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Pavlova-9 [17]
4 years ago
8

A girl is whirling a ball on a string around her head in a horizontal plane. She wants to let go at precisely the right time so

that the ball will hit a target on the other side of the yard. When should she let go of the string?
Physics
1 answer:
hodyreva [135]4 years ago
5 0

Answer:

<u>The girl should leave the string a time when the ball is at a position tangent at that point of the circle passes through the location of target.</u>

Explanation:

The direction of an object moving in circular path changes endlessly with time. The direction of the object at some extent are within the direction of tangential drawn to the circle at that point,  

The  centripetal force on the ball becomes zero once the girl leaves the string. hence, the ball can go into the direction tangential to the circle at the point whenever she left the string.

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The intensity of sunlight incident on the surface of a solar panel on your roof is 1000 W/m^2. Calculate the sunlight energy abs
melisa1 [442]

Answer:

8.64 x 10^7 J

Explanation:

Intensity of sunlight, I = 1000 W/m^2

Length of panel, L = 6 m

Width of panel, W = 4 m

Area of the panel, A = L x W = 6 x 4 = 24 m^2

time, t = 1 hour = 3600 second

Energy = Intensity x area of panel x time

E = 1000 x 24 x 3600 = 8.64 x 10^7 J

5 0
4 years ago
The table shows the charges and the distance between five different pairs of objects.
zmey [24]

Answer:

the list of forces from greatest to least is Y, W, X, Z

Explanation:

Given data:

5 pairs of charged plates with force between each pair as : F, W, X,Y, Z

Also given are the charges q1 and q2 of plates in each pair and the distances between them (refer attachment)

To find: to arrange the force from greatest to least.

Initial step is to calculate the force value using the given data of charges and distance between plates.

We know that the force between two charged plates is given by the formula:

F = \frac{kq1q2}{r^{2} }

where,

q1 and q2 are charge of the two plates

r is the distance between two plates

k is coulomb constant

Since the problem involves comparison of forces, the k values is not necessarily discussed.

Now we have to calculate force values

F = \frac{kq^{2} }{d^{2} }

W = \frac{2kq^{2} }{d^{2} }

X = \frac{kq^{2} }{4d^{2} }

Y = \frac{3kq^{2} }{d^{2} }

Z = \frac{kq^{2} }{9d^{2} }

Assume values of q = 1 and d = 2, we get

F = 0.5k

W = 0.5k

X = 0.0625k

Y = 0.75k

Z =0.025k

F and W are though same, the calculated force value shows W = 2F

Hence W is greater than F.

Therefore, the list of forces from greatest to least is Y, W, F, X, Z

Hence answer is : Y, W, X, Z

7 0
3 years ago
Read 2 more answers
An elaborate pulley consists of four identical balls at the ends of spokes extending out from a rotating drum. A box is connecte
Klio2033 [76]

Answer:

its speed will be less than V

Explanation:

When the ball falls a distance d, its final kinetic energy plus rotational kinetic energy of the drum equals its initial potential energy.

K = U

With its speed V at the end of d, we have

1/2mV² + 1/2Iω² = mgd where I = rotational inertia of drum and balls, ω = angular speed of drum and balls and m = mass of box

1/2mV² + 1/2Iω² = mgd

1/2mV² = mgd - 1/2Iω²

V² = [2(mgd - 1/2Iω²)/m]

V = √[2(mgd - 1/2Iω²)/m]

When the four balls are moved inward closer to the drum, their rotational inertia increases and also its angular speed which thus causes an increase in rotational kinetic energy. But, since the box still falls the same distance of d, its final kinetic energy plus rotational kinetic energy of the drum plus balls still equals its initial potential energy

K = U

I' = new rotational inertia of drum and balls, ω' = new angular speed of drum and balls

With its new speed is now V' at the end of d,

1/2mV'² + 1/2I'ω'² = mgd

1/2mV'² = mgd - 1/2I'ω'²

V² = [2(mgd - 1/2I'ω'²)/m]

V' = √[2(mgd - 1/2I'ω'²)/m]

Since I' and ω' increase, the rotational kinetic energy of the drum and balls (1/2I'ω'²) increases. Thus, the difference (mgd - 1/2I'ω'²) < (mgd - 1/2Iω²) which implies that the kinetic energy of the box decreases. Hence, since its kinetic energy decreases, its speed V' also decreases.

So,  V' < V

So, its speed will be less than V.

3 0
3 years ago
How far (in meters) above the earth's surface will the acceleration of gravity be 85.0 % of what it is on the surface?
suter [353]

Answer:

X = 6910319.7 m

Explanation:

let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.

on the surface of the earth, the gravitational acceleration is given by:

g1 = GM/(r^2) = [(6.67408×10^-11)(5.972×10^24)]/[(6371×10^3)^2] = 9.82 m/s^2

at X meters above the earth's surface, g2 = 85/100(9.82) = 8.35m/s^2

then:

g2 = GM/(X^2)

X^2 = GM/g2

   X =  \sqrt{GM/g2}

       = \sqrt{(6.67408×10^-11)(5.972×10^24)/ 8.35

       = 6910319.7 m

Therefore, the acceleration of gravity becomes 85% of what it is on the surface at  6910319.7 m .

4 0
3 years ago
When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed sh
aev [14]

Answer:

Explanation:

Given

Initial speed is u=V

Maximum height of Pebble is H

Deriving maximum height of Pebble and considering motion in vertical direction

v^2-u^2=2 as

where v=final velocity

u=initial velocity

a=acceleration

s=Displacement

Final velocity will be zero at maximum height

0-(V)^2=2\times (-g)\cdot H

H=\frac{V^2}{2g}

i.e. maximum height is dependent on square of initial velocity

for twice the height

2H=\frac{(V')^2}{2g}

on comparing

V'=\sqrt{2}V

7 0
3 years ago
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