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2 years ago

Calculate the mass (in kg) of 54.3 m³ of granite. The density of granite is 2700 kg/m³. Give your answer to 2 decimal places!!

Physics

1 answer:

MrMuchimi2 years ago

8 0

Answer:

146,610.00 kg

Explanation:

The density of a substance is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

For the sample of granite in this problem, we know:

$V=54.3 m^3$ is the volume

$d=2700 kg/m^3$ is the density

So, we can re-arrange the equation to solve for m, the mass of the sample:

$m=dV=(2700)(54.3)=146,610.00 kg$

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just olya [345]

Answer:

Explanation:

40 divided by 10 then which would equal 4. Add the 1.0 , 2 ,and 15 together. Then multply the 60 by 18.0 after you are done dividing the answer is 3 with a remainder of 6.

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2 years ago

If you are driving 72 km/h along a straight road and you look to the side for 4.0 s, how far do you travel during this inattenti

Ann [662]

We know that speed equals distance between time. Therefore to find the distance we have that d = V * t. Substituting the values d = (72 Km / h) * (1h / 3600s) * (4.0 s) = 0.08Km.Therefore during this inattentive period traveled a distance of 0.08Km

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2 years ago

The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the

otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

$F=qE$

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

$F=5.50\mu N = 5.50\cdot 10^{-6} N$

Therefore, if the electric field magnitude is

$E=8.50 kN/C = 8500 N/C$

Then the charge on each antenna would be

$q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC$

8 0

2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago

In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component

Dimas [21]

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1) (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2) (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

3 0

2 years ago