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3 years ago

5

What else did you change in the Levers lab when you changed the position of the fulcrum?

2 answers:

geniusboy [140]3 years ago

8 0

<h2>Answer:</h2>

<u>By changing the position of the fulcrum, the amount of effort Force also changes</u>

<h2>Explanation:</h2>

The mechanical advantage of a lever is changed when the distance between the pivot point (fulcrum) and the weight is changed. The way levers operate is by an effort applied at a point, which moves a load at another point through a balance point called the fulcrum. If the distance between the fulcrum and the effort is increased, the effort needed to lift a weight decreases proportionally.

konstantin123 [22]3 years ago

6 0

Answer:

the effort force

Explanation:

i just took the test

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How does an increase in greenhouse gases in the atmosphere cause an increase in global temperature?

daser333 [38]

Answer:

<h2>Greenhouse gases absorb and emit radiation and less heat dissipates to space</h2>

Explanation:

when heat is trapped it is conducted to the trapped area and gradually heats up its surrounding

8 0

3 years ago

Un automovil parte del reposo y acelera uniformemente hasta alcanzar una rapidez de 0,255km/h en un tiempo de 3/4 Minutos determ

Elden [556K]

Answer:

a = 1.5*10^-3 m/s^2

x = 0.033m = 3.3cm

Explanation:

To calculate the acceleration and the distance traveled by the car you use the following formulas:

$v=v_o+at$ (1)

$x=v_ot+\frac{1}{2}at^2$ (2)

v: final velocity = 0,255 km/h

vo: initial velocity = 0 m/s

t: time = 3/4 min

a: acceleration = ?

x: distance

In order to use the equations (1) and (2) you first convert the units of the final velocity to m/s, and the time to seconds.

$v=0,255\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}\\\\v=0.07m/s\\\\t=\frac{3}{4}min*\frac{60s}{1min}=45s$

Next, you solve the equation (1) for the acceleration a:

$a=\frac{v}{t}=\frac{0.07m/s}{45s}=1.5*10^{-3}\frac{m}{s^2}$

With this value of a you can calculate the distance traveled by the car, by using the equation (2):

$x=\frac{1}{2}(1.5*10^{-3}m/s^2)(45s)^2=0.033m=3.3cm$

hence, the acceleration of the car is 1.5*10^-3 m/s^2 and the distance traveled in 3/4 min is 0.033m

5 0

3 years ago

Suppose Car 1 collides with Car 2; Car 2 was not moving. In an ideal situation with no friction, according to the Law of Conserv

insens350 [35]

Your answer is C it is slightly reduced

4 0

3 years ago

Read 2 more answers

Cathy is moving into a tenth-floor apartment. She has a grand piano that weighs 570 pounds and is too big to fit in the elevator

Kazeer [188]

She will need pulleys. Nothing else will work

4 0

3 years ago

A 2.10 X 10^3 kg car starts from rest at the top of a 5.0

xeze [42]

Answer:

3.80 m/s

Explanation:

Solution: From the question we have:

A 2.10 x 10^3 kg car starts from rest at the top of a 5.0 m long driveway that is sloped at 20 degrees with the horizontal.

average friction force of 4.0 x 10^3 N impedes the motion

To find out: find the speed of the car at the bottom of the driveway.

=>h = 5×sin20 = 1.71 m.

PE = mgh = 2100×9.8×1.71 = 35,192 J.

KE = PE-work done by frictional force

KE=PE-4000×5

0.5mV^2 = 35192 - 4000×5

1050V^2 = 15,192

V^2 = 14.47

V = 3.80 m/s.

4 0

3 years ago