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Finger [1]
3 years ago
9

A 21.7kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.2m/s. How much thermal energy due to frictio

n was generated in this process?
Physics
1 answer:
nadya68 [22]3 years ago
5 0

Answer:

Subtract the kinetic energy at the bottom from the potential energy loss. The remainder becomes frictional heat.

Potential energy loss:

M g H = 21.7*9.81*3.5 = 745.1 J

Kinetic energy at bottom of slide:

= (1/2) M v^2 = 52.5 J

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

h=11.653\ m

Hence, The maximum height above the point of release is 11.653 m.

3 0
3 years ago
Equipotential surfaces a) make an angle of 45 degrees with the electric field. b) are parallel to the electric field. c) are per
Schach [20]

Answer:

Option c) are perpendicular to the electric field

Explanation:

Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90^{\circ} to the equipotential surface.

Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.

Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.

7 0
3 years ago
If your parents were going through a divorce and you needed to talk to someone, who would be the best professional to see?Clinic
just olya [345]

psychologist counseling would be the correct answer I believe

3 0
3 years ago
Read 2 more answers
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.03 m/s on the horizontal section of a
seraphim [82]

Answer:

2.38 m/s, 4.31 m/s, lower

Explanation:

a)

Initial energy = final energy

½ m v₀² + ½ I ω₀² = mgh + ½ m v₁² + ½ I ω₁²

Since the ball is rolling without slipping, ω = v / r.

For a hollow sphere, I = ⅔ m r².

½ m v₀² + ½ (⅔ m r²) (v₀ / r)² = mgh + ½ m v₁² + ½ (⅔ m r²) (v₁ / r)²

½ m v₀² + ⅓ m v₀² = mgh + ½ m v₁² + ⅓ m v₁²

⅚ m v₀² = mgh + ⅚ m v₁²

⅚ v₀² = gh + ⅚ v₁²

v₀² = 1.2gh + v₁²

v₁ = √(v₀² − 1.2gh)

Given v₀ = 4.03 m/s, g = 9.80 m/s, h = 0.900 m:

v₁ = √((4.03)² − 1.2 (9.80) (0.900))

v₁ ≈ 2.38 m/s

At the top of the loop, the sum of the forces in the radial direction is:

∑F = ma

W + N = m v² / R

N = m v² / R - mg

N = m (v² / R - g)

Given v = 2.38 m/s, R = 0.450 m, and g = 9.80 m/s²:

N = m ((2.38)² / 0.450 - 9.80)

N = 2.77m

N ≥ 0, so the ball stays on the track.

b)

Initial energy = final energy

Borrowing from part a):

v₂ = √(v₀² − 1.2gh)

This time, h = -0.200 m:

v₂ = √((4.03)² − 1.2 (9.80) (-0.200))

v₂ ≈ 4.31 m/s

c)

Without the rotational energy:

½ m v₀² = mgh + ½ m v₁²

½ v₀² = gh + ½ v₁²

v₀² = 2gh + v₁²

v₁ = √(v₀² - 2gh)

This is less than v₁ we calculated earlier.

6 0
3 years ago
1. A 1.5 kg ball moves in a circle that is 0.5 m in radius at a speed of 5.1 m/s.
kolezko [41]

Answer: a= 52.02 m/s²

Fc= 78.03 N

Explanation: Solution attached:

3 0
3 years ago
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