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3 years ago

Describe an object's velocity when an acceleration-time graph is zero?

Physics

2 answers:

svp [43]3 years ago

7 0

Anything times zero is zero

tangare [24]3 years ago

5 0

The principle is that the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (i.e., a horizontal line). If the acceleration is positive, then the slope is positive (i.e., an upward sloping line).

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What is the difference between Earth's magnetic and geographic poles? How do navigators take advantage of this?

Triss [41]

Earth's magnetic poles are located near, but not quite exactly at, our planet's geographic poles (the "spin axis" on a globe). Compasses work as direction-finding devices because Earth has a magnetic field. The needle of a compass, itself a small bar magnet, points in the direction of the magnetic pole.

5 0

3 years ago

What is the IMA of the following pulley system?

Mrac [35]

The IMA of pulley system can be defined as the ratio of output force to input force.

And IMA stands for Ideal Mechanical advantage

4 0

3 years ago

Read 2 more answers

A ball is rolled twice across the same level laboratory table and allowed to roll off the table and strike the floor. In each tr

schepotkina [342]

By definition we know that the distance is equal to the speed by time
d = v * t
Clearing the time we have
t = d / v
for conservation of energy. we have to for every attempt
mgh = (1/2) mv ^ 2
Clearing the speed
v = Root (2gh)
Then, substituting
t = d / v
t = h / (Root (2gh)
We conclude that the time is the same since it depends on the height of the table to the floor.

5 0

3 years ago

You have 2 resistors of unknown values you label Ra and Rb. You have an old battery and a multimeter you bought years ago for 7$

Veseljchak [2.6K]

Answer:

the answers, the correct one is D, Rb₂ = 29.97 ohm

Explanation:

For this exercise we use ohm's law and the equivalent resistance ratio for series and parallel circuits.

Serial circuit

(Ra + Rb) is = V

(Ra + Rb) 0.111 = 10

(Ra + Rb) = 10 / 0.111 = 90.09

parallel circuit

$\frac{1}{R} = \frac{1}{Ra} + \frac{1}{Rb}$

R = $\frac{Ra \ Rb}{Ra + Rb}$

\frac{Ra \ Rb}{Ra + Rb} i_p = V

\frac{Ra \ Rb}{Ra + Rb} 0.5 = 10

\frac{Ra \ Rb}{Ra + Rb} = 10 / 0.5 = 20

we write and solve our system of equations

Ra + Rb = 90.09

\frac{Ra \ Rb}{Ra + Rb} = 20

we solve for Ra in the first equation

Ra = 90.09 - Rb

RaRb = 20 (Ra + Rb)

we substitute Ra in the second equation

(90.09-Rb) Rb = 20 [(90.09-Rb) + Rb]

90.09 Rb - Rb² = 20 90.09

Rb² - 90.09 Rb + 1801.8 = 0

we solve the quadratic equation

Rb = [90.09 ± $\sqrt{90.09^2 - 4 \ 1801.8}$ ] / 2

Rb = [90.09 ± 30.15] / 2

Rb₁ = 60.12 ohm

Rb₂ = 29.97 ohm

the smallest value is Rb = 30 ohm

When checking the answers, the correct one is D

6 0

2 years ago

An 870 N firefighter, F_ff, stands on a ladder that is 8.00 m long and has a weight of 355 N, F_ladder. The weight of the ladder

castortr0y [4]

Answer:

Explanation:

Weight of the ladder is 355N

WL = 355N

The weight of the ladder acts at center of the ladder I.e at 4m from the bottom

Weight of firefighters is 870N

Wf = 870N

The fire fighter is at 6.3m from the bottom of the ladder.

N1 is the normal force exerted by the wall

N2 is the normal force exerted by the ground

Using Newton law

Check attachment

ΣFy = 0, since body is in equilibrium

N₂ - WL - Wf = 0

N₂ = WL + Wf

N₂ = 870 + 355

N₂ = 1225 N

This the normal force exerted by the ground on the wall.

Now to get N₁, let take moment about point A

So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.

ΣMa = 0

Clockwise moment is equal to anti-clockwise moment

Moment Is the produce of force and perpendicular distance.

M = F×r

So,

WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0

355•Cos50 × 4 + 870•Cos50 × 6.3 =

N₁•Sin50 × 8

1825.52 + 3523.12 = 6.13N₁

6.13N₁ = 5348.64

N₁ = 5348.64/6.13

N₁ = 872.54 N.

The normal force exerted by the wall is 872.54N

5 0

3 years ago