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Korolek [52]
3 years ago
11

Which statement best summarizes the theory of evolution by natural selection?

Physics
2 answers:
prisoha [69]3 years ago
8 0

Answer:

c is the one that makes the most sense

Explanation:

eimsori [14]3 years ago
6 0
I’m pretty sure it’s c
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For the simple harmonic motion equation
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Given that : d = 5sin(pi t/4), So, maximum displacement, d = 5*(+1) = 5 Also, maximum displacement, d = 5*(-1) = -5 
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bro I'm sorry I'm only in 7th grade

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In a lab,two pieces of cellphane tape are placed on a lab table. They are quickly ripped off the lab table. Which would you pred
galben [10]

Answer:

they would repel each other

Explanation:

they have gained the same charge

4 0
3 years ago
A 0.42 kg mass is attached to a light spring with a force constant of 34.9 N/m and set into oscillation on a horizontal friction
Whitepunk [10]

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

E=U=\frac{1}{2}kA^2 (1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

E=K=\frac{1}{2}mv_{max}^2 (2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{(34.9 N/m)(0.05 m)^2}{0.42 kg}}=0.456 m/s

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

E=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

E=\frac{1}{2}kA^2=\frac{1}{2}(34.9 N/m)(0.05 m)^2=0.044 J

So, solving (3) for v, we find the speed when x=1.5 cm:

v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

v=\frac{v_{max}}{2}=\frac{0.456 m/s}{2}=0.228 m/s

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

E=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\x=\sqrt{\frac{2E-mv^2}{k}}=\sqrt{\frac{2(0.044 J)-(0.42 kg)(0.228 m/s)^2}{34.9 N/m}}=0.044 m=4.4 cm

4 0
3 years ago
An object travels 8 meters in the first second of travel, 8 meters again during the second second of travel, and 8 meters again
laiz [17]

Answer:

0 m/s²

Explanation:

Acceleration is the change in velocity over change in time.

Velocity is the change in position over change in time.

During the first second, the velocity is 8 m/s.

During the second second, the velocity is 8 m/s.

During the third second, the velocity is 8 m/s.

The velocity stays constant at 8 m/s, so the object is not accelerating.

6 0
3 years ago
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