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2 years ago

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A heater marked 2000W/3000W is switched on for 4 hours. For the first hour it is on the highest setting and for the last 3 hours

it is on the lowest setting
1.1 How much energy does it transfer to the room in kWh (9kWh)

1 answer:

Leni [432]2 years ago

3 0

Answer:

9 kW-hr

Explanation:

3000 W = 3 kW for 1 hr = <u>3 kW - hr </u>

2000 W = 2 kW for 3 hr = <u> 6 kW -hr</u>

<u />

total power = <u> 3 + 6 = 9 kW-hr</u>

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Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?

Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

$A_y=+15 m$

$\theta=32^{\circ}$

We have to find the x-component of A and magnitude of A.

According to question

$A_y=\mid A\mid cos\theta$

Substitute the values then we get

$15=\mid A\mid cos32$

$\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}$

$\mid A\mid=17.7$ m

$tan\theta=\frac{perpendicular\;side}{Base}$

$tan32=\frac{A_x}{A_y}=\frac{A_x}{15}$

$0.62\times 15=A_x$

$A_x=9.3$

The value of x-component of A is negative because the vector A lie in second quadrant.

Hence, the x- component of A=-9.3 m

6 0

3 years ago

Read 2 more answers

The radius of earth is 6,370,000 m. Express this measurement in km in scientific notation with the correct number of significant

AlexFokin [52]

Answer:

6.37 x 10³ Km

Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m = 6,370,000 x 0.001

= 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

= 6.37 x 10³ Km

7 0

3 years ago

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that osc

marin [14]

Answer:

Time period of the osculation will be 0.0671 sec

Explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$

4 0

3 years ago

Read each description below. Choose the force diagram (free-body diagram) that best represents the description. You may neglect

Papessa [141]

A. Diagram A
B. Diagram C & D
C. Diagram B
D. Diagram C & D
E. Diagram B
F. Diagram C & D
These are simplified representations of an object's body and the force vectors acting on it. Some of the main forces that are involve are normal force, friction, push or pull and gravity.

5 0

3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

$a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2$

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

$a'=6.9 m/s^2$

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

$T-m'g=m'a'$

where

$m'$ is the new mass of the rocket

Re-arranging the equation and solving for m', we find

$m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg$

And since the initial mass of the rocket was

$m=1.9 \cdot 10^4 kg$

This means that the mass of fuel burned is

$\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg$

3 0

3 years ago