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3 years ago

The density of a substance is defined as its

Physics

1 answer:

Fantom [35]3 years ago

7 0

<h2>Answer</h2>

<u>Density is the mass per unit volume</u>

<h2>Explanation</h2>

Density is actually is the ratio of the mass of an object to its volume. It is used to compare the different objects on the basis of their weight. You can say that more the weight of an object more will be its density. For example, the density has the role in floating the ships on the water. When we insert the wooden piece inside water it pops up to the surface. The weight of the wood piece is more than the water and volume is greater which affects its density. When the object gets smaller the size on pressing, the density of the object increase.

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Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.

MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0

2 years ago

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w

Zigmanuir [339]

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

$\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm$

8 0

3 years ago

At the same moment from the top of a building 3.0 × 10 2 m tall, one rock is dropped and one is thrown downward with an initial

Bess [88]

The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)

(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s

(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s

The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s

<em>ANSWER: 19.12 s</em>

5 0

3 years ago

Read 2 more answers

In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron

slega [8]

Answer:

a) v_average = 11 m / s, b) t = 0.0627 s

, c) F = 7.37 10⁵ N

, d) F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

v² = v₀² - 2 a x

The fine speed zeroes them

a = v₀² / 2x

a = 22²/2 0.69

a = 350.72 m / s²

The average speed is

v_average = (v + v₀) / 2

v_average = (22 + 0) / 2

v_average = 11 m / s

b) The average time

v = v₀ - a t

t = v₀ / a

t = 22 / 350.72

t = 0.0627 s

c) The force can be found with Newton's second law

F = m a

F = 2100 350.72

F = 7.37 10⁵ N

.d) the ratio of this force to weight

F / W = 7.37 10⁵ / (2100 9.8)

F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

5 0

2 years ago

The escape speed from the moon is much smaller than from earth. True or False

Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

$v_{e} = \sqrt{\frac{2GM}{r}}$ (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is $5.972x10^{24}kg$ and its radius is $6371000m$

Then, replacing those values in equation 1 it is gotten.

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}$

$v_{e} = 11.18m/s$

For the case of the Moon:

$v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}$

$v_{e} = 2.38m/s$

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0

3 years ago