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Svetach [21]
3 years ago
6

26. A solid wheel accelerates at 3.25 rad/s2 when a

Physics
1 answer:
Yanka [14]3 years ago
5 0

Answer:

9.0 N

Explanation:

The location of the mass of the wheel on the wheel = Evenly distributed

The acceleration of the solid wheel, α = 3.25 rad/s²

The applied force on the wheel = 4.5 N

The location mass of the replacement wheel = All on (around) the rim

The moment of inertia of the new wheel, I = m·r² (From an online source)

We have;

The moment of inertia for a solid wheel = 1/2·m·r²

The torque, τ = Moment of inertia, I × Acceleration, α

For the solid wheel, we have;

τ  = 1/2·m·r² × 3.25 rad/s²

τ = r × F = r × m × a

For the replacement wheel, we have;

τ  = m·r² × 3.25 rad/s² = 2 × 1/2·m·r² × 3.25 rad/s²

∴ τ = 2 × r × F

Given that the radius remains the same, the force applied on the replacement wheel needs to be doubled

The force that should be exerted on the strap to give the same angular velocity, F' = 2 × F

The required force, F' = 2 × 4.5 N = 9.0 N.

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makkiz [27]

Answer:

Acceleration of the object is 4\ m/s^2.

Explanation:

It is given that, the position of the object is given by :

r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j

Velocity of the object, v=\dfrac{dr}{dt}

Acceleration of the object is given by :

a=\dfrac{d^2r}{dt^2}

a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)

Using the property of differentiation, we get :

a=\dfrac{d^2r}{dt^2}=-4\ m/s^2

So, the magnitude of the acceleration of the object at time t = 2.00 s is 4\ m/s^2. Hence, this is the required solution.

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3 years ago
What is the most effective means of establishing awareness of hazards in commercial, industrial, and storage facilities with lar
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Answer:

C: Contacting the facilities.

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3 years ago
Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp
mart [117]
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If the wavelength is 10 m and the frequency is 5 Hz, what is the wave speed?
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The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
3 years ago
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