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3 years ago

A Plane crashes into the side of a mountain. If the crash took place over 9

Physics

1 answer:

Julli [10]3 years ago

6 0

Given parameters:

Duration of crash = 9min

Velocity before crash = 2000 yards per second

Unknown:

Acceleration of the plane = ?

Solution:

Acceleration is the change of velocity with time.

Acceleration = $\frac{V - U}{t}$

where v is the final velocity

u is the initial velocity

t is the time taken

We need to convert the time into seconds from min;

60s = 1 min

So, 9min = 60 x 9 = 540s

So input the parameters and solve;

Acceleration = $\frac{2000 - 0 }{540}$ = 3.7yards/s²

The acceleration of the plane is 3.7yards/s²

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two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma

nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force = 3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos $\theta$

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)

35 = √49 x²

x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0

3 years ago

A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn

jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = $\sqrt{Cx^{2}+Cy^{2} }$

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

#SPJ1

6 0

2 years ago

A block with a mass of 9.00 kg is pulled at a constant speed across a horizontal tabletop with a spring scale. The scale reads 6

snow_tiger [21]

Answer:0.69

Explanation:

Coefficient of kinetic friction=f/R=61.8/90=0.69

7 0

3 years ago

When you whisper you produce a 10-dB sound.

masha68 [24]

All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>

4 0

4 years ago

Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope

Mandarinka [93]

Answer:

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s.

the rate at which they raise the temperature of the room.

2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

Explanation:

A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir to the work input to the refrigerator:

k=QC/W

power is defined as work per unit time

1.k=1500/750=2

2. 1200/400=3

3.2000/500=4

4.1000/250=4

5.1500/500=3

6.3000/1000=3

performance coefficient from largest to the smallest

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

5 0

3 years ago