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kykrilka [37]
3 years ago
10

A lamp draws a current of 0.1 A when it is connected to a 122-V source. (

Physics
1 answer:
Setler79 [48]3 years ago
4 0
The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W
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A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin
yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19 km

tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2

\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees north or west or (180 - 72.65) = 107.35 degrees north of east

3 0
3 years ago
Question number 8 <br> Plz help
Dennis_Churaev [7]
To me, that sounds like the "Law of Conservation of Energy".
5 0
3 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
What was your electric potential relative to a metal pipe if a spark jumped 1.1 cm through dry air from your finger to the pipe?
romanna [79]
The electric field is given by volts/distance: E= \frac{V}{d}.  The breakdown voltage of dry air is about 3x10^6V/m.  So solving for V we get
V=Ed
or V=(3e6V/m)(0.011m)=33,000V
6 0
3 years ago
Read 2 more answers
PART ONE
Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

4 0
3 years ago
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