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3 years ago

A jet airliner moving initially at 889 mph

Physics

1 answer:

Eduardwww [97]3 years ago

8 0

Answer:

1500 mph

Explanation:

Take east to be +x and north to be +y.

The x component of the velocity is:

vₓ = 889 cos 0° + 830 cos 59°

vₓ = 1316.5 mph

The y component of the velocity is:

vᵧ = 889 sin 0° + 830 sin 59°

vᵧ = 711.4 mph

The speed is found with Pythagorean theorem:

v² = vₓ² + vᵧ²

v² = (1316.5 mph)² + (711.4 mph)²

v = 1496 mph

Rounded to two significant figures, the jet's speed relative to the ground is 1500 mph.

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An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

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3 years ago

A penny rolls along the table for a distance of 1.3 meters. Jackie pushes it 40 centimeters further in the same direction.

brilliants [131]

40 meters times 1 meter over 100 centimeters equals 0.4 meters. 1.3 meters + 40 centimeters =. 1.3 m + 0.4 m = 1.7 m. The answer is 1.7 meters

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3 years ago

Formula for weight on moon compared to weight on earth

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Weight on moon = (0.16) • Earth weight

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3 years ago

In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor

leonid [27]

Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

$W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}$ .....(I)

Where, $F(x)$ = net force

We know, the net force is the sum of forces.

So, $\sum{F}=ma$

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

$F(x)+(-F(f))=ma$

Here, frictional force is negative because F(f) acts against the F(x)

Now put the value in equation (I)

$W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

We need to find the value of $\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}$

Using newton's second law

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}$ ...(II)

We know that,

Acceleration is rate of change of velocity.

$a=\dfrac{dv}{dt}$

Put the value of a in equation (II)

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}$

$\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

Now, the work done by the net force on the block is,

$W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}$

The work done by the net force on the block is equal to the change in kinetic energy of the block.

Hence, The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

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3 years ago

Figure 3 shows a bicycle of mass 15 kg resting in a vertical position, with the front and back

Vinil7 [7]

Explanation:

There are three forces on the bicycle:

Reaction force Rp pushing up at P,

Reaction force Rq pushing up at Q,

Weight force mg pulling down at O.

There are four equations you can write: sum of the forces in the y direction, sum of the moments at P, sum of the moments at Q, and sum of the moments at O.

Sum of the forces in the y direction:

Rp + Rq − (15)(9.8) = 0

Rp + Rq − 147 = 0

Sum of the moments at P:

(15)(9.8)(0.30) − Rq(1) = 0

44.1 − Rq = 0

Sum of the moments at Q:

Rp(1) − (15)(9.8)(0.70) = 0

Rp − 102.9 = 0

Sum of the moments at O:

Rp(0.30) − Rq(0.70) = 0

0.3 Rp − 0.7 Rq = 0

Any combination of these equations will work.

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2 years ago