What is the Force in Newtons exerted by an object with a mass of 15 kg and an Acceleration of 12 m/s2?
1 answer:
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Answer:
Electric flux in a) , b) and c) is same which is 0.373 × 10 ⁶ N m²/C
Explanation:
given,
surface charge (q) = 3.3 × 10⁻⁶ C
to calculate electric flux = ?
a) radius = 0.76 m
area of sphere = 4 π r²
electric flux =
electric flux =
flux = 0.373 × 10 ⁶ N m²/C
electric flux in the other two cases will also be same as electric flux is independent of area
so, Electric flux in a) , b) and c) is same which is 0.373 × 10 ⁶ N m²/C
Answer:
Velocity from second channel will be 1.6875 m/sec
Explanation:
We have given width of the channel , that is diameter of the channel 1 = 12 m
So radius
Speed through the channel 1
Width , that is diameter of the channel 2
So
From continuity equation
So velocity from smaller channel will be 1.6875 m /sec
Answer:
Sphere, cylinder hoop
Explanation:
To analyze Which student is right it is best to propose the solution of the problem. Let's look for the speed of the center of mass. Let's use the concept of mechanical energy
In the highest part of the ramp
Em₀ = U = mg y
In the lowest part
Here the energy has part of translation and part of rotation
=
+
= ½ m
² + ½ I w²
Where I is the moment of inertia of the body and w the angular velocity that relates to the velocity of the center of mass
= w r
w = / r
Let's replace
= ½ I (
/ r)²
Energy is conserved
mg y = ½ m ² + ½ I
² / r2
½ (m + I / r²) ² = m g y
½ (1 + I / m r²) ² = g y
= √ [2gy / (1 + I / mr²)]
This is the velocity of the center of mass of the bodies, as they all have the same radius with comparing this point is sufficient. Now let's use the speed definition
v = d / t
t = d / v
t = d / (√ [2gy / (1 + I / mr²)])
t = (d / √ 2gy) √(1 + I / m r²)
Therefore we see that time is proportional to the square root. All quantities are constant and the one that varies is the moment of inertia.
The moments of inertia of
Sphere is Is = 2/5 M r²
Cylinder Ic = ½ M r²
Hoop Ih = M r²
Let's replace each one and calculate the time
Sphere
ts = (d / √2gy) √ (1 + 2/5 Mr² / mr²)
ts = (d / √ 2gy) √ (1 +2/5) = (d / √ 2gy) √(1.4)
ts = (d / √ 2gy) 1.1
Cylinder
tc = (d / √2gy) √ (1 + 1/2 Mr² / Mr²)
tc = (d / √2gy) √ (1 + ½) = (d / √ 2gy) √ 1.5
tc = (d / √ 2gy) 1.2
Hoop
th = (d / √2gy) √ (1 + mr² / mr²)
th = (d / √2gy) √(1 + 1) = (d / √ 2gy) √ 2
th = (d / √ 2gy) 1.41
We have the results for the time the body that arrives the fastest is the sphere and the one that is the most hoop. Therefore the correct answer is
ts < tc < th
Sphere, cylinder hoop