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Rzqust [24]
2 years ago
11

Which structures protect the cell? Check all that apply.

Physics
2 answers:
MAXImum [283]2 years ago
8 0

Answer:

cell wall and cell membrane

zimovet [89]2 years ago
6 0

Answer:

Cell wall and Cell membrane.

Explanation:

The cell membrane controls what goes in and out of the cell

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A person’s blood type depends on the type of protein marker, called an , found on the cell membranes of the red blood cells.
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Consider the system shown in fig. 6-26. the rope and pulley have negligible mass, and the pulley is frictionless. the coefficien
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3 years ago
(8c5p80) Imagine a landing craft approaching the surface of Callisto, one of Jupiter's moons. If the engine provides an upward f
Alexus [3.1K]

Answer:

The weight of the landing craft in the vicinity of Callisto's surface is 3480 N.

Explanation:

The engine of the craft provides an upward thrust of 3480 N so that the space craft descends at a constant speed.

This implies that the net force on the space craft is zero.

The upward thrust will be equal to the downward gravitational pull by Callisto.

So the weight of the craft near the vicinity will be 3480 N.

3 0
3 years ago
A basketball player standing under the hoop shoots the ball straight up with an initial velocity of v0Part (a) What is the maxim
Mama L [17]

Answer:

Maximum height will be h=\sqrt{\frac{v_0^2}{2g}}

Explanation:

We have given initial velocity through which basketball is thrown =v_0

Acceleration due to gravity =g\  m/sec^2

At maximum height velocity will be zero

So final velocity v = 0 m /sec

According to third equation of motion v^2=u^2+2gh

0^2=v_0^2-2gh ( Negative sign is due to upward acceleration )

h=\sqrt{\frac{v_0^2}{2g}}

6 0
3 years ago
A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d
Iteru [2.4K]

Answer:

A point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

                         

Explanation:

We can find the distance with the following equation since the acceleration is cero (the disk rotates at a constant rate):

d = v*t

Where:

v: is the tangential speed of the disk

t: is the time = 30 s  

The tangential speed can be found as follows:

v = \omega*r

Where:

ω: is the angular speed = 100 rpm

r: is the radius = 50 cm = 0.50 m

v = \omega*r = 100 \frac{rev}{min}*\frac{2\pi rad}{1 rev}*\frac{1 min}{60 s}*0.50 m = 5.24 m/s    

Now, the distance traveled by the disk is:

d = v*t = 5.24 m/s*30 s = 157.2 m

Therefore, a point on the outside rim will travel 157.2 meters during 30 seconds of rotation.

I hope it helps you!

3 0
2 years ago
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