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NemiM [27]
2 years ago
9

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 130-kg laser sensor that measures the thi

ckness of the ice (see the drawing). The helicopter and the sensor both move only in the horizontal direction and have a horizontal acceleration of magnitude 2.57 m/s2. Ignoring air resistance, find the tension in the cable towing the sensor.
Physics
1 answer:
umka21 [38]2 years ago
8 0

Answer:T=1316.21 N

Explanation:

The tension has two components: Vertical and Horizontal. The

horizontal component is ma, the vertical component is mg. Using

Pythagoras theorem, we can find the tension as:

T=((ma)^2 (mg)^2)^(1/2)

So

T=((129*2.84)^2 (129*9.8)^2)^(1/2)

T=1316.21 N

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According to the first law of thermodynamics, the amount of work done by a heat engine equals the amount of
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C. thermal energy added to the engine minus the waste heat.

The law exemplifies conservation of energy. 
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Many people love roller coasters. Some people call them scream machines. They agree that riding a roller coaster is a thrilling
Digiron [165]

Answer:

gravity & inertia

Explanation:

what goes up must come down & an object in motion will stay in motion until stopped

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3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
nikklg [1K]

Answer:

Part a)

F_n = 306 N

Part B)

v = 12.1 m/s

Explanation:

Part A)

At the top of the hump the force on the rider is

1) Normal force

2) weight

so here we know that

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

F_n = (100)(9.81) - \frac{100(9^2)}{12}

F_n = 306 N

Part B)

At the top of the loop we will have

F_n + mg = \frac{mv^2}{R}

in order to remain in contact the normal force must be just greater than zero

so we will have

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

v = \sqrt{15\times 9.81}

v = 12.1 m/s

5 0
3 years ago
How do I find the tension? its supposed to be in the mid 40s​
babunello [35]

Answer:

a = 2.3 m/s²

T = 45 N

Explanation:

Draw a free body diagram for each mass.

For the mass on the incline, there are four forces:

Weight force mg pulling down.

Normal force N perpendicular to the incline.

Friction force Nμ pushing down the incline.

Tension force T pulling up the incline.

For the hanging mass, there are two forces:

Weight force Mg pulling down.

Tension force T pulling up.

Sum of the forces on the hanging mass in the -y direction:

∑F = ma

Mg − T = Ma

T = Mg − Ma

Sum of the forces on the sliding mass in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces on the sliding mass in the parallel direction:

∑F = ma

T − mg sin θ − Nμ = ma

Substitute:

Mg − Ma − mg sin θ − mgμ cos θ = ma

Mg − mg (sin θ + μ cos θ) = ma + Ma

Mg − mg (sin θ + μ cos θ) = (m + M) a

a = [ Mg − mg (sin θ + μ cos θ) ] / (m + M)

Plug in values:

a = [ 6.0×9.8 − 5.0×9.8 (sin 30° + 0.20 cos 30°) ] / (5.0 + 6.0)

a = 2.3 m/s²

Now find tension:

T = Mg − Ma

T = 6.0×9.8 − 6.0×2.3

T = 45 N

3 0
3 years ago
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