Answer:It's important to listen to your body and what it's really saying. Don't listen, and your exercise could wind up working against you, actually triggering muscle loss and fat gain. Plus, pushing too hard comes with a huge risk of injury-and all of your training won't do you much good if you're sidelined on race day.
Explanation:
Translation of important question part (Google translation used)
Force of 6N causes a displacement of 0.03m. remove the balance and connect a body 0.5 kg to the end, pull it to move it 0.02 m, release it and see how it oscillates.
(a)Determine the spring constant.
(b)Calculate angular velocity, frequency, and oscillation period
Answer:
(a)K=200 N/m
(b) w= 20 rad/s f=3.2 Hz T=0.3125 s
Explanation:
(a)
From Hooke's law, we deduce that F=kx where F is applied force, k is spring constant and x is extension of the spring. Making k the subject of the formula then k=F/x and substituting F with 6 N and x with 0.03 m then k=6/0.03=200 N/m
(b)
Angular velocity, w is given by
where m is the mass and k is spring constant calculated in part a above. Substituting mass with 0.5 kg and k with 200 N/m then

We know that frequency, f is given by
and substituting 20 rad/s for w then
Finally, oscillation period, T is usually the reciprocal of frequency hence T=1/f and substituting f with 3.2 Hz then T=1/3.2=0.3125 s
Answer:
v_oy = 16.33 m/s
Explanation:
To find the vertical velocity of the tiger, you use the information about the horizontal velocity and maximum horizontal distance traveled.
You use the following formula for the range of the trajectory:
( 1 )
v_ox: horizontal initial velocity = 4.5m/s
v_oy: vertical initial velocity = ?
g: gravitational acceleration = 9.8m/s^2
x_max: range of the trajectory = 15 m
You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:

hence, the initial vertical velocity of the tiger is 16.33m/s
Answer:
When a light wave hits a metallic surface, the electrons on the surface are pushed and ... in a process called "total internal reflection", making that surface look shiny too.
Explanation:
metallic shiny nonmetallic dull:)
Answer:
(a). The maximum electric field strength is 0.0133 V/m.
(b). The maximum magnetic field strength in the electromagnetic wave is 
(c). The wavelength of the electromagnetic wave is 
Explanation:
Given that,
Maximum potential = 4.00 mV
Distance = 0.300\ m
Frequency = 1.00 Hz
(a). We need to calculate the maximum electric field strength
Using formula of the potential difference




(b). We need to calculate the maximum magnetic field strength in the electromagnetic wave
Using formula of the maximum magnetic field strength in the electromagnetic wave

Put the value into the formula


(c). We need to calculate the wavelength of the electromagnetic wave
Using formula of wavelength


Put the value into the formula


Hence, (a). The maximum electric field strength is 0.0133 V/m.
(b). The maximum magnetic field strength in the electromagnetic wave is 
(c). The wavelength of the electromagnetic wave is 