Pls help will mark Brainliest

theres two cups on a board - one half full and one empty. which one will fall off quicker while lifting the board up?

34kurt

The half empty will fall quicker

Please leave a thanks <3

6 0

2 years ago

A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv

iren2701 [21]

Explanation:

(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively. Each force must be represented by a distinct arrow starting on and pointing away from the dot.

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

(b) i. Derive an expression for the speed of the sphere at point A.

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

ii. Derive an expression for the normal force the track exerts on the sphere at point A.

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin. The sphere is replaced with a hoop of the same mass and radius. Will the value of Hmin increase, decrease, or stay the same? Justify your answer.

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0. Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5. For a hoop, k = 1. As k increases, H increases.

(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above. The track makes an angle of θ above the horizontal at point C. Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate. Calculate the maximum height above the bottom of the loop that the sphere will reach.

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ. At the maximum height, the vertical velocity is 0. During this time, the sphere is in free fall. The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0

3 years ago

PLEASE HELP 15 POINTS The electric field around a positive charge is shown in the diagram. Describe the nature of these lines. P

DENIUS [597]

The field lines spread apart as we move away from the charge, and they point away from the charge

Explanation:

The electric field produced by a single-point positive charge is a radial field, whose strength is given by the equation

$E=k\frac{Q}{r^2}$

where

k is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge at which the field is calculated

There are two pieces of information given by the field lines shown in the graph:

The spacing between the lines gives an indication of the strength of the field: the closer to each other they are, the stronger the field. In this case, as we move away from the charge, the spacing between the lines increases, and this means that the field becomes weaker (in fact, it follows an inverse square law, $E\propto \frac{1}{r^2}$
The direction of the lines gives the direction of the electric field, which points away from the central charge. This is because the direction of the electric field corresponds to the direction of the force that a positive test charge would feel when immersed in the electric field: in this case, if we place a positive test charge in this field, then it would get repelled away from the central charge (remember that the electric force between two positive charges is repulsive), and therefore, the direction of the electric field is away from the central charge.

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0

3 years ago

Determine the Specific Heat Capacity (Cs) of the liquid. Where: Mass of liquid (M₂) = 0.0012Kg, Initial temperature (₁)=23°C, Fi

alexandr1967 [171]

The Specific Heat Capacity of the liquid is 43.846 KJ/(Kg K).

It is given that Mass of liquid (M) = 0.0012Kg, Initial temperature (T1) = 23°C, Final temperature (T2) = 62°C, Voltage (V) = 8v, Current (I) = 0.95A, Time (t) = 270s

The quantity of heat that must be applied to an object in order to cause a unit change in temperature is known as the heat capacity or thermal capacity of that object.
We know that heat capacity for a substance is :
H = m*C*ΔT - equation (1)

When a conductor is subjected to current flow, the conductor's free electrons are set in motion and collide with one another. Moving electrons experience kinetic energy loss and partial thermal energy conversion as a result of the collision. This impact of current is referred to as its heating effect.
Due to electric current, heat energy is :
H = Power * Time
H = Current * Voltage * Time
H = I*V*t - equation (2)