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4 years ago

Please help! I especially need help with the second question but help with the first one would be most appreciated!

Physics

1 answer:

lara31 [8.8K]4 years ago

4 0

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/ $s^{2}$

Explanation:

Given that Team-A pulls with a force , $F_{1} = 50N$

and Team-B pulls with a force , $F_{1} = 45N$

∵ $F_{1} > F_{2}$

The rope will move in the direction of force $F_{1}$ .

∴ Team-A will win.

b) Considering both the teams as one system of total mass , $m = 246+253 = 499 kg$

Net force on the system , $F = F_{1} - F_{2}$ = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/ $s^{2}$

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

The sensitivity of a measuring instrument is the value of the smallest quantity that can be read or estimated with it. What is t

nirvana33 [79]

Answer:

The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.

6 0

4 years ago

What happens to the iron in the coilgun if the electricity in the coil was turned on

Andrews [41]

<em><u>The piece of iron has become a magnet. Some substances can be magnetized by an electric current. When electricity runs through a coil of wire, it produces a magnetic field. The field around the coil will disappear, however, as soon as the electric current is turned off.</u></em>

7 0

3 years ago

Find the value of currents through each branch

Irina-Kira [14]

Answer:

the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

(I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

(I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

$\left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]$

These equations have the solution ...

$\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]$

This means the branch currents are as follows:

top left: I2 = 0.625 A

middle left: I1 = 2.500 A

bottom left: I1-I2 = 1.875 A

top center: I2+I3 = 2.500 A

bottom center: I2+I3-I1 = 0 A

right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

(I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

It helps to be familiar with the formulas for resistors in series and parallel.

8 0

3 years ago

Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?

Vladimir [108]

Answer:

Explanation:

a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .

First condition of equilibrium is that net force on the body should be zero.

or F net = 0

A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

So first condition of equilibrium is not satisfied.

Second condition is that , net torque acting on the body must be zero.

In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie towards the line joining the body under rotation with the axis .

F is along r

torque = r x F

= r F sinθ

θ = 0 degree

torque = 0

Hence 2nd condition is fulfilled.

7 0

3 years ago