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3 years ago

Speed is velocity in a given direction O True O False

Physics

2 answers:

Zarrin [17]3 years ago

8 0

Answer:

FALSE

Explanation:

Velocity = speed with direction.

Think of speed and direction like rockets and missiles. Rockets are not smart. Missiles are smart. Rockets go in one direction. Missiles can track their targets, they have a specific destination, a specific direction.

Velocity is often used in physics, because its almost useless to know how fast an object is going if you don't know which direction it is going.

Think of it like this. If the Weather man told you a hurricane was traveling at 30 miles an hour, but didn't tell you which direction it was going, you would have no idea where to run, or if it was going to hit you at all. However, if he told you it was going 30 miles an hour to the North, and you were West of it, you would be fine, and wouldn't have to worry.

irinina [24]3 years ago

5 0

False. speed equals velocity

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Anthony walks to the pizza place for lunch. He walk 1km east, then 1km south and then 1km east again. What distance did he cover

Komok [63]

Answer:

Pizza pizzza pizza

Explanation:

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3 years ago

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your an

AveGali [126]

Complete Question

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the wheel starts turning, you are at the very top.

What is your angular position 75 seconds after the wheel starts turning, measured counterclockwise from the top? Express your answer as an angle between 0∘ and 360∘. Express your answer in degrees.

Answer:

$\phi=123.75$

Explanation:

From the question we are told that:

Height $h=27m$

Period $T=32sec$

Time $t=75sec$

Generally the equation for angular velocity is mathematically given by

$\omega=\frac{2 \pi}{T}$

$\omega=\frac{2 \pi}{32}$

$\omega=0.196rad/s$

Therefore

$\theta=\omega t$

$\theta=0.196rad/s*75sec$

$\theta=843.75 \textdegree$

Therefore

$\phi=\theta-2(360)$

$\phi=123.75$

6 0

3 years ago

Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th

Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0

4 years ago

What is the scientific term for rocks formed from lava?

sergeinik [125]

Answer:

extrusive I'm pretty sure that's right

5 0

4 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

(c) -330.8 J

To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

5 0

3 years ago