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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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. Inside a conducting sphere of radius 1.2 m, there is a spherical cavity of radius 0.8 m. At the center of the cavity is a poin
MAVERICK [17]

Answer:

 E = 1,873 10³ N / C

Explanation:

For this exercise we can use Gauss's law

        Ф = E. dA = q_{int} / ε₀

Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.

The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.

        The surface of a sphere is

             A = 4π r²

             E 4π r² = q_{int} /ε₀

  The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is

           q_{int} = q₁ + q₂

           q_{int} = (530 - 200) 10⁻⁹

           q_{int} = 330 10⁻⁹ C

The electric field is

             E = 1 / 4πε₀   q_{int} / r²

            k = 1 / 4πε₀

            E = k q_{int}/ r²

Let's calculate

           E = 8.99 10⁹   330 10⁻⁹/ 1.32²

           E = 1,873 10³ N / C

7 0
3 years ago
An object is weighed at different locations on the
Step2247 [10]

Answer:

C

Explanation:

The weight will always be different while mass is described as the stuff inside an object, and that stays the same.

Such as it weighs differently in space.

6 0
2 years ago
Read 2 more answers
What is the length of an aluminum rod at 65°C if its length at 15°C is 1.2 meters? A. 0.00180 meter B. 1.201386 meters C. 1.2
Deffense [45]

Answer:

Option B is the correct answer.

Explanation:

Thermal expansion

            \Delta L=L\alpha \Delta T

L = 1.2 meter

ΔT = 65 - 15 = 50°C

Thermal Expansion Coefficient for aluminum, α = 24 x 10⁻⁶/°C

We have change in length

          \Delta L=L\alpha \Delta T=1.2\times 24\times 10^{-6}\times 50=1.44\times 10^{-3}m

New length = 1.2 + 1.44 x 10⁻³ = 1.2014 m

Option B is the correct answer.

3 0
3 years ago
Read 2 more answers
The force of gravitation between two spherical bodies is Gm1
malfutka [58]

Answer:

r is the separation between the two spherical bodies

3 0
2 years ago
To make the next delivery, Sam-I-Am goes 45 mi/h for 2 hours. How far did he travel in that time?
andrezito [222]

Hello =D

This problem is about cinematic

So

V = 45 mi/h

t = 2 h

Then

V= X/t

X = V*t

Then

X = (45)*(2)

X = 90 mi

Best regards

5 0
3 years ago
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