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stepan [7]
2 years ago
5

Name two types of evidence used to support the theory of evolution. Explain how scientists use each type of evidence to provide

support for evolution. Give one example of each.
Physics
1 answer:
bearhunter [10]2 years ago
7 0
Well, one of the theories was that Charles darwin proved that theory evolution is real so that one theory. 
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A business letter is informal correspondence and is sent to people within
jarptica [38.1K]

Answer:

FALSE!!!

Explanation:

Business letters are formal and normally given to someone of higher importance.

5 0
3 years ago
I have no idea of how to approach this problem
nataly862011 [7]

Answer:

p=1

Explanation:

Well me know that v=m/s

and that a=m/s^2

so

(m/s)^{2} = (m/s^2)(x^p)\\\\(m^2/s^2)/(m/s^2)=x^p\\\\m^2s^2/ms^2=x^p\\\\(m^2/m)(s^2/s^2)=x^p\\\\m=x^p\\\\p=1

Note: We don't take into account 2 because it's a scalar, it doesn't have units so it doesn't add anything to the equation.

4 0
3 years ago
A motor cycle travelling at 10 m/s accelerates at 4m/s(squared) for 8s.
Sever21 [200]


10 x 4^2 = 160 / 8..

V = 20m/s...

...x 8 = 100 miles,meters, metric what ever m stands for after 8 seconds.

This is my guess since the problem says 4m/s^2

V= distance/ ST (traveled/used)
8 0
3 years ago
11. A student lifts a 25 kg mass a vertical distance of 1.6 m in a time of 2.0 seconds.
goldenfox [79]

Answer:

a) 245 N

b) 392 J

c) 196 W

Explanation:

a) Force needed to lift the 25 kg object is the same as the weight of the object:

Force= m * g = 25kg * 9.8 \frac{m}{s^2} = 245 N

b) Work is the force exerted times the distance moved:

Work = F * d = 245N * 1.6 m = 392 Joules = 392 J

c) The power exerted is the quotient of the work done over the amount of time it took to move the object:

4 0
3 years ago
When an old lp turntable was revolving at 33.3 rpm, it was shut off and uniformly slowed down and stopped in 5.5 seconds. throug
velikii [3]
First, let's convert the initial angular speed from rpm (rev/min) into rad/s, keeping in mind that 1 rev = 2 \pi rad and 1 min=60 s:
\omega_i = 33  \frac{rev}{min}=33\frac{rev}{min} \cdot  \frac{2 \pi rad}{60 s}=3.45 rad/s

Now we can find the angular acceleration of the lp, keeping in mind that the final speed is zero:
\alpha =  \frac{\omega_f - \omega_i}{t}= \frac{0-3.45 rad/s}{5.5 s}=-0.63 rad/s^2
and the acceleration is negative because the LP is decelerating.

Now we can find the angle covered by the LP from the beginning to the end of its motion:
\theta (t)= \omega_i t + \frac{1}{2}\alpha t^2 = (3.45 rad/s)(5.5 s)+ \frac{1}{2}(-0.63 rad/s^2)(5.5 s)^2=
=9.45 rad

And finally, we can convert it into number of revolutions:
1 rev : 2 \pi rad = x: 9.55 rad
x= \frac{9.55}{2 \pi}= 1.52 rev
4 0
3 years ago
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