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deff fn [24]
3 years ago
7

What are 16 stages of a stars life

Physics
1 answer:
Alla [95]3 years ago
8 0
Https://www.slideshare.net/mobile/jan_parker/life-cycle-of-stars-3196871
the answer to your question!
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If the voltage drop across the first resistor is 13.00V, then how many volts remain for the 2nd resistor?
Scorpion4ik [409]

Answer:

2+1

Explanation:

2+1

8 0
2 years ago
Read 2 more answers
In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?
IceJOKER [234]

Answer: a=1016.66 m/s^{2}

Explanation:

Acceleration a is expressed in the following formula:

a=\frac{V_{f}-V_{o}}{t}

Where:

V_{f}=610 m/s is the final velocity of the projectile

V_{o}=0 m/s  is the initial velocity of the projectile

t=0.6 s is the time

Solving:

a=\frac{610 m/s-0 m/s}{0.6 s}

a=1016.66 m/s^{2} This is the acceleration of the projectile

6 0
3 years ago
If in 30 mins Alex runs to a store that is 4 km away, what is her average speed in km/h
masha68 [24]
15 km 30 divided by 4 is 7.5 km in 30 min times that by 2 the answer is 15
6 0
3 years ago
A weightlifter works out at the gym each day. Part of her routine is to lie on her back and lift a 43 kg barbell straight up fro
Tasya [4]

Answer:

A. 231.77 J

B. 5330.71 J

C. 46 donuts

Explanation:

A. To lift the barbell once, she will have to extend it the full length of her arm. The work done will then be:

W = F * d

Where the force is the weight of the barbell.

F = m * g

Hence, the work done in lifting the barbell is:

W = m * g * d

W = 43 * 9.8 * 0.55

W = 231.77 J

B. If she does 23 repetitions, the total energy she expend will be equal to the Potential energy when the barbell is lifted multiplied by 23:

E = 23 * m * g * d

E = 23 * 231.77

E = 5330.71 J

C. 1 Joule = 4.184 calories

5330.71 Joules = 5330.71 * 4.184 = 22303.69

If 1 donut contains 490 calories, the number of donuts she will need will be:

N = 22303.69/490 = 45.5 donuts or 46 donuts

5 0
4 years ago
The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiato
djverab [1.8K]

Answer:571.09 kJ

Explanation:

Given

Temperature of cooling water from engine exit=240^{\circ} F\approx 115.55^{\circ}C

After Passing through the radiator its temperature decreases to 175^{\circ}F\approx 79.44^{\circ}F

specific heat of water=4.184 J/g^{\circ}C

Volume of water = 1 gallon\approx 3.78 L

density of water \rho =1 gm/mL

Thus mass of water=\rho \times V=3.78\times 1=3.78 kg

Heat transferred to the surrounding is equal to heat absorbed by cooling water

Q=m\cdot c\cdot \Delta T

Q=3.78\times 4.184\times 1000\times (115.55-79.44)

Q=3.78\times 4.184\times 1000\times (36.11)

Q=571.09 kJ

4 0
3 years ago
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