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3 years ago

14

What happens to the density if the volume of an object decrease and the mass remains the same?

1 answer:

vekshin13 years ago

7 0

The density would increase because you still have the same amount of weight, but it is just packed more tightly in a smaller object.

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When a 5 kg rock is dropped from a height of 6 m on Planet X, it loses 24 J of GPE. What is the acceleration due to gravity on P

yan [13]

Answer:

g = 1.25m/s²

Explanation:

Given the following data;

Mass = 5kg

Height = 6m

Gravitational potential energy = 24J

To find the acceleration due to gravity;

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

$P.E = mgh$

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

GPE = mgh

Substituting into the equation, we have;

24 = 5*6*g

24 = 30g

g = 30/24

g = 1.25m/s²

Therefore, the acceleration due to gravity on Planet X is 1.25m/s².

8 0

3 years ago

PLEASE HELP also i know this isnt physics but there was no HOPE/Gym subject for me to choose.

Andreyy89

Answer:

Plan B.

Because flexibility is best improved by stretching.

Explanation:

Improving and increasing flexibility is done by having stretching sessions daily which maintains and widens the range of motion in the joints and stretches muscles.

4 0

2 years ago

A 150kg person is pushed from the behind and accelerates at 5 m/s2. What is the net force applied to the person? ANSWER FOR BRAI

kogti [31]

Explanation:

Force = Mass × Acceleration

Mass = 150kg

Acceleration = 5 m/s^2

Force = 150 × 5 = 750N

6 0

3 years ago

A graph labeled velocity versus time with horizontal axis time (seconds) and vertical axis velocity (meters per second). A blue

jekas [21]

Answer:

B

Explanation:

It's because the line keeps going straight meaning it has a constant velocity from beginning to end.

5 0

3 years ago

Linear Thermal Expansion (in one dimension)

Leokris [45]

Answer:

1) $\Delta L= 0.612\ m$

2) a. $\Delta V_G=0.57\ L$

b. $\Delta V_S=0.021\ L$

c. $V_0=0.549\ L$

Explanation:

1)

given initial length, $L=1275\ m$

initial temperature, $T_i=-15^{\circ}C$

final temperature, $T_f=25^{\circ}C$

coefficient of linear expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

<u>∴Change in temperature:</u>

$\Delta T=T_f-T_i$

$\Delta T=25-(-15)$

$\Delta T=40^{\circ}C$

We have the equation for change in length as:

$\Delta L= L.\alpha. \Delta T$

$\Delta L= 1275\times 12\times 10^{-6}\times 40$

$\Delta L= 0.612\ m$

2)

Given relation:

$\Delta V=V.\beta.\Delta T$

where:

$\Delta V$ = change in volume

V= initial volume

$\Delta T$ =change in temperature

initial volume of tank, $V_{Si}=60\ L$

initial volume of gasoline, $V_{Gi}=60\ L$

initial temperature of steel tank, $T_{Si}=15^{\circ}C$

initial temperature of gasoline, $T_{Gi}=15^{\circ}C$

coefficients of volumetric expansion for gasoline, $\beta_G=950\times 10^{-6}\ ^{\circ}C$

coefficients of volumetric expansion for gasoline, $\beta_S=35\times 10^{-6}\ ^{\circ}C$

a)

final temperature of gasoline, $T_{Gf}=25^{\circ}C$

∴Change in temperature of gasoline,

$\Delta T_G=T_{Gf}-T_{Gi}$

$\Delta T_G=25-15$

$\Delta T_G=10^{\circ}C$

Now,

$\Delta V_G= V_G.\beta_G.\Delta T_G$

$\Delta V_G=60\times 950\times 10^{-6}\times 10$

$\Delta V_G=0.57\ L$

b)

final temperature of tank, $T_{Sf}=25^{\circ}C$

∴Change in temperature of tank,

$\Delta T_S=T_{Sf}-T_{Si}$

$\Delta T_S=25-15$

$\Delta T_S=10^{\circ}C$

Now,

$\Delta V_S= V_S.\beta_S.\Delta T_S$

$\Delta V_S=60\times 35\times 10^{-6}\times 10$

$\Delta V_S=0.021\ L$

c)

Quantity of gasoline spilled after the given temperature change:

$V_0=\Delta V_G-\Delta V_S$

$V_0=0.57-0.021$

$V_0=0.549\ L$

8 0

3 years ago