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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.

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