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julia-pushkina [17]
3 years ago
9

A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is

the magnitude
Physics
1 answer:
AveGali [126]3 years ago
3 0

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

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Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
Ymorist [56]

This question involves the concepts of the law of conservation of momentum and velocity.

The velocity of the eight ball is "5.7 m/s".

According to the law of conservation of momentum:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = mass of number three ball = 5 g

m₂ = mass of the eight ball = 6 g

u₁ = velocity of the number three ball = 3 m/s

u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)

v₁ = final velocity of the three number ball = - 5 m/s

v₂ = final velocity of the eight ball = ?

Therefore,

(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)

v_2=\frac{34\ g.m/s}{6\ g}\\\\

<u>v₂ = 5.7 m/s</u>

<u></u>

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

4 0
2 years ago
In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6
Rasek [7]

Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

            E = 2.5 x 10⁻¹⁴ J

8 0
3 years ago
Two pebbles, Pebble A and Pebble B, are thrown horizontally with the same force. Pebble A's mass is 3 times the
Hitman42 [59]

Answer:

Pebble A has 1/3 the acceleration as pebble B.

Explanation:

F = m×a

mass of a = 3 × mass of b (m_a = 3 × m_b)

Same starting force, F

m_a = mass of a

m_b = mass of b

a_a = acceleration of a

a_b = acceleration of b

F = m_a × a_a = m_b × a_b

3 × m_b × a_a = m_b × a_b

3 × a_a = a_b

OR

a_a = a_b / 3

5 0
2 years ago
Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0
3 years ago
Read 2 more answers
A bus hits a bug and the bug splatters on the windshield, which force is greater?
grandymaker [24]

The forces acting on a body and the type of motion that results are given by Newton's three Laws of motion

  • The <u>force </u>of the bus is <u>the same</u> as reactive force of the bug

Reason:

According to Newton's third Law of motion, given that the bug collides

with the bus, the force with which the bus hits the bug is equal to the

reactive force of the bug on the on the bus

According to Newton's second Law of motion, force is equal to the rate of

change of the momentum produced

The impulse of the force of the bus on the bug is given as follows;

F·Δt = m·(v₂ - v₁)

Given that the force of the bus is large, the change in momentum, m·(v₂ - v₁),

is also large such that the parts of the bug are split by the rapid change in

velocity, and the bug splatters on the windshield, and is then carried along

on the trip,

The equally large reactive force of the bug, is such that the bug splatters

due its magnitude

Therefore, the correct response is that <u>the forces are the same</u>

Lean more here:

brainly.com/question/21279060

8 0
2 years ago
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