Someone help me I will mark brainless it’s 2 questions .

Which of the following is an advantage of asexual reproduction compared to sexual reproduction?

azamat

Answer:

Asexual reproduction requires less energy and will produce more offspring over time

Explanation:

6 0

3 years ago

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What is 6.39 times 10 to the 23rd power

Sveta_85 [38]

Answer:

6.39*(10 to the 23rd)=

6.39*10²³

6 0

3 years ago

A child whose weight is 235 N slides down a 4.90 m playground slide that makes an angle of 37.0° with the horizontal. The coeffi

Basile [38]

Answer:

Explanation:

a) First, let's calculate the value of the Friction force, which is given by the formula:

Ff = u*W

As the Friction force has an X component, it would be:

Ff = u*m*g*cosФ

Where m*g is the weight of the child.

Solving for Ff:

Ff = 0.051 * 235cos37 = 9.57 N

Now, to get the energy transferred to thermal energy (or heat) we need to get the Work done, so:

Wf = Ff * d

Wf = 9.57 * 4.9 = 46.9 J

b) We need to get the downslope component of the child weight, which is:

Wy = 235sin37 = 141.43 N

As you can imagine, the gravity also does work, so:

Wg = Wy * d

Wg = 141.43 * 4.9 = 693 J

Now, let's get the kinetic energy, which can be obtained by this expression:

ΔKe = Wg - Wf

ΔKe = 693 - 46.9 = 646.1 J

The formula for kinetic energy is:

Ke = 1/2 m*V^2

We have the innitial speed which is 0.355 m/s, and the mass can be obtained by m*g so:

Fw= m*g ----> m = Fw/g

m = 235 / 9.8 = 23.97 kg

so the innitial energy is:

Ke = 1/2 * 23.97 * (0.355)^2

Ke = 1.51 J

This could be Ke1, so to get Ke2:

ΔKe = Ke2 - Ke1

Ke2 = Δke + Ke1

Ke2 = 646.1 + 1.51 = 647,61 J

Finally, the speed at the bottom would be:

v = √2Ke/m

v = √2*647.61/23.97

v = 7.35 m/s

7 0

3 years ago

What is the coefficient of friction between the skates and the ice?

natta225 [31]

The coefficient of friction is 0.051

Explanation:

The motion of the skater is a uniformly accelerated motion, therefore we can use the following suvat equation:

$v^2 - u^2 = 2as$

where:

v = 0 is the final velocity of the skater (he comes to a stop)

u = 10.0 m/s is his initial velocity

a is the acceleration

$s=1.0\cdot 10^2 m = 100 m$ is the distance he travels before stopping

Solving for a, we find the acceleration of the skater:

$a=\frac{v^2-u^2}{2s}=\frac{0-10.0^2}{2(100)}=-0.5 m/s^2$

We also know that the net force acting on the skater is the force of friction, therefore we can write (Newton's second law of motion):

$F= ma = -\mu mg$

where

$-\mu mg$ is the force of friction

m is the mass of the skater

$\mu$ is the coefficient of friction

$a=-0.5 m/s^2$ is the acceleration

$g=9.8 m/s^2$ is the acceleration of gravity

Solving for $\mu$ , we find the coefficient of friction:

$\mu = -\frac{a}{g}=-\frac{-0.5}{9.8}=0.051$

Learn more about friction:

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#LearnwithBrainly

8 0

3 years ago

An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

Marina86 [1]

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

4 0

3 years ago