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BlackZzzverrR [31]

3 years ago

12

Suppose we wrap a string around the surface of a uniform cylinder of radius 38.0 cm that is supported by an axle passing through

its center. This cylinder rotates without friction and we tie a mass to the end of the string and let it fall under the influence of gravity. As the weight falls, we measure the constant acceleration of the falling mass to be 9.0 m/s2. Find the angular acceleration that the cylinder will experience under these conditions. (Assume the positive direction is the initial direction of rotation of the cylinder. Indicate the direction with the sign of your answer.)

1 answer:

liraira [26]3 years ago

4 0

Answer:

Angular acceleration = 23.68 rad / s²

Explanation:

Given that,

acceleration = 9m/s²

Therefore acceleration of string is 9m/s²

since string is constant in length

cylinder of radius 38.0 cm = 0.38m

Angular acceleration = a / r

Angular acceleration = 9 / 0.38

= 23.68 rad / s²

Angular acceleration = 23.68 rad / s²

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Calculated the measurement uncertainty for Kinetic Energy when :mass = 1.3[kg] +/- 0.4[kg]velocity= 5.2 [m/s] +/- 0.2 [m/s]KE= 1

andriy [413]

Answer:

$\rm KE\pm \Delta KE = 17.6\pm 6.8\ J.$

Explanation:

<u>Given:</u>

Mass, $\rm m\pm\Delta m = 1.3\pm 0.4\ kg.$

Velocity, $\rm v\pm \Delta v = 5.2\pm 0.2\ m/s.$

where,

$\rm \Delta m,\ \Delta v$ are the uncertainties in mass and velocity respectively.

The kinetic energy is given by

$\rm KE = \dfrac 12 mv^2 = \dfrac 12 \times 1.3\times 5.2^2=17.576\approx 17.6\ J.$

The uncertainty in kinetic energy is given as:

$\rm \dfrac{\Delta KE}{KE}=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\dfrac{\Delta KE}{17.6}=\dfrac{0.4}{1.3}+\dfrac{2\times 0.2}{5.2}\\\dfrac{\Delta KE}{17.6}=0.384\\\Rightarrow \Delta KE = 17.6\times 0.384 = 6.7854\ J\approx6.8\ J\\\\Thus,\\\\KE\pm \Delta KE = 17.6\pm 6.8\ J.$

7 0

4 years ago

Vector A has a magnitude of 30 units. Vector B is perpendicular to vector Aand has a magnitude of 40 units. What would the magni

Fudgin [204]

Answer:

$|\vec A + \vec B| = 50 units$

Explanation:

As we know that magnitude of two vectors is given as

$|\vec A + \vec B| = \sqrt{A^2 + B^2 + 2AB cos\theta}$

here we know that

A = magnitude of vector A

B = magnitude of vector B

$\theta$ = angle between two vectors

so here we know that

A = 30 units

B = 40 units

angle = 90 degree

so we have

$|\vec A + \vec B| = \sqrt{30^2 + 40^2 + 2(30)(40)cos90}$

$|\vec A + \vec B| = \sqrt{30^2 + 40^2}$

$|\vec A + \vec B| = 50 units$

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3 years ago

A Sonometer wire of length

Scrat [10]

Answer:

-75 cm

Explanation:

At l ; F = 350 Hz

At l + 15 cm ; F = 280 Hz

I = 350

I + 15 = 280

280I = 350(I + 15)

280I = 350I + 5250

280I - 350I = 5250

-70I = 5250

I = - 75cm

The length is - 75 cm

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3 years ago

A shopper pushes a 5.32 kg grocery cart

Juli2301 [7.4K]

Answer:

$\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

Explanation:

According to “Newton's second law”

“Force” is “mass” times “acceleration”, or F = m× a. This means an object with a larger mass needs a stronger force to be moved along at the same acceleration as an object with a small mass

Force = mass × acceleration

$\text { Acceleration }=\frac{\text { force }}{\text { mass }}$

Given that,

Mass = 5.32 kg

$\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}$

$x=-28.7^{\circ}$

F = 12.7N

Normal force = mg + F sinx,

“m” being the object's "mass",

“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

$\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })$

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

$\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}$

$\text { Acceleration }=\frac{58.232}{5.32}$

$\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}$

$\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}$

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3 years ago

Read 2 more answers

A 30 g horizontal metal bar, 13 cm long, is free to slide up and down between two tall, vertical metal rods that are 13 cm apart

natita [175]

Answer:

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Explanation:

It is given that,

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Length of the bar, l = 13 cm = 0.13 m

Magnetic field, $B=5.5\times 10^{-2}\ T$

Resistance, R = 1.2 ohms

We need to find the terminal speed oat which the bar falls. When terminal speed is reached,

Force of gravity = magnetic force

$mg=ilB$ ..................(1)

i is the current flowing

l is the length of the rod

Due to the motion in rods, an emf is induced in the coil which is given by :

$E=Blv$ , v is the speed of the bar

$iR=Blv$

$i=\dfrac{Blv}{R}$

Equation (1) becomes,

$mg=\dfrac{B^2l^2v}{R}$

$v=\dfrac{mgR}{B^2l^2}$

$v=\dfrac{0.03\times 9.8\times 1.2}{(5.5\times 10^{-2})^2(0.13)^2}$

v = 6901.07 m/s

So, the terminal speed at which the bar falls is 6901.07 m/s. Hence, this is the required solution.

5 0

3 years ago