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3 years ago

What resistance would produce a current of 200A with a potiential difference of 200V?

1 answer:

5 0

Aw, I hate physics, is this on Apex?

Resistance can be calculated with the information given in the question.
Equation for Resistance: R = V/I
V (voltage) = 200 Volts
I (current) = 200 Amps

So 200 divided by 200 = freaking 1

Answer: R = 1 (ohms)

Hope this Helps!

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The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?

Answer:

$\triangle E = 12.79 J$

Explanation:

Sprinters' tendons stretch, $x_s = 43 mm = 0.043 m$

Non athletes' stretch, $x_n = 32 mm = 0.032 m$

Spring constant for the two groups, k = 31 N/mm = 3100 N/m

Maximum Energy stored in the sprinter, $E_s = 0.5kx_s^2$

Maximum energy stored in the non athletes, $E_m = 0.5kx_n^2$

Difference in maximum stored energy between the sprinters and the non-athlethes:

$\triangle E = E_s - E_n = 0.5k(x_s^2 - x_n^2)\\\triangle E = 0.5*3100* (0.043^2 - 0.032^2)\\\triangle E = 0.5*31000*0.000825\\\triangle E = 12.79 J$

4 0

3 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

3 years ago

A rocket ship is accelerating at 200 m/s2, its mas is 135,000,000 kg. What is the force generated by this acceleration?

Rina8888 [55]

Acceleration does NOT "generate" force. Acceleration NEEDS force to make it happen. Without force ... provided by something else ... acceleration can't happen.

The force NEEDED to accelerate a mass with a certain acceleration is

Force needed = (mass) times (acceleration)

For the rocket ship in the question,

Force = (135,000,000 kg) times (200 m/s²)

Force = (135,000,000 x 200) kg-m/s²

Force = 27 Giga-Newtons (27,000,000,000 Newtons)

The gas-generator cycle F-1 rocket engine, developed in the US by Rocketdyne in the late 1950s, was used in the Saturn V rocket, the main launch vehicle of NASA's Apollo moon lander program . Five F-1 engines were used in the first stage of each Saturn V.

==> The thrust of each F-1 engine at full throttle is 7,770 kilo-Newtons.

It would take 3,475 of these F-1 rocket engines, running full-throttle, to provide the force calculated in the answer to this question. If you didn't have 3,475 F-1 rocket engines, then you couldn't accelerate 135,000,000 kg at 200 m/s².

(And by the way ... the mass of each F-1 engine is 8,400 kg. So 3,475 engines alone account for 22% of the mass you're trying to accelerate. And don't even get me started about the mass of the FUEL you'd need to carry.)

5 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$