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suter [353]
4 years ago
7

Is the series equivalent resistor larger than either resistor, or is it smaller?

Physics
1 answer:
VikaD [51]4 years ago
6 0
The equivalent of several resistors in series is always more than the biggest single one, because it's their sum. ... In parallel, the equivalent is always less than the smallest one.
You might be interested in
How far does a car go in 10 seconds at the speed of 44m/s
musickatia [10]

Speed=44m/s

Speed=distance/time

Distance=speed x time

Distance=44 x 10

Distance=440m

7 0
3 years ago
A professor sits on a rotating stool that is spinning at 10.0 rpm while she holds a heavy weight in each of her hands. Her outst
MAVERICK [17]

Answer:

0.5864 m

Explanation:

r_1 = 0.705 m

\omega_1 = 10 rpm

\omega_2 = 20.5 rpm

\left(m r_{1}^{2}\right) \omega_{1} &=\left(m r_{2}^{2}\right) \omega_{2}

r_{2}^{2} &=\frac{\left(m r_{1}^{2}\right) \omega_{1}}{\left(\omega_{2}\right)}

=\frac{\left(r_{1}^{2}\right) \omega_{1}}{\left(\omega_{2}\right)}

=\frac{\left((0.705 \mathrm{m})^{2}\right)(10 \mathrm{rpm})}{(20.5 \mathrm{rpm})}

=\sqrt{\frac{\left((0.705 \mathrm{m})^{2}\right)(10 \mathrm{rpm})}{(20.5 \mathrm{rpm})}}

r_{2} =0.5864\ m

The weights are 0.5864 m apart.

7 0
4 years ago
A 89.3 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.77 r
balu736 [363]

Answer:

\omega_{2} = 1.107\,\frac{rad}{s}

Explanation:

Let assume that circular platform is a solid cylinder. Given the absence of external forces, the situation can be analyzed by applying the Principle of Angular Momentum, which states that:

I_{1}\cdot \omega_{1} = I_{2}\cdot \omega_{2}

The initial moment of inertia is:

I_{1} = \frac{1}{2}\cdot (89.3\,kg)\cdot (1.69\,m)^{2}

I_{1} = 127.525\,kg\cdot m^{2}

Likewise, the final moment of inertia is:

I_{2} = 127.525\,kg\cdot m^{2} + (8.83\,kg)\cdot (1.352\,m)^{2} + (21.1\,kg)\cdot (1.69\,m)^{2}

I_{2} = 203.929\,kg\cdot m^{2}

The final angular speed is:

\omega_{2} = \frac{I_{1}}{I_{2}} \cdot \omega_{1}

\omega_{2} = \frac{127.525\,kg\cdot m^{2}}{203.929\,kg\cdot m^{2}} \cdot (1.77\,\frac{rad}{s} )

\omega_{2} = 1.107\,\frac{rad}{s}

4 0
3 years ago
Read 2 more answers
Reviewing Interactive Solution 5.46 will help in solving this problem. A stone is tied to a string (length = 0.551 m) and whirle
morpeh [17]

Answer:

The speed of the stone is

v = 7.45 m/s

Explanation:

Length, L=0.551m

maximum tension in the spring = 9.6%

So let speed of stone be

Tv = TH + 9.6/ 100 * TH

Tv - m*g = m*v^2/L

TH = m*v^2 / L

Factor mass to cancel in the equation

Solve to v

v^2= L*g*100 / 9.6

Replacing numeric:

v^2=0.551m*9.8m/s^2*100 / 9.6

v = sqrt( 56.24 m^2/s^2)

v = 7.45 m/s  

6 0
4 years ago
An 800 V/m electric field is directed along the +x-axis. If the potential at x = 0 m is 2000 V, what is the potential at x = 2 m
Art [367]

Answer:

400 V

Explanation:

Electric filed , E=800 V/m

V_1=2000 V

x_1=0,x_2=2 m

We know that

E=\frac{V_1-V_2}{x_2-x_1}

Using the formula

800=\frac{2000-V_2}{2-0}

800=\frac{2000-V_2}{2}

2000-V_2=800\times 2=1600

V_2=2000-1600

V_2=400 V

Hence, the potential at x=2 m=400 V

4 0
3 years ago
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