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3 years ago

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The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 mA for 42 continuous hours. During that ti

me the voltage will drop from 1.5 V to 1.0 V. Assume the drop in voltage is linear with time.
How much energy does the battery deliver in this 37h interval?

1 answer:

Sladkaya [172]3 years ago

4 0

Answer:

The manufacturer of a 9V dry-cell flashlight battery says that the battery will deliver 20 mA for 80 continuous hours. During that time the voltage will drop from 9V to 6V. Assume the drop in voltage is linear with time. How much energy does the battery deliver in this 80 h interval?

Explanation:

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Who developed radiometric dating? What age did he assign to the oldest rocks?

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Ernest Rutherford
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3 years ago

a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed

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Mass of the car (m) = 2000 Kg
Initial velocity (u) = 15 m/s
Force (F) = 10000 N
Time (t) = 3 s
Let the acceleration be a.
By using the formula, F = ma, we get,
10000 N = 2000 Kg × a
or, a = 10000 N ÷ 2000 Kg
or, a = 5 m/s^2
Let the final velocity be v.
By using the formula, v = u + at, we get,
v = 15 m/s + 5 m/s^2 × 3 s
or, v = 15 m/s + 15 m/s
or, v = 30 m/s

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<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

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3 years ago

A football player kicks a ball at a 30o angle from the ground with an initial velocity of 15 m/s. What is the final velocity of

navik [9.2K]

Given that,

Angle = 30°

Initial velocity = 15 m/s

We need to calculate the time of flight

Using formula of time of flight

$T=\dfrac{2u\sin\theta}{g}$

Where, u = initial velocity

g = acceleration due to gravity

Put the value into the formula

$T=\dfrac{2\times15\sin30}{9.8}$

$T=1.5\ sec$

We need to calculate the final velocity of the ball

Using equation of motion

$v=u+gt$

$v=15+9.8\times1.5$

$v=29.7\ m/s$

Hence, The final velocity of the ball is 29.7 m/s.

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4 years ago

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

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3 years ago

Which of the following is a good conductor of heat? 1. water 2. glass 3. wood 4. mercury

trapecia [35]

Answer:

So the answer would be water based on the evidence shown below.

Explanation:

Mercury is a poor conductor of heat but good for electricity, water is a good conductor of heat but a poor conductor of electricity, wood is a poor conductor of heat and electricity, and glass is probably the worst conductor of heat.

7 0

3 years ago