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anastassius [24]

2 years ago

10

carrier. In a later maneuver, the jet comes in for a landing on solid ground with a speed of 100 m/s, and its acceleration can h

ave a maximum magnitude of 5.00 m/s2 as it comes to rest. (a) From the instant the jet touches the runway, what is the minimum time interval needed before it can come to rest

1 answer:

GREYUIT [131]2 years ago

6 0

Answer:

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

Explanation:

t = Time taken for jet to stop

u = Initial velocity = 100 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -5.00 m/s² (slowing down)

Equation of motion

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-100}{-5}\\\Rightarrow t=20\ s$

The time required for the plane to slow down from the moment it touches the ground is 20 seconds.

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A rigid container holds 0.30g of hydrogen gas.

Strike441 [17]

Answer:

Part A: $\mathbf{Q =94 \ J}$ to two significant figures

Part B: $\mathbf{Q =160 \ J}$ to two significant figures

Part C: $\mathbf{Q =220 \ J}$ to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = $C_v=\dfrac{3}{2}R$

For Part A:

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K$

$Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)$

$Q=93.5325 \ J$

$\mathbf{Q =94 \ J}$ to two significant figures

Part B. For hot temperature, $C_v=\dfrac{5}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K$

$Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)$

$Q=155.8875 \ J$

$\mathbf{Q =160 \ J}$ to two significant figures

Part C. For an extremely hot temperature, $C_v=\dfrac{7}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K$

$Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)$

$Q=218.2425 \ J$

$\mathbf{Q =220 \ J}$ to two significant figures

6 0

3 years ago

D. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weig

s2008m [1.1K]

Answer:

Wc = 7.84 weight of crown

Ww = 7.84 - 6.86 = .98 weight of water displaced

Density = 7.84 / .98 = 8 crown is 8 X that of water

Since gold has a density of 19.3 that of water the crown is certainly not 100 percent (if any) gold

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2 years ago

John sees Linda Running towards him at 11 m/s. while running, Linda throws a ball at 5m/s. what is the speed of the ball as obse

Ilia_Sergeevich [38]

11m/s Bc of the fact that he sees her running at 11m/s

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2 years ago

What do neon,oxygen and nitrogen have in common?

julsineya [31]

Their Period number is common means their "Principal Quantum Numbers" are same

Hope this helps!

7 0

3 years ago

A 40 g ball rolls around a 30 cm -diameter L-shaped track, shown in the figure, (Figure 1)at 60 rpm . What is the magnitude of t

levacccp [35]

Answer:

0.47 N

Explanation:

Here we have a ball in motion along a circular track.

For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.

This force is called centripetal force, and its magnitude is given by:

$F=m\omega^2 r$

where

m is the mass of the object

$\omega$ is the angular velocity

r is the radius of the circle

For the ball in this problem we have:

m = 40 g = 0.04 kg is the mass of the ball

$\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s$ is the angular velocity

r = 30 cm = 0.30 m is the radius of the circle

Substituting, we find the force:

$F=(0.040)(6.28)^2(0.30)=0.47 N$

3 0

2 years ago