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aliya0001 [1]
2 years ago
10

How do prokaryotes differ from eukaryotes ?

Physics
1 answer:
Ber [7]2 years ago
5 0
I hope I am correct my Science teacher said this was the answer to your question

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An object is shot upwards, from the ground, with an initial velocity of 120
Alex777 [14]

Answer: 30 metres

Explanation:

Initial velocity of object = 120m/s

Time taken = 4.0s

Distance covered by object = ?

Recall that distance = (Change in velocity / Time taken)

Distance = (120m/s)/4.0s

= (120m/s) / 4.0s

= 30m

Thus, the object will be 30 metres high

6 0
2 years ago
Why are there temperature differences on the moon's surface even though there is no atmosphere present?
nika2105 [10]

The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.  

4 0
3 years ago
A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar
horrorfan [7]

Answer:

W= 4.89 KJ

Explanation:

Lets take

temperature of hot water T₁ = 100⁰C

T₁ = 373 K

Temperature of cold ice T₂= 0⁰C

T₂ = 273 K

The latent heat of ice LH= 334 KJ

The heat rejected by the engine Q= m .LH

Q₂= 0.04 x 334

Q₂= 13.36 KJ

Heat gain by engine = Q₁

For Carnot engine

Q_1=\dfrac{T_1}{T_2}Q_2

Q_1=\dfrac{373}{273}\times 13.36

Q₁  = 18.25 KJ

The work W= Q₁  - Q₂

W= 18.25 - 13.36 KJ

W= 4.89 KJ

7 0
3 years ago
Read 2 more answers
The altitude of the International Space Station ttt minutes after its perigee (closest point), in kilometers, is given by \qquad
Dmitriy789 [7]

Answer:

T = 92.8 min

Explanation:

Given:

The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

                               A(t) = 415 - sin(\frac{2*\pi (t+23.2)}{92.8})

Find:

- How long does the International Space Station take to orbit the earth? Give an exact answer.

Solution:

- Using the the expression given we can extract the angular speed of the International Space Station orbit:

                                 A(t) = 415 - sin({\frac{2*\pi*t }{92.8} + \frac{23.2*2*\pi }{92.8} )

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8

- We know that the relation between angular speed w and time period T of an orbit is related by:

                                T = 2*p / w

                                T = 2*p / (2*p / 92.8)

Hence,                     T = 92.8 min

7 0
3 years ago
The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?
Likurg_2 [28]
\frac{7.85g }{1.59 \frac{g}{ cm^{3} } } = 4.937 cm^{3}
4 0
3 years ago
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