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3 years ago

You cant see the cells in your body because most of them are very small true or false?

Physics

2 answers:

tamaranim1 [39]3 years ago

8 0

Answer: The given statement is True.

Cells are the basic fundamental, structural, and functional unit of all living organisms.

Cells are the microscopic structures, which are responsible for performing all the functions of life. Thus, cells present in the body of living organisms can not be visible through naked eyes.

Therefore, the given statement is True.

Zielflug [23.3K]3 years ago

3 0

True you can't see them because most of them are way too small to see.

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1. How do you describe the path that the light passes through when it travels?

Ber [7]

The white light is divided into its component hues - red, orange, yellow, green, blue, and violet - as it passes through the prism. Dispersion is the splitting of visible light into its many hues.

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Dispersion is the splitting of visible light into its many hues. Each hue is associated with a certain wave frequency, as stated in the Light and Color unit, and different frequencies of light waves bend at different rates while passing through a prism.

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The sun and the weather combine to create a rainbow. As it passes from air to denser water, light enters a water droplet, slowing and bending. The light bounces off the interior of the droplet, separating the wavelengths—or colors—that make up the droplet. A rainbow is created as light escapes the droplet.

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2 years ago

A bike travels at a constant speed of 3.1 m/s for 6 s. how far does it go?

IgorC [24]

Ah for this problem you are thinking quite a bit hard on. The problem is actually simpler than it looks. The problem states that a bike travels at a constant speed of 3.1 m/s for 6 s and asks how far will it go?. To figure this out you simply need to take 3.1 times 6 s because every second the bike travels 3.1 m. So the answer to this problem would be 18.6 m

8 0

4 years ago

Can the magnitude of a vector ever (a) be equal to one of its components, or (b) be less than one of its components? 9. Can a pa

Ber [7]

Answer:

a) the other components are zero, in the direction of one of the coordinate axes

b) the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction

9) constant velocity the acceleration must necessarily be zero,

constant speed can be accelerated since it may be changing the direction of the velocity vector

Explanation:

Vectors are quantities that have modulo (scalar) direction and sense.

a) If in a vector its magnitude is equal to one d its components implies that the other components are zero, therefore the vector must be in the direction of one of the coordinate axes

b) if the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction, other than the direction of the axes, that is

R² = x² + y²

where R is the magnitude of the vector e x, and are the components

9) When a particle has a constant velocity, the acceleration must necessarily be zero,

v = vo + a t

The bold letters indicate vectors If a = 0 implies that v = vo

If a particle has constant speed it can be accelerated since it may be changing the direction of the velocity vector, this type of acceleration has the name of centripetal acceleration

6 0

3 years ago

The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i

Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

$T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1$

Block 2

$T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2$

Block 3

$- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3$

Solving Eq1,2,3 simultaneously

Divide 1 and 2

$\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4$

Put Eq 4 into Eq3

$T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5$

Put Eq 5 into Eq2 and solve for a

$a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6$

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

$T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\$

5 0

4 years ago

A ball with a mass of 275 g is dropped from rest, hits the floor, and re-bounds upward. If the ball hits the floor with a speed

disa [49]

Answer:

a) $\Delta p = 1.350\,\frac{kg\cdot m}{s}$ , b) $\Delta p' = -0.454\,\frac{kg\cdot m}{s}$ , c) D. The magnitud of the change in the ball's momentum.

Explanation:

a) The magnitude of the change in the ball's momentum is:

$\Delta p = (0.275\,kg)\cdot \left[\left(1.63\,\frac{m}{s} \right)-\left(-3.28\,\frac{m}{s} \right)\right]$

$\Delta p = 1.350\,\frac{kg\cdot m}{s}$

b) The change in the magnitude of the ball's momentum:

$\Delta p' = (0.275\,kg)\cdot \left[(1.63\,\frac{m}{s} )-(3.28\,\frac{m}{s} ) \right]$

$\Delta p' = -0.454\,\frac{kg\cdot m}{s}$

c) The magnitude of the change in the ball's momentum is more directly related to the net force acting on the ball, as it measures the effect of the force on change in ball's motion at measured time according to the Impact Theorem. So, the right answer is option D.

3 0

3 years ago