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Karolina [17]
2 years ago
11

A star such as the Sun starts out as a protostar with a diameter a hundred times larger than its diameter during its main-sequen

ce lifetime. Accordingly, the protostar's temperature is much less than its main-sequence temperature. Assume that while the Sun was a protostar it had a diameter 95.00 times that of the present-day Sun. How many times greater was the protostar's surface area
Physics
1 answer:
Elan Coil [88]2 years ago
8 0

The protostar's surface area was 1.1  times greater.

<h3>Surface area of the star</h3>

The surface of the star can be assumed be circular and the surface area can determined as follows;

  • initial diameter = 100d
  • final diameter = 95d

Area = πr²

Area = \frac{\pi d^2}{4} \\\\Area = \frac{(100d)^2}{(95d)^2} \\\\

Surface area = 1.1 times greater

Thus, the protostar's surface area was 1.1  times greater.

Learn more about protostars here:  brainly.com/question/891408

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A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0
3 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

7 0
3 years ago
Which has more kinetic energy a bowling ball or a soccer ball
lawyer [7]
A bowling ball because it is heavier  and it has more air force going against it<span />
6 0
3 years ago
Read 2 more answers
A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 m/s2
Harman [31]
<span>When the fuel  of the rocket is consumed, the acceleration would be zero. However, at this phase the rocket would still be going up until all the forces of gravity would dominate and change the direction of the rocket. We need to calculate two distances, one from the ground until the point where the fuel is consumed and from that point to the point where the gravity would change the direction. 

Given:
a = 86 m/s^2 
t = 1.7 s

Solution:

d = vi (t) + 0.5 (a) (t^2) 
d = (0) (1.7) + 0.5 (86) (1.7)^2 
d = 124.27 m 

vf = vi + at 
vf = 0 + 86 (1.7) 
vf = 146.2 m/s (velocity when the fuel is consumed completely) 

Then, we calculate the time it takes until it reaches the maximum height.
vf = vi + at 
0 = 146.2 + (-9.8) (t) 
t = 14.92 s

Then, the second distance
d= vi (t) + 0.5 (a) (t^2) 
d = 146.2 (14.92) + 0.5 (-9.8) (14.92^2) 
d = 1090.53  m

Then, we determine the maximum altitude:
 d1 + d2 = 124.27 m + 1090.53 m = 1214.8 m</span>
5 0
3 years ago
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