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Sergeeva-Olga [200]

3 years ago

15

a boy takes 15min to reach his school on bicycle.if the bicycle have speed of 2m/s the what is the distance between the house an

d school.

1 answer:

11111nata11111 [884]3 years ago

6 0

When looking for distance you multiply speed by time

So 15 x 2 = 30

30 is the distance between his house and school

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The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by

Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

$m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}$

we have

a. for c = 20 cm

$m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,$

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

$\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.$

Evaluated at c = 20 cm, we have

$\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.$

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0

3 years ago

An object is dropped from a height of 25 meters. At what velocity will it hit the ground?

Goryan [66]

You can use Vf^2-Vi^2 = 2ax

Vf^2 - 0 = 2(9.81)(25)

Or you can use energy

mgh = 1/2mv^2

2gh =v^2

Same thing

6 0

3 years ago

Write one example of a physical change and one example chemical change

mojhsa [17]

Physical change = changes the physical properties (more commonly known as it's look)
Chemical change = changes the chemical properties into an entire new chemical form
Examples of physical change would be melting ice cubes or sugar cubes.
Examples of chemical change would be cooking eggs or burning paper because you're changing its chemical properties.

6 0

3 years ago

Which of the following is a good step in learning how to manage time?

Luden [163]

I think is c because your seeing how your using you’re time.

8 0

3 years ago

Read 2 more answers

Very lost please help.

V125BC [204]

A) The acceleration is due to gravity at any given point if you look at it vertically, so $-10 m/s^2$ .

b) $sin(25) = V_y/V$ , so $V_y = V*sin(25)$ . We use $V = V_0 + a t$ and then the final speed must be 0 because it stops at the highest point. So $0 = V_y - 10t$ . Solve for $t$ and you get $t = 32sin(25)/10 = 16sin(25)/5$

c) $Y = Y_0 + V_0t + (1/2)at^2$ , and then we plug the values: $Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2$ and we already have the time from "b)", so $Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2$ ; then we just rearrange it $Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]$ and finally $Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20$

6 0

3 years ago