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2 years ago

Need Help !

Physics

1 answer:

Bad White [126]2 years ago

3 0

Answer: 5.96m/s

Explanation:

Given the following :

Mass of car (m) = 1500kg

Velocity (V) = 5.25m/s

Forward force of engine = 1250N

Diatance moved = 4.8m

Final Velocity =?

Final kinetic energy = Initial kinetic energy + work done by engine

Initial kinetic energy = 0.5 × mass × velocity^2

Initial kinetic energy = 0.5 × 1500 × 5.25^2

Initial kinetic energy = 20671.875 J

Work done by engine = Force × distance

Work done by engine = 1250 × 4.8 = 6000J

Final kinetic energy = (20671.875 + 6000) J

= 26671.875 J

From kinetic energy = 0.5mv^2

26671.875 = 1/2 × 1500 × v^2

53343.75 = 1500v^2

v^2 = 35.5625

v = sqrt(35.5625)

v = 5.96m/s

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QUESTION 10

Elena L [17]

The maximum value of θ of such the ropes (with a maximum tension of 5,479 N) will be able to support the beam without snapping is:

$\theta =37.01^{\circ}$

We can apply the first Newton's law in x and y-direction.

If we do a free body diagram of the system we will have:

x-direction

All the forces acting in this direction are:

$T_{1}sin(\theta)-T_{2}sin(\theta)=0$ (1)

Where:

T(1) is the tension due to the rope 1
T(2) is the tension due to the rope 2

Here we just conclude that T(1) = T(2)

y-direction

The forces in this direction are:

$T_{1}cos(\theta)+T_{2}cos(\theta)-W=0$ (2)

Here W is the weight of the steel beam.

We equal it to zero because we need to find the maximum angle at which the ropes will be able to support the beam without snapping.

Knowing that T(1) = T(2) and W = mg, we have:

$T_{1}cos(\theta)+T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)-m_{steel}g=0$

$2T_{1}cos(\theta)=m_{steel}g$

T(1) must be equal to 5479 N, so we have:

$cos(\theta)=\frac{m_{steel}g}{2T_{1}}$

$cos(\theta)=\frac{892*9.81}{2*5479}$

$cos(\theta)=0.80$

Therefore, the maximum angle allowed is θ = 37.01°.

You can learn more about tension here:

brainly.com/question/12797227

I hope it helps you!

8 0

2 years ago

3 3

olganol [36]

Answer:

I Dont know u answer it

Explanation:

8 0

2 years ago

(1 point) A pilot flies in a straight path for 1 hour and 30 min. She then makes a course correction, heading 10 degrees to the

kati45 [8]

Answer:

The pilot is 2214.22 miles from her starting position

Explanation:

Since the pilot is traveling at a constant speed of 635 mph, the total distance traveled can be easily found as follows:

$1stleg: d_{1} = 1.5*635 = 952.5 mi\\2ndleg: d_{2} = 2*635 = 1270 mi$

There was a 10 degrees deviation, so the angle between the trajectory of both legs is 170 degrees.

The distance we need to find is that from the start of the first leg to the end of the second leg, those three distances form a triangle and since the side we're interested in is opposite to the 170 degrees angle, we can determine its length by the law of cosines:

$d_{3}^{2} =d_{1}^{2} + d_{2}^{2} -2*d_{1}*d_{2}*cos(170)\\d_{3}^{2} =952.5^{2} + 1270^{2} -2*952.5*1270*cos(170)\\d_{3}=2214.22mi$

The pilot is 2214.22 miles from her starting position

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2 years ago

A proton, traveling with a velocity of 3.7 × 106 m/s due east, experiences a magnetic force that has a maximum magnitude of 5.4

Vikki [24]

The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by
$F=qvBsin\theta=qvB$
since $\theta = 90^{\circ}$ , because the velocity v and the force F in this problem are perpendicular, and so also the angle $\theta$ between the velocity and the magnetic field B should be $90^{\circ}$ .

Let's find the magnitude of the magnetic field; this is given by
$B= \frac{F}{qv}= \frac{5.4\cdot 10^{-14}N}{1.6\cdot 10^{-19}C \cdot 3.7\cdot 10^6 m/s}=0.091 T$

To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force

Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T

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3 years ago

The primary colors of light are A. red, yellow, and blue. B. red, white and blue. C. blue, green, and red. D. cyan, magenta, and

zhuklara [117]

I think it is A I hope I helped you

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2 years ago

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