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3 years ago

To convert minutes per second into kilometre per hour we multiply the speed with

Physics

1 answer:

ankoles [38]3 years ago

3 0

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

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A car is traveling with a constant speed when the driver suddenly applies the brakes, giving the 14) car a deceleration of 3.50

sergij07 [2.7K]

To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.

First, we manipulate the one of the kinematic equations

v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped

Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,

<span>we get v0 = sqrt (2(a)(x))

Substituting the known values,

v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
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Therefore, before stopping the car the original speed of the car would be 14.49 m/s

7 0

4 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:

$P_G = 14.03 \ psig$

$h_m = 0.148 \ m$

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

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

$\delta h = 0.802 \ m$

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

Which wave, A or B, has lower energy? A, because it has a higher amplitude B, because it has a higher amplitude A, because it ha

frutty [35]

Energy of wave depends on its amplitude and it is given as

$E = \frac{1}{2} kA^2$

here k = constant

A = amplitude

so energy will increase or decrease depends on the amplitude of the wave

So here if we need to check which wave has lower energy then we need to compare the amplitude.

If the amplitude is less then energy must be less

So please check in the figure that which wave out of A and B has lesser amplitude to find out the wave of lesser energy

5 0

3 years ago

Read 2 more answers

The pressure, P, of a gas varies directly with its temperature, T, and inversely with its volume, V, according to the equation f

lbvjy [14]

The equation formula:
P V = n R T
1,245 * 2 l = n R * 300 K
n R = 1,245 * 2 : 300 = 8.3
P * 2.5 l = n R * 400 K
P * 2.5 = 8.3 * 400
P = 3,320 : 2.5 = 1328 J
Answer: A ) 1,328 joules.

7 0

3 years ago

Read 2 more answers

A star's brightness as if it were a standard distance from Earth (10 parsecs) is known as what? radiation apparent brightness li

aleksley [76]

Answer:

In contrast, the intrinsic brightness of an astronomical object, does not depend on the distance of the observer or any extinction. The absolute magnitude M, of a star or astronomical object is defined as the apparent magnitude it would have as seen from a distance of 10 parsecs (33 ly).

Explanation:

8 0

3 years ago

To convert minutes per second into kilometre per hour we multiply the speed with​

To convert minutes per second into kilometre per hour we multiply the speed with