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-BARSIC- [3]
4 years ago
13

The heaviest piece of equipment ever carried by plane was 12,400.05kg generator built in Germany in 1993.How far above the groun

d was the generator when the GPE was 91,700,000.00J
Physics
1 answer:
nadezda [96]4 years ago
7 0

Answer:

754.6 m

Explanation:

The GPE (Gravitational potential energy) of an object with respect to the ground is given by

GPE = mgh

where

m is the mass of the object

g = 9.8 m/s^2 is the acceleration due to gravity

h is the heigth above the ground

Here we have

m = 12,400.05 kg is the mass

GPE = 91,700,000.00J is the GPE

Solving the formula for h, we find the heigth:

h=\frac{GPE}{mg}=\frac{91,700,000.00J}{(12,400.05 kg)(9.8 m/s^2)}=754.6 m

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In a 2-dimensional coordinate system, which of these is an example of an object's position?
Serjik [45]

' B ' is the only choice on the list that tells you a definite place, where you could go to pick up the object or meet somebody.

But in order to use it, you also have to know where the origin of coordinates is ... the point (0, 0) .

__________

No, I'm wrong about that.  You have to know where SOME other point is, but that doesn't have to be (0, 0).

4 0
3 years ago
A rectangular pane of glass is 91.1 cm wide and 155.9 cm long, and its area is equal to the length multiplied by the width. Usin
Alexeev081 [22]
We are given:

A rectangular glass pane with

W = 91.1 cm
L = 155.9 cm

We are asked to determine the area of the rectangle. The formula of the area of the rectangle is

A = l x w
A = 91.1 cm x 155.9 cm
A = 14202.49 cm^2<span />
7 0
3 years ago
Read 2 more answers
A sound wave of the form s = sm cos(kx - ?t + f) travels at 343 m/s through air in a long horizontal tube. At one instant, air m
Naddik [55]

Answer:

960.24 Hz

Explanation:

Here is the complete question

A sound wave of the form

S=Smcos(kx−ωt+Φ)

travels at 343 m/s through air in a long horizontal tube. At one instant, air molecule A at x = 2.000 m is at its maximum positive displacement of 6.00 nm and air molecule B at x = 2.070 m is at a positive displacement of 2.00 nm. All the molecules between A and B are at intermediate displacements. What is the frequency of the wave?

Solution

Given x₁ = 2.0 m, x₂ = 2.070 m, maximum positive displacement s = 6.00 nm at x₁, positive displacement s = 2.00 nm at x₂, velocity of wave v = 343 m/s, maximum positive displacement s₀ = 6.00 nm

Let t₀ = 0 at x₁ = 2.0 m for maximum displacement.

So, s = s₀cos(kx−ωt+Φ)

     6 = 6cos(2k - 0 + Φ) = 6cos(2k + Φ)⇒ cos(2k + Φ) = 6/6 = 1

cos(2k + Φ) = 1 ⇒ (2k + Φ) = cos⁻¹ (1) = 0 ⇒ 2k + Φ = 0

Let t₀ = 0 at x₂ = 2.070 m for displacement s = 2.00 nm.

So, s = s₀cos(kx−ωt+Φ)

     2 = 6cos(2.070k - 0 + Φ) = 6cos(2.070k + Φ)⇒ cos(2.070k + Φ) = 2/6 = 1/3

cos(2.070k + Φ) = 1/3 ⇒ (2.070k + Φ) = cos⁻¹ (1/3) = 70.53 ⇒ 2.070k + Φ = 70.53.

We now have two simultaneous equations.

2k + Φ = 0   (1)

2.070k + Φ = 70.53.    (2)

Subtracting (2) -(1)

2.070k - 2k = 70.53

0.070k = 70.53

k = 70.53/0.070 = 1007.554π/180 rad/m = 17.59 rad/m

k = 2π/λ ⇒ λ = 2π/k

and frequency, f = v/λ = v/2π/k = kv/2π = 17.59 × 343/2π = 960.24 Hz

4 0
3 years ago
A boy lifted a 50 newton rock 1 meter. How much work was done?
dem82 [27]

Answer:

The answer is 50 Nm

Explanation:

<h3><u>Given</u>;</h3>
  • Applied Force = 50 Newton
  • Total Displacement = 1 meter
<h3><u>To </u><u>Find</u>;</h3>
  • Work done = ?

Here,

W = F • d

W = 50 • 1

W = 50 Nm

Thus, Work done is 50 Nm

<u>-TheUnknownScientist 72</u>

5 0
3 years ago
Read 2 more answers
Suppose a 0.04-kg car traveling at 2.00 m/s can barely break an egg. What is the min
Paha777 [63]

Answer: The correct answer is option (A).

Explanation:

Momentum of the car with 0.04 kg mass , which travelling with velocity of 2.00 m/s

P_1=mass\times velocity=m_1\times v_1=0.04 kg\times 2.00 m/s=8.00 kg m/s

Then the maximum speed of the another car in order to not to break the eggs will be same as first car:

P_1=P_2

0.08 kg m/s=m_2\times v_2=0.08 kg\times v_2

v_2=1 m/s

Speed slightly more than 1 m/s will increase the momentum of second car and the eggs will break. So, from the given options the minimum speed need by the second car will be 1.42m/s.

4 0
4 years ago
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