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3 years ago

What is the mechanical advantage of the lever shown below?

Physics

1 answer:

den301095 [7]3 years ago

3 0

Answer:

It makes the job easier

Explanation:

lt makes the job easier it requires less work than what was originally,This requires not that much of a force need

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Why is it important to know the center of gravity of an object?

Blizzard [7]

Explanation:

The direction of the force of gravity through the body is downward, towards the centre of the earth and through the COG. This line of gravity is important to understand and visualise when determining a person's ability to successfully maintain Balance.

7 0

3 years ago

11. A 6.0‐m wire with a mass of 50 g, is under tension. A transverse wave, for which the frequency is 810 Hz, the wavelength is

Irina-Kira [14]

Answer:

Time will be 19 ms so option (a) is correct option

Explanation:

We have given that mass of wire m = 50 gram = 0.5 kg

Frequency f = 810 Hz

Wavelength = 0.4 m

Velocity is given by

$v=wavelength\times frequency=810\times 0.4=324m/sec$

Amplitude is given as d = 6 m

So time $t=\frac{distance}{velocity}=\frac{6}{324}=0.01851=19ms$

So option (a) is correct option

4 0

3 years ago

The energy that a substance contains is

Mamont248 [21]

The energy that substance contain is. B.heat

8 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

tekilochka [14]

Answer: a) It will take more time to return to the point from which it was released

Explanation: To determine how long it takes for the ball to return to the point of release and considering it is a free fall system, we can use the given formula:

$d=v_{0}.t + \frac{1}{2} .a.t^{2}$ , where:

d is the distance the ball go through;

v₀ is the initial velocity, which is this case is 0 because he releases the ball;

a is acceleration due to gravity;

t is the time necessary for the fall;

Suppose <em>h</em> is the height from where the ball was dropped.

On Earth:

h=0.t + $\frac{1}{2}.10.t^{2}$

h = 5t²

$t_{T}$ = $\sqrt{\frac{h}{5} }$

On the other planet:

h = 0.t + $\frac{1}{2}.30.t^{2}$

h = 15.t²

$t_{P}$ = $\sqrt{\frac{h}{15} }$

Comparing the 2 planets:

$\frac{t_{T} }{t_{P} } = \frac{\sqrt{\frac{h}{5} } }\sqrt{{\frac{h}{15} } }$

$\frac{t_{T} }{t_{P} }$ = $\sqrt{3}$ or $t_{T} = \sqrt{3}.t_{P}$

Comparing the two planets, on the massive planet, it will take more time to fall the height than on Earth. In consequence, it will take more time to return to the initial point, when it was released.

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3 years ago

IF YOU ANSWER THESE 2 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 15 POINTS!!!

lisabon 2012 [21]

Answer:

1. a

2. c [and maybe before]

Explanation:

Have a great day! Byeeee

6 0

3 years ago