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2 years ago

9

Use the drop-down menu to complete the statement. One difference between prokaryotic cells and eukaryotic cells is that a eukary

otic cell contains , while a prokaryotic cell does not.

1 answer:

Ahat [919]2 years ago

8 0

Answer:

Nucleus

Explanation: I did the quiz and got it right. Have an amazing day!

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What parts of an atom determines which element it is?

uysha [10]

Explanation:

The protons determines the atomic number which helps in identifying an atom of an element

4 0

3 years ago

Read 2 more answers

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

Elden [556K]

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

$Mg(r - L/2) = mg(L/2 - 8)$

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

$400,000(r - 10) = 19,600 (10 - 8)$

$r - 10 = 0.098$

$r = 10.098 m$

so pillar is at distance 10.098 m from one end of the slab

3 0

3 years ago

Summarize: Based on what you have learned, how will the sound that the observer hears

Reil [10]

Answer: The sound will change due to changes in frequency and the wavelength of the airplane.

Explanation: Let assume that the observer is at a stationary position. The wavelength of the sound from the airplane reduces and the frequency increases as the plane is moving toward the observer. As the airplane passes by, that is, moving away from the observer, the frequency starts to reduce while the wavelength of the sound starts to increase.

The sound that the observer hears will change base on the illustration above.

3 0

2 years ago

Find the horizontal component and the vertical component

aleksandr82 [10.1K]

Answer:

v=3.66,h-3.66

Explanation:

vertical = 10sin60 - 10sin 30

horizontal =10cos60 + 10cos 30

v = 10×0.8660-10×0.5

h = 10×0.5 + 10 × 0.8660

v=8.660-5.0 = 3.66

h= 5.0-8.660 = -3.66

8 0

2 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago