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jeka57 [31]
2 years ago
11

A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density? 0.081 kg/m3 12.29 g/cm3 132912.00 g/cm3 canno

t be determined
Physics
2 answers:
777dan777 [17]2 years ago
8 0

Answer:

0.081 kg/m3

Explanation:

REY [17]2 years ago
4 0
The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
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A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi
svetlana [45]

Answer:

E=0 at r < R;

E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}} at 2R > r > R;

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

4 0
3 years ago
If your front lawn is 21.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minu
ziro4ka [17]

Answer:

snow is 64.638 kg / hr

Explanation:

Given data

wide w = 21 feet

long L = 20 ft

area A = 1350 square foot

mass of snow m  = 1.90 mg

to find out

snow in kilograms / hour

solution

we will find snow in kg

so we apply formula that is

snow kg / hour  = w × L ×A ×  m × 60/10^6

put all value we get  snow

snow =  21 × 20 × 1350 ×  1.90 × 60/10^6

snow =  420 × 1350 ×  1.90 × 60/10^6

snow =  1077300 × 60/10^6

snow =  64.638

hence snow is 64.638 kg / hr

7 0
3 years ago
If the price of gasoline at a particular station in Europe is 5 euros per liter. An American student in Europe is allowed to use
Mashcka [7]

Answer:

Number of gallons =2 gallon

Explanation:

given data:

rate of gasoline ineurope = 5 euro per liter

total money to buy gasoline =  40 euro

total gasoline an american can buy in europe = \frac{40}{5}

= 8 litres of gasoline

As given in the question 1 ltr is 1 quarts therefore  

Total no. of quarts is 8 quarts

As from question 4 quarts is equal to one gallon, hence

Number of gallons= \frac{8}{4} = 2 gallon

5 0
3 years ago
31. Draw a free body diagram for a 15.5N box that is being pushed to the right with a 18. N force while experiencing 4.30 N of r
posledela

Answer:

See answers below

Explanation:

a.

F = mg,

15.5 N = m(9.8 m/s²)

m = 1.58 kg

b.

Fnet = Applied force - resistance,

Fnet = 18 N - 4.30 N,

Fnet = 13.70 N

Fnet = ma

13.70 N = (1.58 kg)a

a = 8.67 m/s²

For the free body diagram, draw a box with an upward arrow labeled 15.5 N, a downward label labeled 15.5 N, a right label labeled 18 N, and a left label labeled 4.30 N.

7 0
2 years ago
At a certain time a particle had a speed of 48 m/s in the positive x direction, and 4.5 s later its speed was 92 m/s in the oppo
larisa86 [58]

Answer:

-31.1 m/s^2

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is calculated as:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it is important to also take into account the direction of the velocity.

For the particle in this problem, we have:

u = +48 m/s is the initial velocity (positive direction)

v = -92 m/s is the final velocity (negative direction)

t = 4.5 s is the time interval

Therefore, the average acceleration is

a=\frac{v-u}{t}=\frac{-92-(+48)}{4.5}=-31.1 m/s^2

4 0
3 years ago
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