What kind of law of motion A car still moves for a short period even after the brakes

A physician might deal with potential problems associated with the ________ characteristic of services by providing physical cue

asambeis [7]

I believe Intangibility is the answer! :P I hope this helps!

8 0

3 years ago

A gas is placed in a storage tank at a pressure of 49.2 atm at 39.0C . As a safety device, a small metal plug in the tank is mad

Amiraneli [1.4K]

Answer:

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=49.2 atm\\T_1=39.0^oC = 312.15 K\\P_2=?\\T_2=198^oC=471.15 K$

Putting values in above equation, we get:

$\frac{49.2atm }{312.15 K}=\frac{P_2}{471.15 K}\\\\P_2=74.26 atm$

The maximum pressure that will be attained in the tank before the plug melts and releases gas should be less than 74.26 atm.

4 0

3 years ago

According to Bode’s Law a planet is missing between Jupiter and Saturn. True False

defon

Bode law, a planet was believed to exist between .... An Astronomer's Account of the Missing Planet Between Mars and Jupiter as Interpreted Jupiter ·Saturn · Uranus · Neptune.

6 0

3 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

What is meant by the term 'mass movement'?

vitfil [10]

Mass movement is the movement of surface materials caused by gravity. A great example would be a mud slide.

8 0

3 years ago

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