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3 years ago

2. A motor with a resistance of 32 is connected to a voltage source. Four amps of current

Physics

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer:128

Explanation:

solve by V=IR formula

V=32 * 4

V= 128

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Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle 0 between th

erik [133]

Based on the mass of the bob and the angle between the cord and the vertical, the pendulum length is 0.50m.

The maximum kinetic energy can be found to be 9.42 x 10⁻⁴J.

<h3>What is the pendulum length?</h3>

This can be found as:

= g-force / w²

Solving gives:

= 9.8 / 4.43²

= 0.4998 m

= 0.50 m

<h3>What is the maximum kinetic energy?</h3>

This can be found as:

= 0.5 × m × w² × A²

Maximum kinetic energy is:

= 0.5 × 60 × 10⁻³ × (4.43 × 0.4998 x 0.08 rad)²

= 9.42 x 10⁻⁴J

Find out more on maximum kinetic energy at brainly.com/question/24690095.

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1 year ago

What is the range of motion of the elbow if extension is 0° and flexion is 145°?

inna [77]

Answer:

0 to 145 degrees

Explanation:

The normal range of flexion and extension is from 0 to 145 degrees.

6 0

2 years ago

It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

$\epsilon = BvL sin \theta$

where

$B=5\cdot 10^{-5} T$ is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

$\theta=40^{\circ}$ is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

$\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV$

8 0

3 years ago

A child has a toy car on a horizontal platform. The car starts from rest and reaches a maximum speed in 4 s. If the mass of the

dmitriy555 [2]

Answer:

a=4m/s²

Explanation:

F=ma

0.4=0.1a

7 0

3 years ago

Read 2 more answers

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago