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4 years ago

What are the characteristics of prokaryotes?

1 answer:

5 0

Prokaryotes lack an organized nucleus and other membrane-bound organelles. Prokaryotic DNA is found in a central part of the cell called the nucleoid. The cell wall of a prokaryote acts as an extra layer of protection helps maintain cell shape, and prevents dehydration.

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Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth

mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

4 0

3 years ago

Read 2 more answers

A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece

seraphim [82]

Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

4 0

3 years ago

Why does it take more time for a larger sample of water to freeze?

Stels [109]

Answer:The greater the amount of water that there is it will take longer for the water to freeze because more heat has to be dissipated into the environment

Explanation:

3 0

3 years ago

Read 2 more answers

All but one statement applies to electromagnetic radiation

lina2011 [118]

D) Electromagnetic radiation travels in the form of longitudinal waves.

3 0

3 years ago

Needddd helppppppp!!!

yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0

3 years ago