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A jeep-like vehicle has a canvas top that seals around all sides. When the jeep is at rest, the canvas roof is flat and level. I

Alla [95]

When Jeep is in motion the Canvas top is initially flat

so when Jeep is in motion the air at the top of the Jeep is in high speed

while the air inside the jeep is not moving that fast

so as per Bernuolli's Theorem we know

$P + \frac{1}{2}\rho v^2 = constant$

so if we increase the speed the pressure will decrease

so the canvas will experience less pressure at the top of canvas while at the bottom it will have more pressure from inside

so the Canvas will bulge outside or curved outside

8 0

3 years ago

What is the mass of a 4900 N bobsled

tensa zangetsu [6.8K]

A single kilogram is equivalent to 9.8 newtons, meaning that to find the weight of the bobsled, you would simply divide the weight by 9.8. Doing this we find that the bobsled is 500kg

4900 / 9.8 = 500

6 0

3 years ago

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Two clay balls collide and stick. Ball 1 has a mass of 10 kg and an initial velocity of 5 m/s in the positive x direction. Ball

kari74 [83]

Answer:

- 1.67 m/s

Explanation:

mass of first ball, m = 10 kg

mass of second ball, M = 20 kg

initial velocity of first ball, u = 5 m/s

initial velocity of second ball, U = - 5 m/s (direction is opposite to the first ball)

Let the final velocity of the combined mass is V.

Use the principle of conservation of linear momentum

Momentum before collision = Momentum after collision

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

10 x 5 + 20 x (-5) = (10 + 20) x V

50 - 100 = 30 V

- 50 = 30 V

V = - 1.67 m/s

Thus, the combined mass moves in the negative x axis direction with the velocity of 1.67 m/s.

8 0

3 years ago

(a) What is the vapor pressure of water at 20.0ºC ? (b) What percentage of atmospheric pressure does this correspond to? (c) Wha

DENIUS [597]

Answer:

a). Vapour pressure = 2333.141 Pa

b). Percentage of atmospheric pressure = 99 %

c). Percentage of water vapour = 1.4 %

Explanation:

a). From water saturation pressure table,

Vapour pressure of water at 20°C is 17.5 torr

We know, 1 torr = 133.322 Pa

Therefore, 17.5 torr = 2333.141 Pa

Therefore, vapour pressure at 20°C is 2333.141 Pa

b). At 20°C, the pressure at atmospheric unit = 0.0231 atm

= 0.0231 x $10^{5}$ Pa

= 2310 Pa

Therefore, $\frac{2310}{2333.141}$

= 0.99

Therefore percentage of atmospheric pressure = 0.99 x 100

= 99 %

c). We Know

Relative Humidity (Rh)= $\frac{P_{v}}{P_{saturation}}$

where $P_{v}$ is vapour pressure

Rh is given as 100% = 1

∴ 1 = $\frac{P_{v}}{P_{saturation}}$

$P_{v}=P_{saturation}$ = 2333.141 Pa

Now humidity ratio is

$w = \frac{m_{v}}{m_{a}}$ , where $m_{v}$ is mass of vapour

$m_{a}$ is mass of air

$w = 0.622\times \frac{P_{v}}{P_{total}-P_{v}}$

Let the total pressure be $10^{5}$ Pa

$w = 0.622\times \frac{2333.141}{10^{5}-2333.141}$

$w = 0.01485$

Therefore, percentage of water vapour is 0.01485 x 100

= 1.4 %

8 0

3 years ago

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass g

diamong [38]

Answer:

The Astronaut' mass is 55.99 kg.

Explanation:

Given that,

The net external force acting on the Astronaut, F = 50 N

The acceleration of the Astronaut, $a=0.893\ m/s^2$

It is required to find the Astronaut' mass. We know that the force acting on an object is given by :

F = ma

m is mass of Astronaut

$m=\dfrac{F}{a}\\\\m=\dfrac{50\ N}{0.893\ m/s^2}\\\\m=55.99\ kg$

So, the Astronaut' mass is 55.99 kg.

7 0

3 years ago

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