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3 years ago

How many Mars-sized planets can fit in Jupiter

Physics

1 answer:

Maru [420]3 years ago

7 0

Answer:

You could put over six planets the size of Mars inside the Earth. The largest planet in our Solar System, Jupiter's size is astounding. Jupiter has a volume of 1.43 x 1015 cubic kilometers. To show what this number means, you could fit 1321 Earths inside of Jupiter

Explanation:

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The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

galben [10]

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ $\angle BCA = (90-30)$

⇒ $=60^{\circ}$

then,

⇒ $\angle BAC=(180-45-60)$

⇒ $=75^{\circ}$

Now,

The corresponding angle will be:

⇒ $\angle FAC=60^{\circ}$

⇒ $\angle FAB=70+60$

⇒ $=135^{\circ}$

Aspect of F across the AC arm will be:

= $F\times cos(60)$

On putting the values of F, we get

= $500\times (.5)$

= $200 \ Newton$

Component F along the AC (in magnitude) will be:

= $F\times cos(135)$

= $500\times (-.707)$

= $-353.5 \ N \$

4 0

2 years ago

Some earthquake waves, a cat’s purr, and elephant communications consist of ultrasonic waves.

Kobotan [32]

The answer to your question is true.

7 0

3 years ago

Read 2 more answers

PLEASE HELP ME!! What is the difference between direct current and alternating current?

Paha777 [63]

In a direct current the electric charge flows in one direction.
in an alternating the electric charge changes in its direction periodically.

6 0

3 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago

A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the

Nesterboy [21]

Answer:

A.) 1372 N

B.) 1316 N

C.) 1428 N

Explanation:

Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:

a. The load moves downward at a constant velocity

At constant velocity, acceleration = 0

T - mg = ma

T - mg = 0

T = mg

T = 140 × 9.8

T = 1372N

b. The load accelerates downward at a rate 0.4 m/s??

Mg - T = ma

140 × 9.8 - T = 140 × 0.4

1372 - T = 56

-T = 56 - 1372

- T = - 1316

T = 1316N

C. The load accelerates upward at a rate 0.4 m/s??

T - mg = ma

T - 140 × 9.8 = 140 × 0.4

T - 1372 = 56

T = 56 + 1372

T = 1428N

8 0

3 years ago