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Marizza181 [45]
3 years ago
5

An airliner of mass 1.70×105kg1.70×105kg lands at a speed of 75.0 m/sm/s. As it travels along the runway, the combined effects o

f air resistance, friction from the tires, and reverse thrust from the engines produce a constant force of 2.90×105N2.90×105N opposite to the airliner's motion. What distance along the runway does the airliner travel before coming to a halt?
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
4 0

Answer:

The airliner travels 1.65 km along the runway before coming to a halt.

Explanation:

Given

Resistive forces = (2.90 × 10⁵) N = 290000 N

Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg

Velocity of airliner = 75 m/s

Let the distance over moved by the airliner be equal to d

According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.

Work done by the resistive forces = (290000) × d = (290,000d) J

Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J

290000d = 478,125,000

d = (478,125,000/290,000)

d = 1648.7 m = 1.65 km

Hope this helps!!!

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harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

Torque = \vec{r} \times \vec{F}

Where \vec{r} is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), \vec{F} is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench.  By doing this we have that r (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

35cm * \dfrac{1m}{100cm} = 0.35m

Now using the torque formula:

Torque = \vec{r} \times \vec{F}

Torque = rFsin(\theta)}

Where \theta is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector  \theta = 90°, thus :

Torque = rFsin(\theta)}

65 N m = (0.35m)Fsin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{1300} {7}N

so the force is approximately 186 N.

4 0
4 years ago
Is it true that it takes more energy to vaporize 1 kg of saturated liquid water at 100 C than it would at 120 C?​
Vinvika [58]

Explanation:

Yes, it takes more energy to vaporize 1 kg of saturated liquid water at than it would at .

3 0
2 years ago
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A long solenoid has a radius of 4.0 cm and has 800 turns/m. If the current in the solenoid is increasing at the rate of 3.0 A/s,
kolbaska11 [484]

Answer:

Explanation:

Given that,

Radius of solenoid R = 4cm = 0.04m

Turn per length is N/l = 800 turns/m

The rate at which current is increasing di/dt = 3 A/s

Induced electric field?

At r = 2.2cm=0.022m

µo = 4π × 10^-7 Wb/A•m

The magnetic field inside a solenoid is give as

B = µo•N•I

The value of electric field (E) can

only be a function of the distance r from the solenoid’s axis and it give as,

From gauss law

∮E•dA =qenc/εo

We can find the tangential component of the electric field from Faraday’s law

∮E•dl = −dΦB/dt

We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.

E∮dl = −d/dt •(πr²B)

2πrE = −πr²dB/dt

2πrE = −πr² d/dt(µo•N•I)

2πrE = −πr² × µo•N•dI/dt

Divide both sides by 2πr

E =- ½ r•µo•N•dI/dt

Now, substituting the given data

E = -½ × 0.022 × 4π ×10^-7 × 800 × 3

E = —3.32 × 10^-5 V/m

E = —33.2 µV/m

The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m

where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.

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3 years ago
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What does it mean for their to be a net force on an object vs no net force? Which one is described as balanced forces and which
STatiana [176]
It’s like the force put against something or someone✨
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Vector A is in the direction 44.0 degrees clockwise from the y-axis. The x-component of A Ax=-15.0 m. Part A: What is the y-comp
Gnesinka [82]

Answer:

a)  Y component of the vector =15.54 m

b) Vector magnitude = 21.6 m

Explanation:

The given vector makes 44 degree angle with Y axis, as given. This is same as 90 -44 = 46 degrees with the horizontal or X axis.

b) X component of the given vector = A_{x} = A cos 46 =15

⇒ A = 15/cos 16 = 21.6 m = Total vector magnitude.

a) Y component of the vector = 21.6 sin 46 = 15.54 m

b) A = 21.6 m

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