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3 years ago

a Stone is dropped into a deep well and is heard to hit the water 3.41s after being dropped. determine the depth of the well

1 answer:

5 0

Using SUVAT

U = 0 m/s
A = g = 9.8 ms^-2
T = 3.41 s
S = ?

s = ut + 1/2at^2
s = (0)(3.41) + 1/2(9.8)(3.41^)
s = 1/2(9.8)(3.41^2)
s = 1/2(9.8)(11.6281)
s = 56.97769 m

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What will happen two temperatures if you increase particle motion

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If the temperature is increased the particles gain more kinetic energy or vibrate faster. This means that they move faster and take more space.

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3 years ago

Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the

cupoosta [38]

The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

$\mathbf{v_f ^2 = v_i^2 + 2gh}$

$\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}$

$\mathbf{v_f ^2 =151 m/s}$

$\mathbf{v_f =\sqrt{151 m/s}}$

$\mathbf{v_f =12.29 \ m/s}$

From the equation of continuity:

$\mathbf{A_iV_i = A_fV_f}$

$\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}$

$\mathbf{ r^2_iV_i = r^2_fV_f}$

$\mathbf{ (\dfrac{10}{2})^2\times 2.0 = r_f^2 \times 12.29}$

$\mathbf{ 50 = 12.29 \times r_f^2}$

$\mathbf{ r_f= \sqrt{\dfrac{50}{12.29} }}$

$\mathbf{ V_f= 2.02 \ cm }$

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

brainly.com/question/10822213

6 0

2 years ago

A string that is under 54.0 N of tension has linear density 5.20 g/m . A sinusoidal wave with amplitude 2.50 cm and wavelength 1

kicyunya [14]

Answer:

8.89288275 m/s

Explanation:

F = Tension = 54 N

$\mu$ = Linear density of string = 5.2 g/m

A = Amplitude = 2.5 cm

Wave velocity is given by

$v=\sqrt{\frac{F}{\mu}}\\\Rightarrow v=\sqrt{\frac{54}{5.2\times 10^{-3}}}\\\Rightarrow v=101.90493\ m/s$

Frequency is given by

$f=\frac{v}{\lambda}\\\Rightarrow f=\frac{101.90493}{1.8}\\\Rightarrow f=56.61385\ Hz$

Angular frequency is given by

$\omega=2\pi f\\\Rightarrow \omega=2\pi 56.61385\\\Rightarrow \omega=355.71531\ rad/s$

Maximum velocity of a particle is given by

$v_m=A\omega\\\Rightarrow v_m=0.025\times 355.71531\\\Rightarrow v_m=8.89288275\ m/s$

The maximum velocity of a particle on the string is 8.89288275 m/s

5 0

3 years ago

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below: