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3 years ago

Phsyical properties are

2 answers:

8 0

The answer is C. Any physical property can be observed without changing the identity of the substance. That’s one of the main things that separates a physical and chemical property.

aliya0001 [1]3 years ago

5 0

Hello!

I believe physical properties are C) Those properties which can be observed without changing the identity of the substance .

I hope it helps!

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A 5.0 kg wooden block Is 100 meters from the ground. What is the gravitational potential energy for the block?

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Answer:

Explanation:

The gravitational potential energy of a body can be found by using the formula

GPE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 9.8 m/s²

From the question we have

GPE = 5 × 9.8 × 100

We have the final answer as

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3 years ago

A soccer player is running upfield at 10 m/s and comes to a stop in 3 seconds facing the same direction. What is his acceleratio

I am Lyosha [343]

His acceleration would be 3.34 m/s

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3 years ago

Where does diffusion occur in the excretory system

k0ka [10]

Answer:

Diffusion of water, salts, and waste products occurs in the kidneys. Diffusion of calcium from food into cells occurs in the intestines. Molecules are not the only things that can diffuse. Heat from the body diffuses away in the form of sweat that evaporates.

Explanation:

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3 years ago

5) A person holds a 1.7 kg bucket and lets it move down at

Elena L [17]

Answer:

12.5J

Explanation:

Given parameters:

Mass of bucket = 1.7kg

Height = 75cm = 0.75m

Unknown;

Work done on the bucket by the person = ?

Solution:

To solve this problem, we use the work done equation;

Work done = force x distance = mgh

m is the mass

g is the acceleration due to gravity

h is the height

Now, insert parameters and solve ;

Work done = 1.7 x 9.8 x 0.75 = 12.5J

7 0

3 years ago

Suppose a rocket ship accelerates upwards with acceleration equal in magnitude to twice the magnitude of g (we say that the rock

pashok25 [27]

Answer:

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

c) $H=294300\ m$

d) $t_T=544.95\ s$

e) Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

Explanation:

Given:

acceleration of rocket, $a=2g=2\times 9.81=19.62\ m.s^{-2}$

time for which the rocket accelerates, $t_a=100\ s$

<u>For the course of upward acceleration:</u>

using eq. of motion,

$s_a=ut+\frac{1}{2}at_a^2$

where:

$u=$ initial velocity of the rocket at the launch $=0$

$s_a=$ height the rocket travels just before its fuel finishes off

so,

$s_a=0+\frac{1}{2}\times 19.62\times 100^2$

a) $s_a=98100\ m$ is the height where the rocket stops accelerating and its fuel is finished and starts decelerating while it still continues to move in the upward direction.

<u>Now the velocity of the rocket just after the fuel is finished:</u>

$v_a=u+at_a$

$v_a=0+19.62\times 100$

b) $v_a=1962\ m.s^{-1}$ is speed of the rocket going when it stops accelerating.

After the fuel is finished the rocket starts to decelerates. So, we find the height of the rocket before it begins to fall back towards the earth.

Now the additional height the rocket ascends before it begins to fall back on the earth after the fuel is consumed completely, at this point its instantaneous velocity is zero:

using equation of motion,

$v^2=v_a^2-2gh$

where:

$g=$ acceleration due to gravity

$v=$ final velocity of the rocket at the top height

$0^2=1962^2-2\times 9.81\times h$

$h=196200\ m$

c) So the total height at which the rocket gets:

$H=h+s$

$H=196200+98100$

$H=294300\ m$

Time taken by the rocket to reach the top height after the fuel is over:

$v=v_a+g.t$

$0=1962-9.81t$

$t=200\ s$

Now the time taken to fall from the total height:

$H=v.t'+\frac{1}{2}\times gt'^2$

$294300=0+0.5\times 9.81\times t'^2$

$t'=244.95\ s$

Hence the total time taken by the rocket to strike back on the earth:

$t_T=t_a+t+t'$

$t_T=100+200+244.95$

$t_T=544.95\ s$

Zero, since the average velocity is the net displacement per unit time and when the rocket strikes back the earth surface the net displacement is zero.

8 0

3 years ago