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3 years ago

I need help again!!!!!! Due today!!!! NO LINK PLEASE

Physics

1 answer:

Elden [556K]3 years ago

4 0

Answer:

The Acceleration of the object = 6.4 m/s²

Explanation:

Mass of block (m) = 5 kg

Action force on block, (F₁) = 40 N

To Find:

Acceleration of the object (a) = ?

Required solution:

Frictional force opposing the motion (F₂) = 8 N

Here in this question we have to find Acceleration of the object. So, firstly we have to find Net force of block after that we will find Acceleration of the object on the basis of conditions given above

⇒ Net force = Action force on block - Opposing friction force

⇒ F = F₁ - F₂

⇒ F = 40 - 8

⇒ F = 32 N

Now, we have to two elements that used in formula, Net force and Mass of block.

Net force of the block (F) = 32 N

Mass of block (m) = 5 kg

And we have to find Acceleration of the object.

We can find Acceleration of the object by using the Second law of Newton which says F = ma

Here,

F is the Force in N.

m is the Mass in kg.

a is the Acceleration in m/s².

So let's find Acceleration (a) !

† From second law of Newton

⇛ F = ma

⇛ a = F/m

⇛ a = 32/5

⇛ a = 6.4 m/s²

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To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

$V_d = \frac{I}{nAq}$

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

$I = 25 A$

$A= 1.2*20 *10^{-6} m^2$

$q= 1.6*10^{-19}C$

$N = 8.47*10^{19} mm^{-3}$

$V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}$

$V_d = 7.68*10^{-5}m/s$

The hall voltage is given by

$V=\frac{IB}{ned}$

Where

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n = number of free electrons

d = distance

e = charge of electron

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$V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}$

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3 years ago

Compute the power output (watts) during one minute of treadmill exercise, given the following: Treadmill grade-10% Horizontal sp

erma4kov [3.2K]

Answer:

c. 981 watts

$P=981\ W$

Explanation:

Given:

horizontal speed of treadmill, $v=100\ m.min^{-1}=\frac{5}{3} \ m.s^{-1}$

weight carried, $w=588.6\ N$

grade of the treadmill, $G\%=10\%$

Now the power can be given by:

$P=v.w$

$P=588.6\times\frac{5}{3}$ (where grade is the rise of the front edge per 100 m of the horizontal length)

$P=981\ W$

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3 years ago

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A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

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