2 years ago

If a scale on Earth reads 650 N, what is your mass?

1 answer:

4 0

If the scale reads 650N, then the mass of whoever it is standing on the scale is

(weight) / (gravity) = (650N) / (9.8 m/s²) = 66.3 kilograms .

It's not MY mass, even if I'm the one standing on the scale.
If I stand on a scale and it reads 650 N, the scale is broken.

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A rubber ball with a mass 0.20 kg is dropped vertically from a height of 1.5 m above the floor. The ball bounces off of the floo

Digiron [165]

Potential Energy = mass * Hight * acceleration of gravity
PE=hmg
PE = 1.5 * .2 * 9.81
PE = 2.943
it lost .6 so 2.943 - .6 = 2.343
now your new energy is 2.343 so solve for height
2.343 = mhg
2.334 = .2 * h * 9.81
h = 1.194
the ball after the bounce only went up 1.194m

8 0

2 years ago

On a guitar, the lowest toned string is usually strung to the E note, which produces sound at 82.4 82.4 Hz. The diameter of E gu

Vsevolod [243]

The complete and comprehensive solution is attached.

8 0

2 years ago

What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate

Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec$

We know that angular frequency is given by

$\omega =\sqrt{\frac{k}{m}}$

$1.256 =\sqrt{\frac{7}{m}}$

Squaring both side

$1.577 =\frac{7}{m}$

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6 0

2 years ago

A fan is to accelerate quiescent air to a velocity of 12.5 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

Reika [66]

Answer:

= 829.69 Watt

≅ 830 Watt

Explanation:

Given that,

Velocity of air flow = 12.5m/s

Rate of flow of air = 9m³/s

Density of air = 1.18kg/m³

power by kinetic energy = 1/2(mv²)

mass = density × volume

m = 1.18 × 9

= 10.62 kg/s

power = 1/2 mV²

= 1/2 (10.62 × 12.5²)

= 829.69 Watt

≅ 830 Watt

Flow rate

Velocity of the air

m/s

Density of the air

1.18

5 0

2 years ago

A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

$\frac{dA}{dt}$ = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B $\frac{dA}{dt}$

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0

2 years ago