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Anettt [7]
3 years ago
9

Call 7744882898 ::::::

Physics
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

what does that mean?

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EDGE 2021 WILL GIVE BRAINLIEST
defon

Answer:

B A new compound is formed

8 0
3 years ago
Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an
sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The <em><u>work </u></em>is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the <em>two tugboats "pull the tanker a distance 0.83km toward the north"</em>, that is the displacement, and you have to calculate the net force toward the north.

3. <u>Tugboat #1</u>.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N

4. <u>Tugboat #2</u>:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle:  = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) =  1.8×10⁶N  × cos(11°) = 1.77×10⁶N =

5. <u>Total net force, Fn</u>:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. <u>Work, W</u>:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0
3 years ago
A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the ar
kramer
<span>If there isn't any force then the normal contact force will be 


N=m*g=7.5*9.81=73.58N 

which is 73.58-23=50.58N less 

so, there the person must pull at 23 degree upward 

break down the tension in two components, vertical and horizontal. 


vertical tension= 50.58=T*sin23 

T=50.58/sin23=129.45N</span>
7 0
3 years ago
Read 2 more answers
A 6.00 A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5×1028 free elect
navik [9.2K]

Answer:

Explanation:

Current, I = 6 A

diameter of wire, d = 2.05 mm

number of electrons per unit volume, n = 8.5 x 10^28

If the diameter is doubled,

The resistance of the wire is inversely proportional to the square of the diameter of the wire, so the resistance is  one forth an the current is directly proportional to the diameter of the wire so the current is four times the initial value.  

8 0
3 years ago
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
3 years ago
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