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Masja [62]
3 years ago
14

The speed of an object and the direction in which it moves constitute a vector quantity known as the velocity. an ostrich is run

ning at a speed of 14.0 m/s in a direction of 54.0 ° north of west. what is the magnitude of the ostrich's velocity component that is directed (a) due north and (b) due west?
Physics
1 answer:
Orlov [11]3 years ago
6 0
Velocity of an object is its rate of change of the object's position per interval of time. Velocity is a vector quantity which means that it consists of a magnitude and a direction. Magnitude is represented by the speed and the direction is represented by the angle. To determine the velocity components, we use trigonometric functions to determine the angle of the components. For the north component we, use the sine function while, for the west component, we use the cosine function. We calculate as follows:
north velocity component = (16.8 m/s) (sin 54°) = 16.4 m/s
<span>west velocity component = (16.8 m/s) (cos 54°) = 3.49 m/s</span>



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During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ
zavuch27 [327]

Answer: angular displacement in rad = 3038.45 rad

angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

3038.45/2π = 483.599 rev

7 0
3 years ago
Read 2 more answers
A uniformly charged rod (length = 2.0 m, charge per unit length = 3.0 nc/m) is bent to form a semicircle. What is the magnitude
Artist 52 [7]

Answer:

84.82N/C.

Explanation:

The x-components of the electric field cancel; therefore, we only care about the y-components.

The y-component of the differential electric field at the center is

$dE = \frac{kdQ }{R^2} sin(\theta )$.

Now, let us call \lambda the charge per unit length, then we know that

dQ = \lambda Rd\theta;

therefore,

$dE = \frac{k \lambda R d\theta }{R^2} sin(\theta )$

$dE = \frac{k \lambda  d\theta }{R} sin(\theta )$

Integrating

$E = \frac{k \lambda   }{R}\int_0^\pi sin(\theta )d\theta$

$E = \frac{k \lambda   }{R}*[-cos(\pi )+cos(0) ]$

$E = \frac{2k \lambda   }{R}.$

Now, we know that

\lambda = 3.0*10^{-9}C/m,

k = 9*10^9kg\cdot m^3\cdot s^{-4}\cdot A^{-2},

and the radius of the semicircle is

\pi R = 2.0m,\\\\R = \dfrac{2.0m}{\pi };

therefore,

$E = \frac{2(9*10^9) (3.0*10^{-9})   }{\dfrac{2.0}{\pi } }.$

$\boxed{E = 84.82N/C.}$

7 0
3 years ago
URGENT. Please help.
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6 0
3 years ago
Sunspots are _______ and _________ than the gases surrounding them.
Scrat [10]
The answer is B. darker; cooler
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Hope this helps

3 0
3 years ago
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A block of mass 22 kg is sliding along the ice at constant speed 5.0 m/s just ahead of it is q block of mass 29 kg sliding in th
mars1129 [50]

Answer:

Answer:

New speed of the 22-kg block is 1.57 m/s

Explanation:

Mass of block  

Mass of another block  

Initial speed of the block  

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For conservation of momentum, we have

Substitute all the values and solving for final speed of the 22kg block is

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Couldnt write the answer so check picture

8 0
3 years ago
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