Answer:
D
Explanation:
<em>The correct answer would be in the axle of the wheels while you ride your bicycle.</em>
Options A, B, and C requires that the forces of friction is increased in order to have more control.
However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.
<u>The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.</u>
Answer:
Explanation:
Here by ideal gas equation we can say
now we know that pressure is kept constant here
so we will have
since we know that number of moles and pressure is constant here
so we have
now we know that initial temperature is 17.8 degree C
and finally volume is doubled
So we have
so final temperature will be
Answer:
<h3>The answer is 7 Pa</h3>
Explanation:
The pressure transmitted in the hydraulic system can be found by using the formula
f is the force
a is the area
From the question we have
We have the final answer as
<h3>7 Pa</h3>
Hope this helps you
Answer:
(a) 272.73 m
(b) 0.338 N/C
Explanation:
frequency, f = 1100 kHz = 1100 x 1000 Hz
E(t) = Eo Sin(2πft)
Eo = 0.62 N/C
(a) Velocity of light, c = 3 x 10^8 m/s
wavelength, λ = c / f = (3 x 10^8) / (1100000) = 272.73 m
Thus, the wavelength is 272.73 m.
(b) at t = 3.1 microsecond = 3.1 x 10^-6 s
E = Eo Sin (2 π ft)
E = 0.62 Sin (2 x 3.14 x 1100 x 10^3 x 3.1 x 10^-6)
E = 0.62 Sin (21.4148)
E = 0.62 x 0.5449 = 0.338 N/C
Thus, the electric field at t = 3.1 microsecond s 0.338 N/C.
Answer:
1.84 m
Explanation:
For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.
So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,
a = g + rω²
= 9.8 m/s² + 1.10 m × (18.85 rad/s)²
= 9.8 m/s² + 390.85 m/s²
= 400.65 m/s²
Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.
v² = u² + 2a(h₂ - h₁)
So, v² - u² = 2a(h₂ - h₁)
h₂ - h₁ = (v² - u²)/2a
h₂ = h₁ + (v² - u²)/2a
substituting the values of the variables into the equation, we have
h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)
h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²
h₂ = 1.3 m + 0.54 m
h₂ = 1.84 m
