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Nonamiya [84]
3 years ago
9

How can a satellite sense the contours of the ocean bottom from space?

Physics
1 answer:
melomori [17]3 years ago
6 0
<span>The climate is brutal in the African Savanna. There is a rainy season and a dry season. It rains about 15-30 inches in the rainy season and not more than 4 in the dry season. The rainy season is December-April and the Dry season is May-November.</span><span>Nov 17, 2005</span>
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A girl standing upright exerts a pressure of 15000 N/m2 on the floor. Given that the total area of contact of shoes and the floo
lapo4ka [179]

Explanation:

F=15000/0.02

=300N

total area contact of the shoes and floor when stood in one foot=0.02m^2/2=0.01m^2

P=F/A

=300/0.01

=30 000Pa/30 000Nm^-2

in another way=15000×2

=30000N/m^2

5 0
4 years ago
The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current
Komok [63]

Answer:

I=0.047A

Explanation:

Let's use Ohm's law:

V=IR  

or

I=\frac{V}{R}   (1)

Where:

V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

R=\rho*\frac{l}{A}    (2)

Where:

R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}

Keep in mind that the electrical resistivity of the gold is a known constant which is \rho_g_o_l_d=2.35*10^{-8} and the cross sectional area of the conductor is calculated as:

A=\pi *(r^{2})=\pi  *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} }  =14.96056465\Omega

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

I=\frac{0.7}{14.96056465}\approx0.047A

7 0
3 years ago
A car uses ____________________________ to move. Select one: a. energy b. pulls c. motion d. work
Fynjy0 [20]

Answer:

Option A.

Explanation:

The correct answer is Option A.

The car uses energy to move.

A car is a machine that converts energy locked in fuel like petrol or diesel and turn it into mechanical energy.

The energy produced from the combustion of gasoline is then used to move the shaft, which sends the power to the rear axle and the wheel starts to move.

4 0
4 years ago
The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on
Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0
4 years ago
Why would you want a gold ring rather than a copper ring? (I know gold is more valuable)
rodikova [14]

one leaves green marks on your finger and one doesnt

7 0
3 years ago
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