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vampirchik [111]

3 years ago

5

A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. Find the magnitude of the

force on the wire if it carries a current of 2.2 A.

2 answers:

Evgesh-ka [11]3 years ago

8 0

Explanation:

Below is an attachment containing the solution.

Fittoniya [83]3 years ago

4 0

Answer : 0.814 newton

Explanation:

force (magnetic) acting on the wire is given by

F= ? , I=2.2amp , B = 0.37 T

F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N

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A 6.9-kg wheel with geometric radius m has radius of gyration computed about its mass center given by m. A massless bar at angle

Vilka [71]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is $\alpha =2.538 \ rad/s^2$

Explanation:

From the question we are told that

The mass of the wheel is m = 6.9 kg

The radius is $r = 0.69 \ m$

The radius of gyration is $k_G = 0.4\ m$

The angle is $\theta = 47^o$

The force which the massless bar is subjected to $F = 22.5 \ N$

Generally given that the wheels rolls without slipping on the flat stationary ground surface, it implies that point A is the center of rotation.

Generally the moment of inertia about A is mathematically represented as

$I_a = I_G + M* r^2$

Here $I_G$ is the moment of inertia about G with respect to the radius of gyration which is mathematically represented as

$I_G = M * k_G$

=> $I_a = k_G* M + M* r^2$

=> $I_a =0.4 * 6.9 + 6.9 * 0.69^2$

=> $I_a =6.045 \ kg \cdot m^2$

Generally the torque experienced by the wheel is mathematically represented as

$\tau = F * cos (47)$

=> $\tau = 22.5 * cos (47)$

=> $\tau = 15.34 \ kg \cdot m^2 \cdot s^{-2}$

Generally this torque is also mathematically represented as

$\tau = I_a * \alpha$

=> $15.34 = 6.045 * \alpha$

=> $\alpha =2.538 \ rad/s^2$

4 0

3 years ago

A 70.0-kilogram man is walking at a speed if 2.0 m/s. What is the kinetic energy?

Paraphin [41]

Answer:

140 J

Explanation:

From the the question, the mass of the man =70.0 kg and the speed at which the man is walking =2.0 m/s.

$K.E = \frac{1}{2}m {v}^{2}$

where K.E = the kinetic energy, m=mass and v= speed.

By substitution,

$K.E = \frac{1}{2} \times 70 \times {2}^{2}$

$\implies \: K.E = \frac{280}{2}$

$\implies K.E =140$

Hence the kinetic energy of the man is 140J

3 0

2 years ago

4. For a series of experimentally measured values give the definition of (a) average or mean value (b) the deviation from the me

Naya [18.7K]

Answer:

a) x_average = ∑ $x_{i}$ / n , b) Δx_{i} = x_{i} –x_average,

d) σ = √(1/n-1 ∑ Dx_{i}² )

Explanation:

Some definitions are requested

a) the average value is the sum of all the values divided by the number of them, if the uncertainties are random, this is the closest value to the real one

x_average = ∑ $x_{i}$ / n

b) The deviation from the mean value or absolute error is the measured value minus the average value

Δx_{i} = x_{i} –x_average

c) is the average value of the deviations

Δx_average = ∑ Δx_{i} / n

d) It is a measure of the dispersion of the values with respect to their average value, it takes the worst of all cases, widely used for large numbers of data

σ = √(1/n-1 ∑ Dx_{i}² )

Experimental results should be given as follows

Average value ± uncertainty and the standard deviation

(x_average + - Δx_average)

σ

3 0

2 years ago

"a grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. the angular speed is then increased to 1

kotegsom [21]

The average angular speed of the grindstone is 10 rad/s

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>Angular Speed</em><em> formula as follows:</em>

$\boxed{ \omega = \omega_o + \alpha t }$

$\boxed{ \theta = \omega_o t + \frac{1}{2} \alpha t^2 }$

$\boxed{ \omega^2 = \omega_o^2 + 2 \alpha \theta }$

$\boxed{ \theta = \frac{( \omega + \omega_o )}{2} t }$

<em>where :</em>

<em>ω = final angular speed ( rad/s )</em>

<em>ω₀ = initial angular speed ( rad/s )</em>

<em>α = angular acceleration ( rad/s² )</em>

<em>t = elapsed time ( s )</em>

<em>θ = angular displacement ( rad )</em>

$\texttt{ }$

<u>Given:</u>

radius of the grindstone = R = 4.0 m

initial angular speed = ω₀ = 8.0 rad/s

final angular speed = ω = 12 rad/s

elapsed time = t = 4.0 seconds

<u>Asked:</u>

average angular speed = ?

<u>Solution:</u>

<em>Firstly , we will calculate </em><em>angular displacement </em><em>as follows:</em>

$\theta = \frac{( \omega + \omega_o )}{2} t$

$\theta = \frac{ ( 12 + 8.0 ) }{2} \times 4.0$

$\theta = 10 \times 4.0$

$\boxed {\theta = 40 \texttt{ rad}}$

$\texttt{ }$

<em>Next , we could calculate the </em><em>average angular speed</em><em> as follows:</em>

$\texttt{average angular speed} = \theta \div t$

$\texttt{average angular speed} = 40 \div 4.0$

$\boxed{\texttt{average angular speed} = 10 \texttt{ rad/s}}$

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Moment of Inertia : brainly.com/question/13796477
The Ratio of the Moments of Inertia : brainly.com/question/2176655

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Rotational Dynamics

4 0

3 years ago

Read 2 more answers

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

Ilya [14]

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
$m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0$

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0

2 years ago