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alekssr [168]
3 years ago
10

You have a flashlight that uses 0.75 watts of power and requires 1.5 volt battery.How many joules of energy are used by the flas

hlight in 15 minutes? Hint: Joules are watts and seconds and you are computing the difference between the electric energy at two points A and B.What the total charge q0? Hint: the statement of the problem provides delta V.How many particles, each having a charge of 1.60 x 10^-6, are needed to produce the total charge at q0?
Physics
1 answer:
bija089 [108]3 years ago
4 0

Answer:

(a) Energy will be 675 J

(B) charge will be 450 C

(C) Total number of particles will be 281.25\times 10^6

Explanation:

We have given that a flashlight uses 0.75 watts of power

So power P = 0.75 watt

Voltage is given as V = 1.5 volt

Time is given as t = 15 minutes

We know that 1 minute = 60 sec

So 15 minutes = 15\times 60=900sec

(A) We know that energy is given by E=P\times T=0.75\times 900=675j

(b) We know that energy is also given by E=QV

So 675=Q\times 1.5

Q=450C

Now we have given charge on each particle =1.6\times 10^{-6}C

So number of charge particle n=\frac{450}{1.6\times 10^{-6}}=281.25\times 10^6

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AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
A tennis ball is dropped from a roof. If it takes 38.9 seconds to reach the ground, how fast is the ball moving just before it h
Ilya [14]

Answer:

The velocity of the ball before it hits the ground is 381.2 m/s

Explanation:

Given;

time taken to reach the ground, t = 38.9 s

The height of fall is given by;

h = ¹/₂gt²

h = ¹/₂(9.8)(38.9)²

h = 7414.73 m

The velocity of the ball before it hits the ground is given as;

v² = u² + 2gh

where;

u is the initial velocity of the on the root = 0

v is the final velocity of the ball before it hits the ground

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 7414.73 )

v = 381.2 m/s

Therefore, the velocity of the ball before it hits the ground is 381.2 m/s

5 0
2 years ago
The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built?
Sonja [21]
Neglecting friction and air resistance, the first hill must be built 4 times higher than it is now.
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3 years ago
A truck accidentally rolls down a driveway for 8.0\,\text m8.0m8, point, 0, start text, m, end text while a person pushes agains
gavmur [86]

Answer:

Explanation:

According to work energy theorem

change in kinetic energy of truck = work done against it

work done against it = force x displacement

= - 850 x 8 = 6800 J

change in kinetic energy of truck = - 6800 J .

energy will be reduced by 6800 J

5 0
3 years ago
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Answer:

D,B,C,A,C

Explanation:

I believe that is the correct answers but it is unclear. I don't think the key for the second last question would let the current flowing so the bulb would be off.

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