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Elenna [48]
3 years ago
11

Carol drops a stone in a mine shaft 122.5 metres deep.How

Physics
1 answer:
Verizon [17]3 years ago
7 0

Answer:

 T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken  = ?

time taken by the stone to reach at the bottom

using equation of motion

s = u t + \dfrac{1}{2}gt^2

initial speed , u = 0 m/s

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 122.5}{9.8}}

       t = 5 s

time taken by the sound to travel

    d =v x t

 t = \dfrac{d}{v}

 t = \dfrac{122.5}{340}

    t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

 T = 5.36 s

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A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark
Digiron [165]

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, <u>16 cm</u> is the total distance.

3 0
3 years ago
Which statement describes the relationship between energy and entropy in the universe?
rusak2 [61]

In fact, entropy of an isolated system never decreases (2nd law of thermodynamics), unless some external energy is provided in order to "restore" order in the system and decrease its entropy.

(note that when external energy is added to the system, it is no longer "isolated").


*This is only true if the question is referring to a certain system within the universe. If we are considering the universe itself as the system, then this option is no longer correct, because no external energy can be provided to the universe, and since the universe is an isolated system, its entropy can never decrease. If we are considering the universe itself as the system, none of the options is true.




8 0
3 years ago
Read 2 more answers
A motorcycle traveling at 25 m/s accelerates ya a rate of 7.0 m/s2 for 6.0 seconds. What is the final velocity of the motorcycle
frutty [35]

First write down all your known variables:

vi = 25m/s

a = 7.0m/s^2

t = 6.0s

vf = ?

Then choose the kinematic equation that relates all the variables and solve for the unknown variable:

vf = vi + at

vf = (25) + (7.0)(6.0)

vf = 67m/s

The final velocity of the motorcycle is 67m/s.

4 0
3 years ago
Find the electric field a distance z above a circular ring carrying a constant line charge. For extra credit you may derive the
lora16 [44]

Answer:

F = 8.23 × 10⁻⁸N

Explanation:

F= kq²/r²

k= 9×10⁹Nm²/C²

q= 1.6×10⁻¹⁹C

r= 5.29×10⁻¹¹m

F= 9×10⁹× (1.6×10⁻¹⁹)²/  (5.29×10⁻¹¹)²

F = 8.23 × 10⁻⁸N

4 0
3 years ago
A block is attached to a spring, with spring constant k, which is attached to a wall. it is initially moved to the left a distan
Ghella [55]

Answer:

x(t) = d*cos ( wt )

w = √(k/m)

Explanation:

Given:-

- The mass of block = m

- The spring constant = k

- The initial displacement = xi = d

Find:-

- The expression for displacement (x) as function of time (t).

Solution:-

- Consider the block as system which is initially displaced with amount (x = d) to left and then released from rest over a frictionless surface and undergoes SHM. There is only one force acting on the block i.e restoring force of the spring F = -kx in opposite direction to the motion.

- We apply the Newton's equation of motion in horizontal direction.

                             F = ma

                             -kx = ma

                             -kx = mx''

                              mx'' + kx = 0

- Solve the Auxiliary equation for the ODE above:

                              ms^2 + k = 0

                              s^2 + (k/m) = 0

                              s = +/- √(k/m) i = +/- w i

- The complementary solution for complex roots is:

                              x(t) = [ A*cos ( wt ) + B*sin ( wt ) ]

- The given initial conditions are:

                              x(0) = d

                              d = [ A*cos ( 0 ) + B*sin ( 0 ) ]

                              d = A

                              x'(0) = 0

                              x'(t) = -Aw*sin (wt) + Bw*cos(wt)

                              0 = -Aw*sin (0) + Bw*cos(0)

                              B = 0

- The required displacement-time relationship for SHM:

                               x(t) = d*cos ( wt )

                               w = √(k/m)

3 0
3 years ago
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