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swat32
3 years ago
5

8. A 40.0kg block of metal is suspended from a scale and is immersed in water. The dimensions of the block are 12.0cm x 10.0cm x

10.0cm. The 12.0cm dimension is vertical, and the top of the block is 5.00cm below the surface of the water. (a) What are the forces exerted by the water on the top and bottom of the block? (Take P0 = 1.0130 X 105 N/m2 .) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.
Physics
1 answer:
Alja [10]3 years ago
5 0

Answer:

a.1017.9N

b.1029.7N

c.T=380.6N

d.11.8 N

Explanation:

(a) The absolute pressure at the level of the top of the block is

Po=atmospheric pressure

g=gravity

h1=height

rh0=density of water

P1=Po+rho*g*h1

1.013*10^5+1000(9.81)(0.05)

1.0179*10^5Pa

at the level of the bottom of the block we have

P2=Po+rho*g*h2

1.013*10^5+1000(9.81)(0.17)

1.0297* 10^5 Pa

the downward force exerted on the top by the water is

Ftop=P*A== × = 1.0179* 10^5* 0.100

= 1017.9 N

and the upward force the water exerts on the bottom of the block , which is the buoyant force

Fbot== × = 1.0297 10^5 * 0.100 m

= 1029.7 N

(b) The scale reading is the tension, T, in the cord supporting the block.

if the block is at equilibrium, then sum of vertical forces

EFy=T+Fbot-Ftop-mg=0

T=mg+Ftop-Fbot

T=40*9.81-(1029.7-1017.9)

T=380.6N

(c)  Archimedes principle state that, the buoyant force on the block equals the weight of the displaced water. Thus,

Buoyant force=rho *g*h

= = 1000* 0.100^2 * 0.120 m *9.80 m s=11.8 N

from the answer a Ftop-Fbot

1029.7-1017.9=11.8N, the same as the buoyant force

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aniked [119]

Answer:

a) 46.5º  b) 64.4º

Explanation:

To solve this problem we will use the laws of geometric optics

a) For this part we will use the law of reflection that states that the reflected and incident angle are equal

     θ = 43.5º

This angle measured from the surface is

     θ_r = 90 -43.5

     θ_s = 46.5º

b) In this part the law of refraction must be used

     n₁ sin θ₁ = n₂. Sin θ₂

     sin θ₂ = n₁ / n₂ sin θ₁

The index of air refraction is n₁ = 1

The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface

     θ_s = 90 - θ

     θ_s = 90- 43.5

     θ_s = 46.5º

     sin θ₂ = 1 / 1.68  sin 46.5

     sin θ₂ = 0.4318

     θ₂ = 25.6º

The angle with respect to the surface is

     θ₂_s = 90 - 25.6

     θ₂_s = 64.4º

measured in the fourth quadrant

3 0
3 years ago
This is for physics, i need help fast!!
jarptica [38.1K]
Speed = distance / time
S= 40 000m / 5400s
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A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect fricti
topjm [15]

Answer:

1.66 m/s

Explanation:

Work or kinetic energy = \frac{1}{2} mv^{2}

4864=\frac{1}{2} (3540)v^{2}

v = 1.66 m/s

6 0
3 years ago
The properties of elements can be predicted by which of the following?
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Answer:

1 Answer. The Periodic Table can predict the properties of new elements, because it organizes the elements according to their atomic numbers. Creating new elements is not a simple process. Scientists use a particle accelerator to smash light atoms into a thin metallic foil that contains heavier atoms.

Explanation:

6 0
3 years ago
The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

4 0
3 years ago
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