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stepan [7]
3 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]3 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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A tuning fork with a frequency of 404 Hertz is struck with a soft hammer. The fork is struck a second time by the
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3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
Three joules of work is needed to shift 10 C of charge from one place to another. The potential difference between the places is
dimaraw [331]

Answer:

The potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

Explanation:

Given

Work done W = 3J

Amount of Charge q = 10C

To determine

We need to determine the potential difference V between the places.

The potential difference between the two points can be determined using the formula

Potential Difference (V) = Work Done (W) / Amount of Charge (q)

or

\:V\:=\:\frac{W}{q}

substituting W = 3 and q = 10 in the formula

V=\frac{3}{10}

V=0.3 V

Therefore, the potential difference between the places is 0.3 V.

∴ 1st option i.e. 0.3V is the correct option.

4 0
2 years ago
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