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Serhud [2]
3 years ago
10

A bike rider pedals with constant acceleration to reach a velocity of 7.5 (vf)m/s over a time of 4.5 s(t). during the period of

acceleration, the bike's displacement is 19 m(d). what was the initial velocity of the bike?
Physics
1 answer:
BigorU [14]3 years ago
6 0
<span>The initial velocity of the bike was 1.67 (vf)m/s. This is found by evaluating 7.5/4.5 which yields the velocity per unit of time which is equivalent to initial velocity.</span>
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A charged particle is moving in a uniform magnetic field at a speed of 8.2Ã10^3 m/s in a direction 87° from the direction of th
77julia77 [94]

Answer:

Magnetic field, B = 0.042 T

Explanation:

It is given that,

Speed of charged particle, v=8.2\times 10^3\ m/s

Angle between velocity and the magnetic field, \theta=87

Charge, q=5.7\ \mu C=5.7\times 10^{-6}\ C

Magnetic force, F = 0.002 N

The magnetic force is given by :

F=qvB\ sin\theta

B is the magnetic field  

B=\dfrac{F}{qv\ sin\theta}

B=\dfrac{0.002}{5.7\times 10^{-6}\times 8.2\times 10^3\times sin(87)}

B = 0.042 T

So, the strength of the magnetic field is 0.042 Tesla. Hence, this is the required solution.

5 0
3 years ago
The effect produced when two or more sound waves pass through the same point simultaneously is called?
navik [9.2K]
The answer to the question is sound
7 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
Two 125 kg bumper cars are moving toward each other in opposite directions. Car X is moving at 10 m/s and Car Z at −12 m/s when
nignag [31]
<h2>Given that,</h2>

Mass of two bumper cars, m₁ = m₂ = 125 kg

Initial speed of car X is, u₁ = 10 m/s

Initial speed of car Z is, u₂ = -12 m/s

Final speed of car Z, v₂ = 10 m/s

We need to find the final speed of car X after the collision. Let v₁ is its final speed. Using the conservation of momentum to find it as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

v₁ is the final speed of car X.

m_1u_1+m_2u_2-m_1v_1=m_2v_2\\\\m_2v_2=m_1u_1+m_2u_2-m_2v_2\\\\m_1v_1=125\times 10+125\times (-12)-125\times 10\\\\v_1=\dfrac{-1500}{125}\\\\v_1=-12\ m/s

So, car X will move with a velocity of -12 m/s.

3 0
3 years ago
What effect does the vertical acceleration have on the horizontal velocity of the projectile?
KengaRu [80]
Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity.. 
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
3 0
3 years ago
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