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3 years ago

12

Action-reaction forces : Select one:

1 answer:

ladessa [460]3 years ago

4 0

i think its B)... im not sure so dont take it seriously

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When the area of block increases for constant force, the

Lemur [1.5K]

Answer: B

adding force will add accesion

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2 years ago

Read 2 more answers

solniwko [45]

Option E, Fiat money includes currency, checking deposits and credit cards .

<u>Explanation: </u>

Fiat money has been the currency issued by the government which is not sponsored by actual resources like gold or silver, but by the country that approved it.

Instead of the price of a product, the valuation of fiat money is extracted from the connection between production and consumption and stability of the authorizing state. Fiat currencies, including that of the U.S. dollar, euro, and other major international currencies seem to be the most common paper currencies.

One risk for fiat money is to print too many of those by regimes that contribute to hyperinflation.

Fiat money is government-supported monetary money and is treated as a legal tender. The capital is provided by physical goods such as valuable metals or instruments including checks and credit cards. The world currencies, backed by gold, were symbolic until 1971.

7 0

2 years ago

Please help :)<br><br><br> Microwaves are transmitted by ....... ?

Dima020 [189]

Microwaves are transmitted by Radio Waves.

4 0

2 years ago

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

Tpy6a [65]

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

7 0

3 years ago

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

2 years ago