Answer:
The acceleration of the both masses is 0.0244 m/s².
Explanation:
Given that,
Mass of one block = 602.0 g
Mass of other block = 717.0 g
Radius = 1.70 cm
Height = 60.6 cm
Time = 7.00 s
Suppose we find the magnitude of the acceleration of the 602.0-g block
We need to calculate the acceleration
Using equation of motion
Where, s = distance
t = time
a = acceleration
Put the value into the formula
Hence, The acceleration of the both masses is 0.0244 m/s².
Answer:
maximum amplitude = 0.08 m
Explanation:
Given that
Time period T= 0.58 s
acceleration of gravity g= 9.8 m/s²
We know that time period of simple harmonic motion given as
T = 2π/ω
0.58 = 2π/ω
ω = 10.83rad/s
ω=angular frequency
Lets take amplitude = A
The maximum acceleration given as
a= ω² A
The maximum acceleration should be equal to g ,then block does not separate
a= ω² A
9.8 = 10.83² A
A = 0.08m
maximum amplitude = 0.08 m
<span>The diver is heading downwards at 12 m/s
Ignoring air resistance, the formula for the distance under constant acceleration is
d = VT - 0.5AT^2
where
V = initial velocity
T = time
A = acceleration (9.8 m/s^2 on Earth)
In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m)
So let's substitute the known values and solve for T
d = VT - 0.5AT^2
-7 = 2.5T - 0.5*9.8T^2
-7 = 2.5T - 4.9T^2
0 = 2.5T - 4.9T^2 + 7
We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164.
Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So
V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s
V = 2.5 m/s - 14.47706141 m/s
V = -11.97706141 m/s
So the diver is going down at a velocity of 11.98 m/s
Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point.
V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s)
V = 2.5 m/s - (-9.477061409 m/s)
V = 2.5 m/s + 9.477061409 m/s
V = 11.97706141 m/s
And you get the exact same velocity, except it's the opposite sign.
In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
Answer:
W = 222 N.
Explanation:
The qiestion says" If the acceleration of gravity on the surface of the planet Mercury is 3.7 m / s2, then what would be the weight of a person with mass 60 kg on its surface?
"
Mass of the person, m = 60 kg
The acceleration due to gravity on the surface of gravity is 3.7 m/s²
We need to find the weight of a person on the surface of Mercury.
Weight of an object is given by :
W = mg
So,
W = 60 kg × 3.7 m/s²
W = 222 N
Hence, the person will weigh 2222 N on the surface of Moon.
