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3 years ago

Which of the following is always a part of speed, velocity, and acceleration?

1 answer:

3 0

Speed = distance/time
Velocity = displacement/time
Acceleration = change in velocity/time

Time is a part of all three quantities.

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uniform ladder of length 6.0 m and weight 300 N leans against a frictionless vertical wall. The foot of the ladder isplaced 3.0

olganol [36]

Answer:

Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor

Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor

This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.

Fx1 = Fx2 = 86.6 N

3 0

3 years ago

after singing for a few minutes the soloist rejoins the choir a second choir, consist of 90 additional singers then joins in eac

vredina [299]

If the soloist produces "x" decibels and the 10-person choir produces "y" decibels, combined they will produce "x+y" decibels.

The second choir has 90 additional singers, we base our description on the first choir. If a 10-person choir produces "x+y" decibels, then the 90 person choir produces 10 (x+y) decibels.

7 0

3 years ago

A particle traveling in a circular path of radius 300 m has an instantaneous velocity of 30 m/s and its velocity is increasing a

hodyreva [135]

Answer:

5 m/s2

Explanation:

The total acceleration of the circular motion is made of 2 components: centripetal acceleration and linear acceleration of 4 m/s2. They are perpendicular to each other.

The centripetal acceleration is the ratio of instant velocity squared and the radius of the circle

$a_c = \frac{v^2}{r} = \frac{30^2}{300} = \frac{900}{300} = 3 m/s^2$

So the magnitude of the total acceleration is

$a = \sqrt{a_c^2 + a_l^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 m/s^2$

4 0

3 years ago

A car, starting from the origin, travels

sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0

3 years ago

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The sun is directly overhead at the equator on what day of the year

Phoenix [80]

Both equinoxes ... September 21 and March 21.

7 0

3 years ago

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