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vladimir2022 [97]
1 year ago
14

when a metal sphere is dropped in to a tall cylinder containing liquid its acceleration is g÷2 (gravity over 2) show that : dens

ity of metal =2density of liquid​
Physics
1 answer:
Kryger [21]1 year ago
6 0

The density of the metal sphere is 2 times the density of the liquid as proved.

<h3>Net upward force acting on the metal sphere</h3>

The net upward force acting on the sphere as it is dropped into the liquid is calculated as follows;

F =  σVg - ρVg

ma =  σVg - ρVg  

where;

  • ρ is density of the liquid
  • σ is the density of the metal
  • a is acceleration of the metal

σV(a) =  σVg  - ρVg

σ(a) = σg  - ρg

σ(g/2) = σg -  ρg  

g(σ/2) = g(σ -  ρ)

σ/2 = σ -  ρ

σ/2 - σ = -  ρ

-σ/2 = -  ρ

σ = 2ρ

Thus, the density of the metal sphere is 2 times the density of the liquid as proved.

Learn more about density here: brainly.com/question/1354972
#SPJ1

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How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav
JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

so maximum height on moon is 6 times more than the maximum height on Earth

8 0
2 years ago
A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u
kenny6666 [7]

Answer:

induced emf =  28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt  = Adb/dt

induced emf =  A × ΔB / Δt

induced emf =  πr² × ΔB / Δt

induced emf =  π (0..65)² × ( 0.61 - (-0.28))  / 0.13

induced emf =  0.0286538 V

so induced emf =  28.65 mV

3 0
3 years ago
A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe
KIM [24]

Answer: V = 15 m/s

Explanation:

As  stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately.

8 0
2 years ago
A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is ex
Delvig [45]

Answer:

Explanation:

Impulse results in a change of momentum

FΔt = mΔV

F = mΔV/Δt

The impulse acting on the hammer will equal the impulse acting on the nail

If we assume upward is the positive direction

F = m(vf - vi)/t

F = 1.2(1.0 - (-1.5)) / 0.001

F = 3000 N

7 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
3 years ago
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