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3 years ago

what is the cost to run a single 100 w light bulb for 30 days if the electric company charges $0.18 for every KW-hr?

Physics

1 answer:

xxTIMURxx [149]3 years ago

8 0

Explanation:

power of the bulb = 100w=0.1kw

power consumption when the bulb is used for 30 days ( 30 *24 = 720 hours) =0.1*720=72kwh

cost to run the bulb for 30 days = 72*0.18

= $12.96

Answer is $12.96..................

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Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c

jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

$=-17850 - 30018.6$

$=-47868.6 \ Pa$

$=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

4 0

2 years ago

Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e

Grace [21]

Answer:

The duration is $T =72 \ years /tex]Explanation:From the question we are told that The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m$

Generally the time it would take for the message to get the the other civilization is mathematically represented as

$t = \frac{D}{c}$

Here c is the speed of light with the value $c = 3.08 *10^{8} \ m/s$

=> $t = \frac{3.311 *10^{17} }{3.08 *10^{8}}$

=> $t = 1.075 *10^9 \ s$

converting to years

$t = 1.075 *10^9 * 3.17 *10^{-8}$

$t = 34 \ years$

Now the total time taken is mathematically represented as

$T = 2* t + 2 + 2$

=> $T = 2* 34 + 2 + 2$

=> [tex]T =72 \ years /tex]

4 0

3 years ago

SP: Calculate the moment

ipn [44]

Answer:

Moment of the force is 20 N-m.

Explanation:

Given:

Force exerted by the person is, $F=80\ N$

Distance of application of force from the point about which moment is needed is, $d=25\ cm=\frac{25}{100}\ m=0.25\ m$

Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.

Therefore, the moment of the force about the end of the claw hammer is given as:

$M=F\times d\\\\M=(80\ N)(0.25\ m)\\\\M=20\textrm{ N-m}$

Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.

6 0

3 years ago

It takes at least 4 people to push a refrigerator up a 10-meter ramp. How many people would be needed to push the same refrigera

vaieri [72.5K]

Double the amount of people because the ramp went from 10-meter to 20-meter, so times by 2. 4 times 2 is 8.
The answer is 8 people.
Hope this helps!
Please give Brainliest!

8 0

1 year ago

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo

timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

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

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0

3 years ago

what is the cost to run a single 100 w light bulb for 30 days if the electric company charges $0.18 for every KW-hr?​

what is the cost to run a single 100 w light bulb for 30 days if the electric company charges $0.18 for every KW-hr?