Explanation:
It is given that,
Frequency of vibration, f = 215 Hz
Amplitude, A = 0.832 mm
(a) Let T is the period of this motion. It is given by the following relation as :
(b) Speed of sound in air, v = 343 m/s
It can be given by :
Hence, this is the required solution.
Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like
given that
here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2
now from above equation
So force applied by the person must be 150 N
Wow ! I understand your shock. I shook and vibrated a little
when I looked at this one too.
The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.
"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.
The real question is:
What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?
Acceleration of gravity is
G · M / R²
= (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²
= (6.67 x 10⁻¹¹ · 1.1 x 10³¹ / 4 x 10⁶) (N) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ (kg · m / s²) · m² · kg / kg² · m²
= 1.83 x 10¹⁴ m / s²
That's about 1.87 x 10¹³ times the acceleration of gravity on
Earth's surface.
In other words, if I were standing on the surface of that neutron star,
I would weigh 1.82 x 10¹² tons, give or take.
Answer:
Hey
Yes, this is true.
As some people have it wrong, waves in the water (ocean) are not waves of moving water, rather the wave is moving through the water. A wave is a disturbance of a medium not the meduim moving.
Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
