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juin [17]
3 years ago
7

A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate its centripetal acceleration. (40 m/s2)

.
Physics
1 answer:
Mkey [24]3 years ago
8 0
Centripetal acceleration is the acceleration of an object that is moving in a circular motion. It is calculated by dividing the square of the velocity to the radius of the circular path. It is calculated as follows:

a = v^2 / r = 4.0^2 / 0.4  = 40 m/s^2  
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A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0
3 years ago
2. Find the time taken by the bus to reach the stop. need only group B, 2 answer
Molodets [167]

Answer:

t = 2 seconds

Explanation:

In 2nd question, the question is given the attached figure.

Initial speed of the bus, u = 0

Acceleration of the bus, a = 8 m/s²

Final speed, v = 16 m/s

We need to find the time taken by the car to reach the stop. Acceleration of an object is given by :

a=\dfrac{v-u}{t}

t is time taken

t=\dfrac{v-u}{a}\\\\t=\dfrac{16-0}{8}\\\\t=2\ s

The bus will take 2 seconds to reach the stop.

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8 0
3 years ago
Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of
viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

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I wrote Force and acceleration in bold letters because
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The equation is saying that the Force and the acceleration
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