If the area is 2 square meters (2 m²), what's the force of gravity acting on the column of water?

A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right

V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

g: Gravitational Acceleration

θ: Angle with the vertical

N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

m*g*cos(θ) - 0 = m*vf^2 / R

g*cos(θ) = vf^2 / R

vf^2 = R*g*cos(θ)

vf^2 = 1.45*32.2*cos(34)

vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

ΔK.E = ΔP.E

0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))

vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

vi^2 = 22.744

vi = 4.77 ft/s

4 0

3 years ago

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils, then listen to the frequency of the sound reflected from the

Triss [41]

Answer:

Check the explanation

Explanation:

This is the step by step explanation to the above question:

$v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]$

= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]

v = 344 m/s

vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))

= 0.74 m/s

8 0

3 years ago

Consider a golf ball with a mass of 45.9 grams traveling at 200 km/hr. If an experiment is designed to measure the position of t

Kipish [7]

Answer:

speed of golf ball is 1.15 × $10^{-30}$ m/s

and % of uncertainty in speed = 2.07 × $10^{-30}$ %

Explanation:

given data

mass = 45.9 gram = 0.0459 kg

speed = 200 km/hr = 55.5 m/s

uncertainty position Δx = 1 mm = $10^{-3}$ m

to find out

speed of the golf ball and % of speed of the golf ball

solution

we will apply here heisenberg uncertainty principle that is

uncertainty position ×uncertainty momentum ≥ $\frac{h}{4\pi }$ ......1

Δx × ΔPx ≥ $\frac{h}{4\pi }$

here uncertainty momentum ΔPx = mΔVx

and uncertainty velocity = ΔVx

and h = 6.626 × $10^{-34}$ Js

so put here all these value in equation 1

$10^{-3}$ × 0.0459 × ΔVx = $\frac{6.626*10^{-34}}{4\pi }$

ΔVx = 1.15 × $10^{-30}$ m/s

and

so % of uncertainty in speed = ΔV / m

% of uncertainty in speed = 1.15 × $10^{-30}$ / 55.5

% of uncertainty in speed = 2.07 × $10^{-30}$ %

3 0

3 years ago

How many stars are there in an Open Cluster?*

Step2247 [10]

Answer:

Very few

Most open clusters form with at least 100 stars

Brainly, please.

4 0

3 years ago

A box of mass 1 kg is hung from a spring scale. The reading on the spring scale is approximately 10N. What would be the reading

8090 [49]

The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N

8 0

3 years ago

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