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xenn [34]
4 years ago
7

An objects speed is the distance it travels____the amount of times it takes?

Physics
1 answer:
beks73 [17]4 years ago
8 0
Multiplied by; speed = distance x time
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A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
3 years ago
Which feature of a longitudinal wave corresponds to a trough in a transverse wave?
Rama09 [41]
It’s the crest, the crest is the top part of the wave and the trough is the bottom so they correspond
3 0
3 years ago
Read 2 more answers
In still​ water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles ​downstream,
Vladimir79 [104]
<span>The current is 6 miles per hour.
   Let's create a few equations:
 Traveling with the current:
 (18 + c)*t = 16

   Traveling against the current:
 (18 - c)*t = 8

   Let's multiply the 2nd equation by 2
 (18 - c)*t*2 = 16

   Now subtract the 1st equation from the equation we just doubled.
 (18 - c)*t*2 = 16
 (18 + c)*t = 16

   (18 - c)*t*2 - (18 + c)*t = 0
 Divide both sides by t
 (18 - c)*2 - (18 + c) = 0

   Now solve for c
 (18 - c)*2 - (18 + c) = 0
 36 - 2c - 18 - c = 0
 36 - 2c - 18 - c = 0
 18 - 3c = 0
 18 = 3c
  6 = c

   So the current is 6 mph.
   Let's verify that.
 (18 + 6)*t = 16
 24*t = 16
 t = 16/24 = 2/3

   (18 - 6)*t = 8
 12*t = 8
 t = 8/12 = 2/3

   And it's verified.</span>
4 0
3 years ago
Please help (choose a b or c)
Dafna11 [192]

I GEUSS A and COM is correct answer to this question

3 0
3 years ago
Read 2 more answers
An Earth satellite is orbiting at a distance from the Earth's surface equal to one Earth radius (4 000 miles). At this location,
Taya2010 [7]

Answer:0.25 times

Explanation:

Given

Distance of satellite from earth surface=Radius of earth

Force on the satellite is F=mg'

where g'=acceleration due to gravity at that point

Distance from center of Earth=R+R=2R

Gravitational Force is given by

F=\frac{GM_1M_2}{r^2}

Force F=mg'=\frac{GMm}{4R^2}-----1

Force on earth surface F=mg=\frac{GMm}{R^2}------2

Divide 1 and 2 we get

\frac{g'}{g}=\frac{R^2}{4R^2}

g'=\frac{g}{4}  

7 0
3 years ago
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