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natita [175]
3 years ago
7

The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci

ty of 5.88 m/s.
What is the angle of the plane with respect to tThe velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a velocity of 5.88 m/s.

What is the angle of the plane with respect to the horizontal (in degrees)? he horizontal (in degrees)?
Physics
1 answer:
nasty-shy [4]3 years ago
7 0

Answer:

\theta = 25.3^\circ

Explanation:

The acceleration of the block can be found by the kinematics equations:

v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ

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A simple pendulum consists of a ball connected to one end of a thin brass wire. The period of the pendulum "is 2.51 s". The temp
Artyom0805 [142]

Answer:

0.0034 sec

Explanation:

L = initial length

T = initial time period = 2.51 s

Time period is given as

T = 2\pi \sqrt{\frac{L}{g}}

2.51 = 2\pi \sqrt{\frac{L}{9.8}}

L = 1.56392 m

L' = new length

ΔT = Rise in temperature = 142 °C

α = coefficient of linear expansion = 19 x 10⁻⁶ °C

New length due to rise of temperature is given as

L' = L + LαΔT

L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)

L' = 1.56814 m

T' = New time period

New time period is given as

T' = 2\pi \sqrt{\frac{L'}{g}}

T' = 2\pi \sqrt{\frac{1.56814}{9.8}}

T' = 2.5134 sec

Change in time period is given as

ΔT = T' - T

ΔT = 2.5134 - 2.51

ΔT = 0.0034 sec

5 0
3 years ago
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

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3 years ago
A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw
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26.54 m/s  is the magnitude of its velocity just before it strikes the ground

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time it takes to reach the ground,

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v= 26.54 m/s

The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.

To know more about  velocity visit : brainly.com/question/16379705

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Answer:

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