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1 year ago

Study the diagram and calculate the effort required to balance the load

Physics

1 answer:

Lisa [10]1 year ago

6 0

Answer:

Solution given;

load =600N

effort=?

load distance: 0.6m

effort distance;2.6m

we have

load *load distance :effort *effort distance

0.6*600=effort *2.6

360/2.6=effort

effort:138.46N

Required effort is 138.5N.

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Answer: a)

Explanation:

The buoyant force, as stated by Archimedes’ principle, is equal to the weight of the liquid that occupies the same volumen as the submerged object, as follows:

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Instead, silver density is larger than water density, which explains why the pure silver ingot sinks.

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2 years ago

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

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As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

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so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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1 year ago

Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot

Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

Explanation:

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If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0$

$(m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0$

Taking out ‘mg’ as common and we get

$2 x-50=0$

$2 x=50$

$x=\frac{50}{2}=25 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}$

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

$4 m g(x)+(m g)(x-50 c m)=0$

$4 m g x+m g x-50 m g=0$

Taking out ‘mg’ as common and we get

$5 x=50$

$x=\frac{50}{5}=10 \mathrm{cm}$

Hence, the stick should be pivoted at a distance of,

$x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}$

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

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2 years ago

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At medium temperature above 10 degrees Celsius to below 40 degrees Celsius (e.g 27 degrees Celsius), the photosynthetic enzymes work at their optimum levels, so photosynthesis proceeds.

At a temperature above 40 degrees Celsius, the enzymes that carry out photosynthesis lose their shape and functionality, and the photosynthetic rate declines rapidly.

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1 year ago

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arsen [322]

Answer:

Explanation:

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2 years ago

Study the diagram and calculate the effort required to balance the load​

Study the diagram and calculate the effort required to balance the load