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Delvig [45]
3 years ago
11

At an air show, Julia saw a fighter jet fly by. Three seconds later she heard the sonic boom of the sound barrier being broken.

Did the jet break the sound barrier as it passed Julia?
Physics
2 answers:
Tpy6a [65]3 years ago
8 0
The jet did not break the sound barrier.
Anarel [89]3 years ago
3 0

The jet did not break the sound barrier at Julia's location. The jet broke the barrier before it reached Julia's location. This is proven by the delay between seeing the jet and hearing the sonic boom because the speed of light is faster than the speed of sound.

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A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi
Alex_Xolod [135]
  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0
3 years ago
It takes 6400 years for one gram of radium to decay away to only 1/16 (one-sixteenth) of a gram. The half-life of radium is
Vsevolod [243]
1/16........................................
4 0
3 years ago
Which formula can be used to find the tangential speed of an orbiting object?
Luden [163]
The correct formula for calculating the tangential speed of an orbiting object is V(t)=wr.
V(t)= Tangential Speed
w= Angular Velocity
r= Radius of the Path

Hope this helps.
6 0
3 years ago
Read 2 more answers
A 66-kg base runner begins his slide into second base when he is moving at a speed of 3.4 m/s. The coefficient of friction betwe
Helen [10]

Answer:

a ) = 381.48 J

b )=  84.25 cm

Explanation:

Kinetic energy of the runner

= 1/2 m v²

= .5 x 66 x 3.4²

= 381.48 J

The final kinetic energy of the runner is zero .

Loss of mechanical energy

= 381.48 J

This  loss in  mechanical energy is due to action of frictional force .

b )

Let s be the distance of slide

deceleration due to frictional force

= μmg/m

.7 x 66 x 9.8 / 66

a = - 6.86 m s⁻¹

v² = u² - 2 a s

0 = 3.4² - 2x6.86 s

s = 3.4² / 2x6.86

= .8425 m

84.25 cm

5 0
3 years ago
What is the value of work done on an object when a 50–newton force moves it 15 meters in the same direction as the force?
laila [671]

Answer:

Work done, W = 750 joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

It moves to a distance of, d = 15 meters

We need to find the work done on an object. We know that the product of force and distance covered is called the work done. As the force and the displacement are in same direction. So,

W=F\times d

W=50\ N\times 15\ m

W = 750 joules

So, the work done on an object is 750 joules. Hence, this is the required solution.

4 0
3 years ago
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