Answer:
8.77×10⁻⁵ m
Explanation:
ω = angular velocity = 3.49 rad/s = 33 rpm
r = Radius of the groove = 0.1 m
f = Frequency = 3.98 kHz
Tangential Velocity of record
v = ωr
⇒v = 3.49×0.1
⇒v = 0.349 m/s
The velocity of the wave is 0.349 m/s
v = fλ
![\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{0.349}{3980}=8.77\times 10^{-5}\ m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bv%7D%7Bf%7D%5C%5C%5CRightarrow%20%5Clambda%3D%5Cfrac%7B0.349%7D%7B3980%7D%3D8.77%5Ctimes%2010%5E%7B-5%7D%5C%20m)
∴ Wavelength in the groove is 8.77×10⁻⁵ m
Answer:
a) 60kg
Explanation:
mass is not affected by gravity
Answer:
A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.
Explanation:
A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.
This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.
It is the energy an object has because of its motion.
Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
![v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s](https://tex.z-dn.net/?f=v_x%20%3D%20v%20cos%20%5Ctheta%20%3D%20%2825%29%28cos%2035.9%5E%7B%5Ccirc%7D%29%3D20.3%20m%2Fs)
and the time taken to cover the horizontal distance d is
![t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bd%7D%7Bv_x%7D%3D%5Cfrac%7B52%7D%7B20.3%7D%3D2.56%20s)
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
![y=u_y t - \frac{1}{2}gt^2](https://tex.z-dn.net/?f=y%3Du_y%20t%20-%20%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
where
is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
![y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m](https://tex.z-dn.net/?f=y%3D%2814.7%29%282.56%29%20-%20%5Cfrac%7B1%7D%7B2%7D%289.8%29%282.56%29%5E2%3D5.52%20m)
The height of the crossbar is h = 3.05 m, so the ball passes
![h' = 5.52- 3.05 = 2.47 m](https://tex.z-dn.net/?f=h%27%20%3D%205.52-%203.05%20%3D%202.47%20m)
above the crossbar.