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klio [65]
3 years ago
14

A 0.50-kg object moves on a horizontal frictionless circular track with a radius of 2.5 m. An external force of 3.0 N, always ta

ngent to the track, causes the object to speed up as it goes around. If it starts from rest, then at the end of one revolution the radial component of the force of the track on it is ____________.
Physics
1 answer:
Allisa [31]3 years ago
3 0

Answer:

Radial force component of force = 37.68 N

Explanation:

By Newton's 2 nd law of motion,

F = ma

F = 3.0 N, m = 0.5 Kg, a (Linear Acceleration ) = ?

3 = 0.5 a

a = 6 m/sec^{2}

Now, a = 6 m/sec^{2} , r = 2.5 m ,α(angular acceleration )  = ?

a = r α

6= 2.5 α

hence α = 2.4 rad/sec^{2}

w= ?, w_0= 0, \alpha =2.4\ rad/sec^{2},\theta =2\ \pi (One\ Revolution)

we know that,

w^{2} =w_0^2 +2 \alpha\cdot s\\          =0 + 2\cdot 2.4\cdot 2\pi\\=30.14 \\w=\sqrt{30.144} = 5.49\ Rad/ sec

Radial force component of force  = m r w^{2}

since m =0.5 kg, r =2.5 m w =5.49 rad/sec

Radial force component of force =0.5\cdot\ 2.5\cdot\ 5.49^{2} = 37.68 N

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A parallel plate capacitor with air between the plates has a potential difference of 71.0 V. Determine the potential difference
Gwar [14]

Answer:

15.8 V

Explanation:

The relationship between capacitance and potential difference across a capacitor is:

q=CV

where

q is the charge stored on the capacitor

C is the capacitance

V is the potential difference

Here we call C and V the initial capacitance and potential difference across the capacitor, so that the initial charge stored is q.

Later, a dielectric material is inserted between the two plates, so the capacitance changes according to

C'=kC

where k is the dielectric constant of the material. As a result, the potential difference will change (V'). Since the charge stored by the capacitor remains constant,

q=C'V'

So we can combine the two equations:

CV=CV'\\CV=(kC)V'\\V'=\frac{V}{k}

and since we have

V = 71.0 V

k = 4.50

We find the new potential difference:

V'=\frac{71.0}{4.50}=15.8 V

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2 years ago
A person has a gravitational force (weight) of 600 N on Earth. Suppose the mass of the Earth is double and the radius shrinks to
Alenkasestr [34]

Explanation:

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2 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

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iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

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then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

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Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

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