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Mariulka [41]
3 years ago
8

The cars on an amusement-park ride travel at a constant velocity of 4.0 m/s on a circular track

Physics
1 answer:
lys-0071 [83]3 years ago
7 0
V^2=u^2+2as
V=0
a =-u^2/2s
a=[4]^2/2[4]
a=-2m/s^2
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Answer:

C: equal to mg

Explanation:

in free-fall, gravity is always the net force on an object

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3 years ago
A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release
Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

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b.

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Initial energy of spring=Kinetic energy  of object

37.8=\frac{1}{2}(3)v^2

v^2=\frac{37.8\times 2}{3}

v=\sqrt{\frac{37.8\times 2}{3}}

v=5.02 m/s

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W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J

8 0
3 years ago
You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a
gtnhenbr [62]

Answer:

The necessary separation between  the two parallel plates is 0.104 mm

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length of each side of the square plate, L = 6.5 cm = 0.065 m

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potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}

Where;

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d = \frac{VL^2 \epsilon_o}{Q} \\\\d =  \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm

Therefore, the necessary separation between  the two parallel plates is 0.104 mm

6 0
3 years ago
We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o
Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T  ...(i)

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Resultant force m₁a = m₁g sin α+T

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Solving both the equations by adding them,

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18a=11gsin 65°+7g sin 15°=115.45

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Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

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