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alexira [117]
3 years ago
10

Formula One race cars are capable of remarkable accelerations when speeding up, slowing down, and turning corners. At one track,

cars round a corner that is a segment of a circle of radius 95 m at a speed of 68 m/s. What is the approximate magnitude of the centripetal acceleration, in units of g?
Physics
1 answer:
anygoal [31]3 years ago
3 0

Answer:

Centripetal acceleration of the car is (4.96 g) m/s²

Explanation:

It is given that,

Radius of circle, r = 95 m

Speed of the car, v = 68 m/s

We need to find the centripetal acceleration. It is given by :

a_c=\dfrac{v^2}{r}

So, a_c=\dfrac{(68\ m/s)^2}{95\ m}

a_c=48.67\ m/s^2

Since, g = 9.8 m/s²

So,

a_c=(4.96\ g)\ m/s^2

So, the magnitude of the centripetal acceleration is (4.96 g) m/s². Hence, this is the required solution.

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melisa1 [442]

answer:

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  2. v \times \frac{x}{t}
4 0
2 years ago
Can there be displacement of an object in the absence of any force acting on it? Think, Discuss it with your friends and teacher
Fudgin [204]

Answer:

An object can have a displacement in the absence of any external force acting on it

Explanation:

When a object moves with a constant velocity (v), then it gets displaced in the direction of motion but the net external force experienced by the object is zero.

F  external  =ma

If object moves with constant velocity, acceleration is zero.

Since, a=0  ⟹F  external  =0

Using  s=ut+  1/2 at  ^2

 ⟹    Displacement    s=ut    (∵a=0)

Hence, an object can have a displacement in the absence of any external force acting on it

Hope this helped you:)

5 0
3 years ago
How many photons are absorbed during the dental x-ray?
MariettaO [177]
<span>E=hc/wav. len
E = (6.62 x 10^-34 x 3 x 10^8)/0.0275 x 10^-9
E = 7.22182 x 10^-15 J
To convert to eV divide by 1.6 x 10^-19
E = 7.22182 x 10^-15/1.6 x 10^-19 eV
E =45.36 x 10^3 eV
Th energy, E, of a single x-ray photon in eV is = 45.36keV.

Number of photons, n = total energy/ energy of photon
n = 3.85 x 10^-6/7.22182 x 10^-15
n = 5.33 x 10^8 photons </span>
8 0
3 years ago
The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin
Masja [62]

Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage v_p=20000volt

Output voltage, that is secondary voltage v_s=355000volt

Number of turns in primary N_p=2250

For transformer we know that \frac{V_p}{V_s}=\frac{N_p}{N_s}

\frac{20000}{355000}=\frac{2250}{N_s}

N_s=39937.5

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0
3 years ago
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Zolol [24]
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5 0
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