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Valentin [98]
3 years ago
7

The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.What power is absorbed by the bulb?

Physics
1 answer:
Verizon [17]3 years ago
3 0

Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

Power = Voltage × Current

Given:

Voltage = 3 V

Current = 200 mA = 0.200 A

Conversion used : (1 mA = 0.001 A)

Now put all the given values in the above formula, we get:

Power = Voltage × Current

Power = 3V × 0.200 A

Power = 0.600 W

Thus, the power absorbed by the bulb is, 0.600 W

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Robert dropped his new iPhone from his balcony. It hit the ground 3.5 seconds later. What was the height of his balcony?
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its B. 60 meters

Explanation:

cause I looked up a calculator and solved it

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3 years ago
Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea
motikmotik

Explanation:

Work done by gravity is given by the formula,

           W = \rho A (h_{1} - h)g (h - h_{2}) ......... (1)

It is known that when levels are same then height of the liquid is as follows.

           h = \frac{h_{1} + h_{2}}{2} ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

       W = \frac{1}{4} \rho gA(h_{1} - h_{2})^{2})

           = \frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})

           = 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

3 0
3 years ago
A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its
Savatey [412]

Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

c). ΔE_{t}=8.4635 J

Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}

c).

net work= EkA+EkB

E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J

3 0
3 years ago
A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a
stira [4]

Answer:

Explanation:

Given

acceleration is given by

a=-g-c_Dv^2

where \ddot{y}=a

\dot{y}=v

Also acceleration is given by

a=v\frac{\mathrm{d} v}{\mathrm{d} s}

ds=\frac{v}{a}dv

\int ds=\int \frac{v}{-g-0.001v^2}dv

\Rightarrow Let -g-0.001v^2=t

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at\ v_0=50\ m/s,\ t=-g-0.001(50)^2

t=-g-2.5

at v=0,\ t=-g

\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}

\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}

s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}

s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})

s=113.608\ m

when air drag is neglected maximum height reached is

h=\frac{v_0^2}{2g}

h=\frac{50^2}{2\times 9.8}

h=127.55\ m

3 0
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Sound is an example of a ______ wave.
tester [92]
Sound is a longitudinal wave.
7 0
3 years ago
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