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Pepsi [2]
3 years ago
10

Powerful sports cars can go from zero to 25 m/s (about 60 mph) in 6 seconds. What is the magnitude of the acceleration, includin

g units
Physics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

The magnitude of the acceleration is 4.2 m/s²

Explanation:

Given;

initial velocity of the powerful sports car, u = 0

final velocity of the powerful sports car, v = 25 m/s

time of motion, t = 6 seconds

Acceleration = Δv / Δt

Δv is change in velocity

Δt is change in time

The magnitude of the acceleration of the powerful sports car during the motion is given by;

a = \frac{v-u}{t}\\\\a = \frac{25-0}{6}\\\\a = 4.2 \ m/s^2

Therefore, the magnitude of the acceleration is 4.2 m/s²

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-- For the first 3 hours, he travels at an average speed of 60 km/hr.  During that time, he covers (60 km/hr) x (3 hrs) = 180 km.

-- At that point, he still has (460km - 180km) = 280 km to go.

-- If he drives that 280km at an average speed of 70 km/hr, it'll take him (280km)/(70km/hr) = 4 hrs to cover it.

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Three parallel wires of length l each carry current Iin the same direction. They’re positioned at the vertices of an equilateral
cluponka [151]

Answer:

F = μi²l/πa

Explanation:

The magnetic force F on a length of wire, l carrying a current i in a magnetic field B is given by

F = Bilsinθ      

The magnetic field due to one wire of length, l carrying a current, i at a distance a from it is given by B = μi/2πa

So, the force on the first wire due to the second wire is F₁ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

the force on the first wire due to the third wire is F₂ = Bilsinθ = μiilsinθ/2πa = μi²lsinθ/2πa

Since the magnetic field due to the one wire is perpendicular to the length of the other wire its field acts upon, θ = 90

So, F₁ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa and

F₂ = μi²lsinθ/2πa = μi²lsin90/2πa = μi²l/2πa

Since the angle between F₁ and F₂ is 60° (since it is an equilateral triangle)

The resultant force F is thus

F = √(F₁² + F₂² + 2F₁F₂cos60°)

F = √(F₁² + F₂² + 2F₁F₂ × 0.5)

F = √(F₁² + F₂² + 2F₁F₂)   (since F₁ = F₂)

F = √(2F₁² + 2F₁²) = √(4F₁²)

F = 2F₁

F = 2μi²l/2πa

F = μi²l/πa

6 0
3 years ago
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Oduvanchick [21]

Lead-207 can change into Lead-208.

When Lead-207 is bombarded with neutrons, the atom acquires the neutron. Adding a neutron only changes the mass number of the atom. This does not involve change in the identity of the atom. So, Lead-207 can change into Lead-208 without change in its chemical properties.

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The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:

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Given data:

Length of the fixed string = 264cms = 2.64 meters

The wavelength for standing waves is given by:

λ = 2L/n

where,

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For n = 1,

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= 5.2 meters

For n = 2,

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= 2.6 meters

For n = 3,

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= 1.7 meters

To learn more about standing waves: brainly.com/question/14151246

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Answer:

B

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