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ella [17]
2 years ago
9

Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t

o slide down from rest.
a) It takes the heavier object 10 times longer to reach the bottom of the incline than the lighter object.
b) It takes the heavier object 5 times longer to reach the bottom of the incline than the lighter object.
c) It takes the lighter object 10 times longer to reach the bottom of the incline than the heavier object.
d) It takes the lighter object 5 times longer to reach the bottom of the incline than the heavier object.
e) The two objects reach the bottom of the incline at the same time.
Physics
1 answer:
poizon [28]2 years ago
5 0

Answer:

(E) The two objects reach the bottom of the incline at the same time.

Explanation:

Given;

first object with mass, m

second object with mass, 5m

The acceleration of gravity for both object is the same = 9.8 m/s²

Since both objects have the same acceleration of gravity, and no external force due friction (frictionless inclined plane), they will reach bottom of the inclined at the time.

Thus, the acceleration due to gravity is constant for all objects regardless of their masses.

Therefore, the correct option is E;

(E) The two objects reach the bottom of the incline at the same time.

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23. While sliding a couch across a floor, Andrea and Jennifer exert forces F → A and F → J on the couch. Andrea’s force is due n
PSYCHO15rus [73]

Answer:

a)  (95.4 i^ + 282.6 j^) N , b) 298.27 N  71.3º and c)   F' = 298.27 N   θ = 251.4º

Explanation:

a) Let's use trigonometry to break down Jennifer's strength

      sin θ = Fjy / Fj

      cos θ = Fjx / Fj

Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be

          T = 90 -32 = 58º

         Fjy = Fj sin 58

         Fjx = FJ cos 58

         Fjx = 180 cos 58 = 95.4 N

         Fjy = 180 sin 58 = 152.6 N

Andrea's force is

         Fa = 130.0 j ^

We perform the summary of force on each axis

X axis

       Fx = Fjx

       Fx = 95.4 N

Axis y

       Fy = Fjy + Fa

       Fy = 152.6 + 130

       Fy = 282.6 N

       F = (95.4 i ^ + 282.6 j ^) N

b) Let's use the Pythagorean theorem and trigonometry

       F² = Fx² + Fy²

       F = √ (95.4² + 282.6²)

       F = √ (88963)

       F = 298.27 N

       tan θ = Fy / Fx

       θ = tan-1 (282.6 / 95.4)

       θ = tan-1 (2,962)

       θ = 71.3º

c) To avoid the movement they must apply a force of equal magnitude, but opposite direction

       F' = 298.27 N

       θ' = 180 + 71.3

       θ = 251.4º

4 0
2 years ago
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T
atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole = \rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.

<em>where</em>,

k is the Coulomb's constant, having value = \dfrac{1}{4\pi \epsilon_o}

\epsilon_o is the electrical permittivity of free space.

Also, r>>s, therefore, \rm r^2+s^2\approx r^2.

\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

\rm F=qE

Therefore, the electric force on the charge Q due to the dipole is given by

\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.

The magnitude of the torque on the dipole is given by

\rm \tau = Fs\ \sin\theta

Since the charge Q is placed in the plane that bisects the dipole, therefore, \theta = 90^\circ.

\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.

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