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Svetradugi [14.3K]

2 years ago

8

In a house, 3 bulbs of 60 watt each are lighted for 3 hours daily, 4 fans of 100 watt each are used for 8 hours daily and an ele

ctric heater of 2 kwh used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 4/kwh.

1 answer:

Virty [35]2 years ago

5 0

Answer:

Total Energy consumed = 146.94 kwh

Total cost = rs 587.76

Explanation:

Given:

3 bulbs each 60 watt [for 3 hours daily]

4 fans each 100 watt [for 8 hours daily

1 electric heater of 2 kWh (2,000 watt) [for half hour daily ]

Rate = rs 4/Kwh

Find:

Total Energy consumed

Total cost

Computation:

Energy used = Power × time

3 bulbs each 60 watt [for 3 hours daily] = 3 x 60 x 3 x 31 = 16,740 watt = 16.74 kw

4 fans each 100 watt [for 8 hours daily = 4 x 100 x 8 x 31 = 99,200 watt = 99.2 kw

1 electric heater of 2 kWh [for half hour daily ] = 1 x 2000 x (1/2) x 31 = 31,000 watt = 31 kw

Total Energy consumed = 16.74 + 99.2 + 31

Total Energy consumed = 146.94 kwh

Total cost = 146.94 x 4

Total cost = rs 587.76

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A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe

KIM [24]

Answer: V = 15 m/s

Explanation:

As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately.

8 0

2 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago

A 980 kg roller coaster cart is traveling along a track at 17 m/s before it rolls down a 30 m tall hill (Point A). What will be

MrRissso [65]

The kinetic energy halfway the hill is $2.86\cdot 10^5 J$

Explanation:

If there are no friction forces acting on the cart, we can apply the law of conservation of energy: the mechanical energy of the cart (which is the sum of potential energy + kinetic energy) must be conserved. So we can write:

$U_A +K_A = U_B + K_B$

where

$U_A=mgh_A$ is the initial potential energy, at point A, with

m = 980 kg (mass of the cart)

$g=9.8 m/s^2$ (acceleration of gravity)

$h_A = 30 m$ (height at point A)

$K_A=\frac{1}{2}mv_A^2$ is the initial kinetic energy, at point A , with

$v_A=17 m/s$ (velocity at point A)

$U_B=mgh_B$ is the final potential energy, at point B, where

$h_B = 15 m$ (height at point B)

$K_B=\frac{1}{2}mv_B^2$ is the final kinetic energy, at point B, where

$v_B$ is the velocity at point B

Here we are interested in finding $K_B$ , so by re-arranging the equation and substituting we find:

$K_B = U_A+K_B-U_B = mg(h_A-h_B)+\frac{1}{2}mv_A^2=(980)(9.8)(30-15)+\frac{1}{2}(980)(17)^2=2.86\cdot 10^5 J$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

8 0

2 years ago

Two children are throwing a ball back-and-forth straight across the back seat of a car. The ball is being thrown 7 mph relative

Evgesh-ka [11]

Answer:

the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Explanation:

Given the data in the question and as illustrated in the diagram below;

Now, Lets assume line AB represent the movement of the car,

AC is the movement of the ball been thrown back and forth in the back seat

Ax is the motion of the ball it flies off the window

so from the diagram, We can see triangle ABC

where AB is 45 mph and AC = 7 mph

and angle ∠CAB = 90°

using SOH CAH TOA

TOA; tanθ = Opposite / Adjacent

tanθ = Opposite / Adjacent

tan( ∠ ABC ) = AC / AB

we substitute

tan( ∠ ABC ) = 7 / 45

tan( ∠ ABC ) = 0.15555

( ∠ ABC ) = tan⁻¹ 0.15555

( ∠ ABC ) = 8.84°

Therefor, angle ( ∠ ABC ) is 8.84°

Meaning angle ( ∠ XAA' ) is also 8.84°

Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car

4 0

2 years ago

(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance

Komok [63]

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity F=W= mg [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

=132 kg ×9.8 m/s^2

=1293.6 kg m/s^2

=1293.6 N [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

$=\frac{1}{4} *1293.6N$

$=323.4 N$

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

$F_{net} =1293.6N -323.4N$

$=970.2 N$

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e

$F_{net} =ma$ [Here a is the acceleration]

$a =\frac{F_{net} }{m}$

$= \frac{970.2}{132} m/s^2$

$=7.35 m/s^2$

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

we know that F= ma

=m×0

=0 N

Hence they will not get any force when they will descend with a uniform speed.

4 0

2 years ago