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4 years ago

12.

Physics

2 answers:

KATRIN_1 [288]4 years ago

7 0

Answer:

31.25 is the correct answer

AveGali [126]4 years ago

7 0

Let t seconds be the time taken for the first stone to reach the ground.

Using s = ut + 0.5at^2, with s = H, u = 0, a = 10:

H = 0.5 x 10 x t^2

H = 5t^2

The time taken for the second stone to reach the ground will be (t - 1) seconds. The stone will fall from a height of H - 20 metres.

Use s = ut + 0.5at^2, with s = H - 20, u = 0, a = 10, t = (t - 1):

H - 20 = 0.5 x 10 (t - 1)^2

H - 20 = 5(t - 1)^2

We know that H = 5t^2:

5t^2 - 20 = 5(t - 1)^2

5t^2 - 20 = 5(t^2 - 2t +1)

5t^2 - 20 = 5t^2 - 10t + 5

-20 = -10t +5

10t = 25

t = 2.5

Substituting for H:

H = 5 x 2.5^2 = 31.25 metres

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A very fine thread is placed between two glass plates on one side and the other side is touching to form a wedge. A beam of mono

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Answer:

53.3micro meters

Explanation:

See attached file

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4 years ago

A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

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3 years ago

Which of these planets has characteristics most similar to Neptune?

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D. Saturn, Jupiter and Uranus

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3 years ago

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

Nesterboy [21]

Answer:

A.) 1372 N

B.) 1316 N

C.) 1428 N

Explanation:

Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:

a. The load moves downward at a constant velocity

At constant velocity, acceleration = 0

T - mg = ma

T - mg = 0

T = mg

T = 140 × 9.8

T = 1372N

b. The load accelerates downward at a rate 0.4 m/s??

Mg - T = ma

140 × 9.8 - T = 140 × 0.4

1372 - T = 56

-T = 56 - 1372

- T = - 1316

T = 1316N

C. The load accelerates upward at a rate 0.4 m/s??

T - mg = ma

T - 140 × 9.8 = 140 × 0.4

T - 1372 = 56

T = 56 + 1372

T = 1428N

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3 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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4 years ago