

diameter of the nozzle = d = 5.79 mm = 0.00579 m

flow rate = 0.111 liters/sec

Required Information:

Velocity = v = ?

Answer:

Velocity = 4.21 m/s

Explanation:

As we know flow rate is given by

Flow rate = Velocity*Area of nozzle

Where

Area of nozzle = πr²

where

r = d/2

r = 0.00579/2

r = 0.002895 m

Area of nozzle = πr²

Area of nozzle = π(0.002895)²

Area of nozzle = 2.6329x10⁻⁵ m²

Velocity = Flow rate/area of nozzle

Divide the litters/s by 1000 to convert into m³/s

0.111/1000 = 1.11x10⁻⁴ m³/s

Velocity = 1.11x10⁻⁴/2.6329x10⁻⁵

Velocity = 4.21 m/s

Therefore, the water exit the nozzle at a speed of 4.21 m/s

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