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stepladder [879]
2 years ago
13

Below are four statements about acceleration. Which statement is not correct? A Acceleration always involves changing speed. B C

hanging direction always involves acceleration. C Changing speed always involves acceleration. D Circular motion always involves acceleration.
Physics
1 answer:
True [87]2 years ago
6 0

Answer:

A. Acceleration always involves changing speed.

Explanation:

Acceleration is defined as the rate of change of velocity.

Velocity is a vector quantity and is defined as speed in a given direction. Therefore, when any or both of the components of velocity is changing with time, a body is said to be accelerating i.e. acceleration involves a change of speed or change in direction, or a change in both speed and direction.

Considering the statements:

A. Acceleration always involves changing speed is false because acceleration can occur when speed is constant but direction is changing.

B. Changing direction always involves acceleration is true because, a change in direction results in a change in velocity.

C Changing speed always involves acceleration is true because, changing speed results in a change in velocity.

D. Circular motion always involves acceleration is true because in circular motion the direction is always changing, and this results in a change in velocity

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A 4.50-kg wheel that is 34.5 cm in diameter rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1
Mila [183]

Answer:

-10.9 rad/s²

Explanation:

ω² = ω₀² + 2α(θ - θ₀)

Given:

ω = 13.5 rad/s

ω₀ = 22.0 rad/s

θ - θ₀ = 13.8 rad

(13.5)² = (22.0)² + 2α (13.8)

α = -10.9 rad/s²

6 0
2 years ago
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Your 64-cm-diameter car tire is rotating at 3.5 rev/s when suddenly you press down hard on the accelerator. After traveling 200
andreyandreev [35.5K]

Answer: angular acceleration = 0.748rad/s²

Explanation: according to the question, our answer needs to be in rad/s², thus all units in rev/s will be converted to rad/s

Assuming the motion of the object is of a constant angular acceleration, then newton's laws of motion is applicable.

The formulae below is used

v² = u² + 2αθ

v = final angular speed =6rev/s = 6*2π = 12π rad/s

u =initial angular speed =3.5rev/s = 3.5 *2π = 7π rad/s

Note 1 rev = 2π rad.

α = angular acceleration.

θ = angular displacement.

Diameter = 64cm = 0.64m, radius = 64/2 = 32cm = 0.32m

The angular displacement can be gotten using the formulae below

S = rθ, where s= linear distance covered = 200m, r = radius = 0.32m

θ = S/r = 200/0.32=625 rad.

By substituting the parameter we have that

(12π)² = (7π)² + 2α(625)

1421.22 = 486.31 + 1250α

1421.22 - 486.31 = 1250α

934.91 = 1250α

α = 934.91/1250

α= 0.748 rad/s²

4 0
2 years ago
A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along
storchak [24]

Answer:

v = \frac{kQ}{a}  

Explanation:

We define the linear density of charge as:

\lambda = \frac{Q}{L}

     Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

dv = \frac{kdq}{r}

          where k is the coulomb's constant

                     r is the distance from dq to center of the circle

Thus.

v = \int_{}^{}\frac{kdq}{a}  

v = \frac{k}{a}\int_{}^{}dq

v = \frac{kQ}{a}     Potential at the center of the semicircle

4 0
2 years ago
An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin
fgiga [73]

Given that,

The electric field is given by,

\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of (\vec{E}\times\vec{B}) vector is the direction of propagation of the wave .

Direction of magnetic field = \hat{j}

B=B_{0}\sin(kx-\omega t)\hat{k}

We need to calculate the poynting vector

Using formula of poynting

\vec{S}=\dfrac{E\times B}{\mu_{0}}

Put the value into the formula

\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}

\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}

Hence, The poynting vector is \dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}

7 0
2 years ago
Suppose we replace the mass in the video with one that is four times heavier. How far from the free end must we place the pivot
Llana [10]

We must place the pivot to keep the meter stick in balance at 90 cm (10 cm from the weight) from the free end.

Answer: Option B

<u>Explanation:</u>

In initial stage, the meter stick’s mass and mass hanged in meter stick at one end are same. Refer figure 1, the mater stick’s weight acts at the stick’s mid-point.

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                  m \times g \times(x)+((m \times g)(x-50 \mathrm{cm}))=0

                  (m \times g \times x)-(50 \times m \times g)+(m \times g \times x)=0

Taking out ‘mg’ as common and we get

                  2 x-50=0

                  2 x=50

                  x=\frac{50}{2}=25 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                 x^{\prime}=100 \mathrm{cm}-25 \mathrm{cm}=75 \mathrm{cm}

So, the stick should be pivoted at a distance of 75 cm at the free end

Now, replace mass with another mass. i.e., four times the initial mass (as given)

If in case, the meter stick is to be at balanced form, then the acting torques sum would be zero. So,

                   4 m g(x)+(m g)(x-50 c m)=0

                   4 m g x+m g x-50 m g=0

Taking out ‘mg’ as common and we get

                   5 x=50

                   x=\frac{50}{5}=10 \mathrm{cm}

Hence, the stick should be pivoted at a distance of,

                   x^{\prime}=100 \mathrm{cm}-10 \mathrm{cm}=10 \mathrm{cm}

So, the stick should be pivoted at a distance of 10 cm from the free end.

Therefore, the option B is correct 90 cm (10 cm from the weight).

3 0
2 years ago
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