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mel-nik [20]
3 years ago
12

Calculate the power of a 120V drill that uses 15A of electricity.

Physics
1 answer:
pishuonlain [190]3 years ago
8 0

Answer:

1800W

Explanation:

Power=Current×Potential difference

=15A×120V

=1800W

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Which is not true about the pH scale?
pochemuha

Answer:

The pH scale was invented by Svante Arrhenius is false.

Explanation :

because the pH scale was invented by Soren Sorensen

4 0
3 years ago
Read 2 more answers
When are all the forces acting upon an object balanced?
Rudik [331]

Answer:

When two forces acting on an object are equal in size but act in opposite directions, we say that they are balanced forces.

Explanation:

7 0
3 years ago
A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?
iren2701 [21]

Answer:

K.E=29.403125J

Explanation:

From the question we are told that

Mass M=100

Height 50-20=30m

Generally the equation for velocity before impact is is is mathematically given by

v=\sqrt{2gh}

v=\sqrt{2*9.8*30}

v=24.25

Generally the equation for Kinetic Energy is is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*100*(24.25)^2\\

K.E=29403.125J

K.E=29.403125J

8 0
3 years ago
The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee
lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is -27.3\hat{i}

y component of vector is 43.6\hat{j}

so position vector is

r=-27.3\hat{i}+43.6\hat{j}

Magnitude of vector is

|r|=\sqrt{27.3^2+43.6^2}

|r|=\sqrt{2646.25}

|r|=51.44 units

Direction

tan\theta =\frac{43.6}{-27.3}=-1.597

vector is in 2nd quadrant thus

180-\theta =57.94

\theta =122.06^{\circ}

4 0
3 years ago
A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
3 years ago
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