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3 years ago

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An airplane in the process of taking off travels with a speed of 80 m/s at an angle of 15° above the horizontal. What is the gro

und speed of the airplane? O 80 m/s O 21 m/s O 77 m/s O 2.6 m/s

1 answer:

9966 [12]3 years ago

5 0

Answer:

Option C

Explanation:

given,

velocity of airplane = 80 m/s

angle with the horizontal = 15°

speed of the ground= ?

when the plane is taking off the horizontal component of the velocity is v cosθ

so,

ground speed of the airplane is = $v\times cos\theta$

= $80 \times cos 15^0$

v = 77.27 m/s

horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s

so, the correct option is Option C

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Answer:

The percentage error is given by 99.9 %

Explanation:

Given:

Mass of object $m = 50$ kg

Height $h = 110$ km

From the formula of potential energy,

$U = mgh$

Where $g = 9.8 \frac{m}{s}$

$U = 50\times 9.8 \times 110000$

Here true value of potential energy,

$U = 50 \times 9.8 \times 11 \times 10^{4}$

Approximate value of student,

$U = 50 \times 9.8 \times 110$

Absolute error is given by

= true value - approximate value

= $50 \times 9.8 \times 11 \times 10^{4} - 50 \times 9.8 \times 110$

= $53846100$

Hence percentage error,

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$= 0.999 \times 100$ %

$= 99.9$ %

Therefore, the percentage error is given by 99.9 %

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3 years ago

Consider a system to be one train car moving toward another train car at rest. When the trains collide, the two cars stick toget

pickupchik [31]

As per the question, the system consists of two cars.

The the masses of two cars are denoted as- $m_{1} \ and\ m_{2} \ respectively$

The two cars undergoes collision with each other. Here collision must be inelastic in nature.

Let the velocities of each car is denoted as- $v_{1} \ and\ v_{2} \ respectively$

It is given that masses of two cars and initial velocities-

$m_{1} =600 kg$ $v_{1} =4m/s$

$m_{2} = 400kg$ $v_{2} =0 m/s$

First we have to calculate the total momentum of the system before collision.

The momentum of a body is the product of its mass with velocity.

Momentum of car 1:

$p_{1} =m_{1} v_{1}$

$=600kg*4m/s$

$=2400\ kg\ m\ s^-1$

The momentum of car 2:

$p_{2} =m_{2} v_{2}$

$=400\ kg*0\ m/s$

$=\ 0\ kg\ m\ s^-1$

Now the total momentum of the system is :

$p_{1} +p_{2}$

$=[2400+0]\ kg\ m\ s^-1$

$=2400\ kg\ m\ s^-1$

Now we are asked to calculate the total momentum of the system after collision.

The basic condition in case of any type of collision is that it is in accordance with law of conservation of linear momentum.

Hence, the momentum will be conserved here.

It means the total momentum before collision must be equal to the total momentum after collision.

Hence, the correct answer to the question will be 2400 kg m/s.

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2 years ago

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If you and your friend are riding your bikes side by side and your reference points are each other, are you in motion? Explain y

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If your ref points are each other and not the surrounding environment, at the same speed you would both appear to be stationary

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Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

$i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

also, given at time 't' the current in the circuit is 55.0% of the peak current

thus

$i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}$

thus,

$0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

or

$0.55=sin(2\pi \times 16.9\times t)}$

or

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or

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4.53 billion years

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