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3 years ago

A vector points -43.0 units

Physics

1 answer:

Naddika [18.5K]3 years ago

8 0

Answer:

Explanation:

To find the direction of this vector we need o find the angle that has a tangent of the y-component over the x-component:

$tan^{-1}(\frac{11.1}{-43.0})=-14.5$ but since we are in Q2 we have to add 180 degrees to that angle giving us 165.5 degrees

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If an object travels at a constant speed in a circular path, the acceleration of the object is:

Vilka [71]

Answer:

1- The acceleration of the object is larger in magnitude the smaller the radius of the circle.

Explanation:

The acceleration of an object in a circular path is

$a = \frac{v^2}{r}$

As can be seen from the equation, if the radius of the circle is decreases, the magnitude of the acceleration increases.

As for the direction of the acceleration, it is always towards the center, and it is always perpendicular to the direction of the velocity.

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3 years ago

Cleo stated that light travels through air in straight paths, and when it moves from air to water, light changes direction, spee

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Answer:

Light slows down when it moves from air into water.

Explanation:

It is the property of light that travels faster in a less dense medium.

In a more dense medium, the speed of the light slows down and bends towards the normal.

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3 years ago

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The brightness of a star is determined

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100% C . By size and distance

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Experiments and investigations must be ____. a. approved b. repeatable c. not reproducible d. accepted

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Experiments and investigations must be B. Repeatable.

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A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway

Andru [333]

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s

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3 years ago