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2 years ago

(1) Arectangularmetal block with mass 18 kg has

Physics

1 answer:

sleet_krkn [62]2 years ago

8 0

Answer:

Volume=0.003m^3 density=6000kg/m^3

Explanation:

mass=18kg

volume=length x width x height

Volume of block=0.1x0.15x0.2

volume of block=0.003m^3

Density=mass ➗ volume

Density=18 ➗ 0.003

Density=6000kg/m^3

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Question 2.

choli [55]

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

$1\ \text{inch}=\dfrac{1}{12}\ \text{feet}$

So,

$5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}$

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

3 0

3 years ago

The area of the united states that has the highest level of average daily solar radiation by month would be

Vlada [557]

The area of the United States that has the highest level of Average Daily Solar radiation by month would be:

A) the Gulf Coast.
B) the Great Lakes.
C) the Southwest.
<span>D) the Pacific Northwest.

If this is your question the answer is C. the southwest.</span>

7 0

2 years ago

PLEASE HELP!

Kobotan [32]

Answer:

Explanation:

Cause a monster truck don

3 0

2 years ago

Calculat the acceleration of a person at latitude<br> 40degreesowing to the rotation of the earth.

beks73 [17]

Answer:

acceleration of person = 9.77 m/s²

Explanation:

given data

latitude = 40 degree

to find out

Calculate the acceleration of a person

solution

we know that here 40 degree = 0.698 rad

acceleration of person = g - ω²R ...............1

and 1 rotation complete in 24 hours = 360 degree

here g is 9.81

so we know Earth angular speed ω = 7.27 × $10^{-5}$ rad/s and R is earth radius that is 6.37 × $10^{6}$ m

put here value in equation 1 we get

acceleration of person = g - ω²R

acceleration of person = 9.81 - (7.27 × $10^{-5}$ )² × 6.37 × $10^{6}$

acceleration of person = 9.77 m/s²

6 0

3 years ago

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal

galben [10]

Hello

1) First of all, since we know the radius of the wire ( $r=1.2~mm=0.0012~m$ ), we can calculate its cross-sectional area
$A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2$

2) Then, we can calculate the current density J inside the wire. Since we know the current, $I=3~A$ , and the area calculated at the previous step, we have
$J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2$

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity $\sigma=3.6\cdot10^7~ \frac{A}{Vm}$ of the aluminium, the electric field is given by
$E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m$

4 0

3 years ago