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stepan [7]
2 years ago
10

A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firml

y in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g

Physics
1 answer:
lapo4ka [179]2 years ago
3 0

Answer:

d) greater than g

Explanation:

You have to apply Newton's Second Law to have an idea about the value of <em>the block's aceleration.</em>

∑F = ma

Where m is the mass, a is the acceleration and F represents all the forces involved.

You have to draw a free body diagram for the block at the top of the loop. (See the attachment)

Applying Newton's Second Law:

y: N+mg=ma

Where N is the <em>normal force</em> (the contact force that the loop's surface applies to the block) and mg is the <em>weight.</em>

Therefore a= \frac{N}{m} + g

So, you can notice that <em>the magnitude of the block's acceleration</em> is greater than the acceleration of gravity because of the effect of the magnitude N/m.

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If a displacement vector has a negative x component and a positive y component,
bezimeni [28]

Answer:

The direction will be NE i.e North east.

Explanation:

From the question, it will be travelling in North east direction because the the negative X components in vector is in horizontal direction which is the Eastern direction when using the four cardinal points and the Y components, the positive is in North direction if you use the four cardinal points. Therefore, the vector with both X components and Y components will be travelling in Northeast direction.

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How much power is used by a hair dryer if it does 40,000 J of work in 40 seconds?
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Answer:

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6 0
3 years ago
We would like to place an object 45.0cm in front of a lens and have its image appear on a screen 90.0cm behind the lens. What mu
liubo4ka [24]

Answer:

the radii of curvature is 30 cm.

Explanation:

given,

object is place at = 45 cm

image appears at = 90 cm

focal length = ?

refractive index = 1.5

radii of curvature = ?

\dfrac{1}{f} = \dfrac{1}{u} +\dfrac{1}{v}

\dfrac{1}{f} = \dfrac{1}{45} +\dfrac{1}{90}

f = 30 cm

using lens formula

\dfrac{1}{f} = (n-1)(\dfrac{1}{R_1} -\dfrac{1}{R_2})

R_1 = R\ and\ R_2 = -R

\dfrac{1}{f} = (n-1)(\dfrac{1}{R} +\dfrac{1}{R})

R = (n -1)\ f

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R = 30 cm

hence, the radii of curvature is 30 cm.

3 0
3 years ago
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