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2 years ago

Mass, size and color are examples of ___

1 answer:

6 0

The short answer (and the long one for that matter) is physical properties of chemicals. If you are being marked by a machine, likely the answer is going to be physical properties.

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The thrust of a certain boat’s engine generates a power of 10kW as the boat moves at constant speed 10ms through the water of a

Lunna [17]

Answer:

The change in power is 4400 W.

Explanation:

Given that,

Power = 10 kW

Speed = 10 m/s

Increases speed = 12 m/s

Given equation is,

$F=kv$

We know that,

The power is,

$P=Fv$

Put the value of F into the formula

$P=(kv)v$

$P=kv^2$

$P\propto v^2$

We need to calculate the new power

Using formula for power

$\dfrac{P}{P'}=\dfrac{v^2}{v'^2}$

Put the value into the formula

$\dfrac{10}{P'}=(\dfrac{10}{12})^2$

$P'=(\dfrac{12}{10})^2\times10$

$P'=14.4\ kW$

We need to calculate the change in power

Using formula of change in power

$\Delta P=P'-P$

Put the value into the formula

$\Delta P=14.4-10$

$\Delta P=4.4\ kW$

$\Delta P=4.4\times1000$

$\Delta P=4400\ W$

Hence, The change in power is 4400 W.

6 0

3 years ago

1. A girl drops a ball off the roof of her house. It takes the ball 2 seconds to hit the ground. How tall

Masja [62]

If the acceleration constant..

you can use the formula s = ut + 1/2at²

Known that :

s = ?

u = 0

t = 2s

a = 10ms-²

Then you can apply the formula

s = ut + 1/2at²

s = 0 + 1/2(10)(2)²

s = 5 × 4

s = 20m

Answer : 20m

Explanation :

The gravity can be 9,8 or 10. Also im not sure how people teach you but in my school, if the ball goes down the gravity is positive and not negative thats why i put 10ms-² and not -10ms-²

s = displacement/distance

u = initial speed

a = acceleration

t = time

sorry if im wrong

5 0

2 years ago

A man can jump 1.5 m on earth. calculate the approximate

Olegator [25]

Answer:

18 m

Explanation:

G = Gravitational constant

m = Mass of planet = $\rho V$

$\rho$ = Density of planet

V = Volume of planet assuming it is a sphere = $\dfrac{4}{3}\pi r^3$

r = Radius of planet

Acceleration due to gravity on a planet is given by

$g=\dfrac{Gm}{r^2}\\\Rightarrow g=\dfrac{G\rho V}{r^2}\\\Rightarrow g=\dfrac{G\rho \dfrac{4}{3}\pi r^3}{r^2}\\\Rightarrow g=\dfrac{4G\rho\pi r}{3}$

So,

$g\propto \rho r$

Density of other planet = $\rho_p=\dfrac{1}{4}\rho_e$

Radius of other planet = $r_p=\dfrac{1}{3}r_e$

$\dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\rho_p r_p}\\\Rightarrow \dfrac{g_e}{g_p}=\dfrac{\rho_e r_e}{\dfrac{1}{4}\rho_e\times \dfrac{1}{3}r_e}\\\Rightarrow \dfrac{g_e}{g_p}=12\\\Rightarrow g_p=\dfrac{g_e}{12}\\\Rightarrow g_p=\dfrac{9.8}{12}$

Since the person is jumping up the acceleration due to gravity will be negative.

From kinematic equations we have

$v^2-u^2=2g_es\\\Rightarrow u^2=v^2-2g_es\\\Rightarrow u^2=0-2\times -9.8\times 1.5\\\Rightarrow u^2=2\times 9.8\times 1.5$

On the other planet

$v^2-u^2=2g_ps\\\Rightarrow s=\dfrac{v^2-u^2}{2g_p}\\\Rightarrow s=\dfrac{0-(2\times 9.8\times 1.5)}{2\times -\dfrac{9.8}{12}}\\\Rightarrow s=18\ \text{m}$