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3 years ago

14

The Moon is responsible for what % of the tides that we experience?

1 answer:

Ivenika [448]3 years ago

7 0

Answer: High tides and low tides are caused by the Moon. The Moon's gravitational pull generates something called the tidal force. The tidal force causes Earth—and its water—to bulge out on the side closest to the Moon and the side farthest from the Moon. These bulges of water are high tides.

Explanation:

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What is the velocity of the car, in meters per second, just after it hits a 135-kg deer initially running at 10.5 m/s in the sam

Gnom [1K]

Answer:

V = 27.46 m/s

Explanation:

given,

mass of the deer(m) = 135 Kg

speed of the deer (u) = 10.5 m/s

assuming,

mass of the car(M) = 900 Kg

initial velocity of car (v) = 30 m/s

using conservation of momentum

m u + M v = (M + m )V

V is the velocity of the car as deer is on the car

135 x 10.5 + 900 x 30 = (900 + 135 ) V

28417.5 = 1035 V

V = 27.46 m/s

so, the velocity of car is equal to V = 27.46 m/s

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4 years ago

There are two types of body waves that travel out from the epicenter of an earthquake they have unique characteristics choose ea

Komok [63]

Answer:

d

Explanation:

distorts or shears rock as it travels through it

4 0

3 years ago

A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.

Nataliya [291]

Answer:

c

Explanation:

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3 years ago

What does the phrase “constant velocity” indicate?

pickupchik [31]

The answer is C is the velocity is the same then the acceleration is also the same.

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3 years ago

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spi

Svetllana [295]

Answer:

a

$P_G = 14.03 \ psig$

b

$h_m = 0.148 \ m$

Explanation:

From the question we are told that

The pressure of the manometer when there is no gas flow is $P_{m} = 15.5 \ psig = 15.5 * 6894.76 = 106868.78 \ N/m^2$

The level of mercury is $h = 950 \ mm = 0.950 \ m$

The drop in the mercury level at the visible arm is $d = 39.0 = 0.039 \ m$

Generally when there is no gas flow the pressure of the manometer is equal to the gauge pressure which is mathematically represented as

$P_g = P_m = g * \delta h * \rho$

Here $\rho$ is the density of mercury with value $\rho = 13.6 *10^{3} kg/m^3$

and $\delta h$ is the difference in the level of gas in arm one and two

So

$\delta h = \frac{106868.78}{ 13.6 *10^{3} * 9.8 }$

$\delta h = 0.802 \ m$

Generally the height of the mercury at the arm connected to the pipe is mathematically represented as

$h_m = 0.950 - 0.802$

=> $h_m = 0.148 \ m$

Generally from manometry principle we have that

$P_G + \rho * g * d - \rho * g * [h - (h_m + d)] = 0$

Here $P_G$ is the pressure of the gas

$P_G +13.6 *10^{3} * 9.8 * 0.039 - 13.6 *10^{3} * 9.8 * [0.950 - (0.148 + 0.039)] = 0$

$P_G = 9.6724 04 *10^{4} \ N/m^2$

converting to psig

$P_G = \frac{ 9.6724 04 *10^{4} }{6894.76}$

$P_G = 14.03 \ psig$

6 0

3 years ago