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3 years ago

An airplane travels 4000m in 16 seconds on a heading of 35 whats is its velocity

1 answer:

6 0

In the question, you just gave a complete and detailed
description of the plane's velocity vector:

4,000/16 meters/second , heading 35 degrees .

You might want to simplify the speed and make it a unit rate,
but otherwise, it's perfect.

250 meters/second, heading 35 degrees .

You might be interested in

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts help.

GrogVix [38]

Answer:

$H = 532 m$

Explanation:

When teacher falls into the cliff then she shout for Help at the same time

so here we know that sound will go down and reflect back up

so here in 3 s distance traveled by the sound

$d = vt$

$d = 340 \times 3$

$d = 1020 m$

now in the same time the distance that teacher will fall down is given as

$d_1 = \frac{1}{2}gt^2$

$d_1 = \frac{1}{2}(9.81)(3^2)$

$d_1 = 44.1 m$

now total distance traveled by teacher and sound in 3 s

$d_{total} = d + d_1$

$d_{total} = 1020 + 44.1$

this total distance must be equal to twice the height of the cliff

$2H = 1064.1 m$

$H = 532 m$

7 0

3 years ago

Read 2 more answers

A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago

Which of the following mathematical operations are and are not not allowed on two quantities with different unit dimensions?

Katyanochek1 [597]

The following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

To answer the question, we have to know what mathematical operations are.

<h3>What are Mathematical operations?</h3>

Mathematical operations are operations which are performed to change the value of a variable. The arithmetic mathematical operations we have are

addition,
subtraction,
multiplication,
subtraction and
equality which shows the relationship between two quantities.

Now, for two quantities with different unit dimensions, addition, subtraction and equality are not allowed because these mathematical operations require the quantities to be in the same unit dimensions

Also, for two different quantities with differnt unit dimensions, multiplication and division are allowed because these mathematical operations do not require the quantities to be in the same unit dimensions.

So, the mathematical operations that are allowed are multiplication and division while the mathematical operations that are not allowed are addition, subtraction and equality.

So, the following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

Learn more about mathematical operations here:

brainly.com/question/26112472

4 0

2 years ago

A 269-turn solenoid is 102 cm long and has a radius of 2.3 cm. It carries a current of 3.9 A. What is the magnetic field inside

RUDIKE [14]

Answer:

Magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

Explanation:

Given;

number of turns of solenoid, N = 269 turn

length of the solenoid, L = 102 cm = 1.02 m

radius of the solenoid, r = 2.3 cm = 0.023 m

current in the solenoid, I = 3.9 A

Magnitude of the magnetic field inside the solenoid near its centre is calculated as;

$B = \frac{\mu_o NI}{l} \\\\$

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

$B = \frac{4\pi*10^{-7} *269*3.9}{1.02} \\\\B = 1.293 *10^{-3} \ T$

Therefore, magnitude of the magnetic field inside the solenoid near its centre is 1.293 x 10⁻³ T

8 0

3 years ago

A cross-country skier slides horizontally along the snow and comes to rest after sliding a distance of 11 m. If the coefficient

Basile [38]

Answer:

v_o = 4.54 m/s

Explanation:

<u>Knowns </u>

From equation, the work done on an object by a constant force F is given by:

W = (F cos Ф)S (1)

Where S is the displacement and Ф is the angle between the force and the displacement.

From equation, the kinetic energy of an object of mass m moving with velocity v is given by:

K.E=1/2m*v^2 (2)

From The work- energy theorem , the net work done W on an object equals the difference between the initial and the find kinetic energy of that object:

W = K.E_f-K.E_o (3)

<u>Given </u>

The displacement that the sled undergoes before coming to rest is s = 11.0 m and the coefficient of the kinetic friction between the sled and the snow is μ_k = 0.020

<u>Calculations</u>

We know that the kinetic friction force is given by:

f_k=μ_k*N

And we can get the normal force N by applying Newton's second law to the sled along the vertical direction, where there is no acceleration along this direction, so we get:

∑F_y=N-mg

N=mg

Thus, the kinetic friction force is:

f_k = μ_k*N

Since the friction force is always acting in the opposite direction to the motion, the angle between the force and the displacement is Ф = 180°.

Now, we substitute f_k and Ф into equation (1), so we get the work done by the friction force:

W_f=(f_k*cos(180) s

=-μ_k*mg*s

Since the sled eventually comes to rest, K.E_f= 0 So, from equation (3), the net work done on the sled is:

W= -K.E_o

Since the kinetic friction force is the only force acting on the sled, so the net work on the sled is that of the kinetic friction force

W_f= -K.E_o

From equation (2), the work done by the friction force in terms of the initial speed is:

W_f=-1/2m*v^2

Now, we substitute for W_f= -μ_k*mg*s, and solving for v_o so we get:

-μ_k*mg*s = -1/2m*v^2

v_o = √ 2μ_kg*s

Finally, we plug our values for s and μ_k, so we get:

v_o = √2 x (0.020) x (9.8 m/s^2) x (11.0 m) = 4.54 m/s

v_o = 4.54 m/s

6 0

3 years ago

Read 2 more answers