<span>So we want to know what kind of wave is the scientist is studying while studying earthquakes. Waves that produce earthquakes are mechanical waves. Gamma rays and radiowaves are both electromagnetic waves and don't require a medium but mechanical do. So the correct answer is mechanical waves.</span>
Using the principle of floatation.
u = w............(a)
Upthrust of fluid is equal to the weight of the object.
Let the volume of the wood be V.
The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V
Formula for upthrust, u = vdg
where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity
weight, w = mg
where m = mass
g = acceleration due to gravity
From (a)
u = w
vdg = mg Cancel out g
vd = m
The v is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.
d is density of fluid which is water in this case, 1000 kg/m³
0.2V * 1000 = m
200V = m
Hence the mass of the object is 200V kg.
But Density of solid = Mass of solid / Volume of solid
= 200V / V
= 200 kg/m³
Density of solid = 200 kg/m³
(a) The plane makes 4.3 revolutions per minute, so it makes a single revolution in
(1 min) / (4.3 rev) ≈ 0.2326 min ≈ 13.95 s ≈ 14 s
(b) The plane completes 1 revolution in about 14 s, so that in this time it travels a distance equal to the circumference of the path:
(2<em>π</em> (23 m)) / (14 s) ≈ 10.3568 m/s ≈ 10 m/s
(c) The plane accelerates toward the center of the path with magnitude
<em>a</em> = (10 m/s)² / (23 m) ≈ 4.6636 m/s² ≈ 4.7 m/s²
(d) By Newton's second law, the tension in the line is
<em>F</em> = (1.3 kg) (4.7 m/s²) ≈ 6.0627 N ≈ 6.1 N
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct
