(01.06 MC)

PtichkaEL [24]

Acceleration is the way the motion is changing.

4 0

3 years ago

Suppose that you're facing a straight current-carrying conductor, and the current is flowing toward you. The lines of magnetic f

alekssr [168]

Answer:c

Explanation:

When the direction of current is towards the observer then the magnetic field around it will be in the form of concentric circles and its direction will be anti-clockwise when viewed from the observer side.

Whenever current is flowing in a current-carrying conductor then the magnetic field is associated with it and direction of the magnetic field is given by right-hand thumb rule according to which if thumb represents the direction of current then wrapping of fingers will give the direction of the magnetic field

8 0

3 years ago

A 1500 kg car is moving on a flat, horizontal flat road. If the radius of the curve is 35 m and

Troyanec [42]

The net force on the car is the friction that keeps it on the road, which points toward the center of the circle of the curve. Then by Newton's second law, we have

• net vertical force:

∑ F = N - W = 0

• net horizontal force:

∑ F = Fs = m a

where

N = magnitude of normal force

W = car's weight

Fs = mag. of static friction

m = car's mass

a = v ²/R = mag. of the centripetal acceleration

v = car's speed

R = radius of curve

Now,

• compute the car's weight:

W = m g = (1500 kg) (9.8 m/s²) = 14,700 N

• solve for the mag. of the normal force:

N = 14,700 N

• solve for the mag. of the friction force, using the given friction coefficient:

Fs = 0.5 N = 7350 N

• solve for the (maximum) acceleration:

7350 N = (1500 kg) a → a = 4.9 m/s²

• solve for the (maximum) speed:

4.9 m/s² = v ²/ (35 m) → v ≈ 13 m/s

4 0

3 years ago

While punting a football, a kicker rotates his leg about the hip joint. The moment of inertia of the leg is 3.75 kg⋅m² and its r

forsale [732]

Answer:

9.6609 rad/s

10.143945 m/s

Explanation:

I = Moment of inertia = 3.75 kgm²

K = Kinetic energy = 175 J

r = Radius = 1.05 m

Kinetic energy is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow \omega=\sqrt{\dfrac{2K}{I}}\\\Rightarrow \omega=\sqrt{\dfrac{2\times 175}{3.75}}\\\Rightarrow \omega=9.6609\ rad/s$

The angular velocity of the leg is 9.6609 rad/s

Velocity is given by

$v=r\omega\\\Rightarrow v=1.05\times 9.6609\\\Rightarrow v=10.143945\ m/s$

The velocity of the tip of the punters shoe is 10.143945 m/s

4 0

3 years ago

A horizontal air diffuser operates with inlet velocity and specific enthalpy of 250 m/s and 270.11 kj/kg, repectively, and exit

Ivahew [28]

Answer: c) 90 m/s

Explanation:

Given

Invest velocity, v1 = 250 m/s

Inlet specific enthalpy, h1 = 270.11 kJ/kg = 270110 J/kg

Outlet specific enthalpy, h2 = 297.31 kJ/kg = 297310 J/kg

Outlet velocity, v2 = ?

0 = Q(cv) - W(cv) + m[(h1 - h2) + 1/2(v1² - v2²) + g(z1 - z2)]

0 = Q(cv) + m[(h1 - h2) + 1/2(v1² - v2²)]

0 = [(h1 - h2) + 1/2(v1² - v2²)]

Substituting the values of the above, we get

0 = [(270110 - 297310) + 1/2 ( 250² - v²)

0 = [-27200 + 1/2 (62500 - v²)]

27200 = 1/2 (62500 - v²)

54400 = 62500 - v²

v² = 62500 - 54400

v² = 8100

v = √8100

v = 90 m/s

4 0

3 years ago