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3 years ago

Which famous astronomer discovered jupiter’s largest moons?.

Physics

1 answer:

balu736 [363]3 years ago

4 0

Answer:

Galileo Galilei

Explanation:

The famous astronomer Galileo Galilei was the one who discovered Jupiter’s large moons.

I hope it helps! Have a great day!

Muffin ^^

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Which waves have wavelengths longer than those of visible light? Give an example of how each kind of wave is used.

Kazeer [188]

1. Radio Waves
ex. Wi-Fi
2. Microwaves
ex. Mobile Phones
3. Infrared Radiation
ex. Heat Lamps

6 0

3 years ago

Read 2 more answers

A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 36.5 m/s. At the same time, it h

sammy [17]

Answer:

66.26 m/s

Explanation:

Horizontal velocity, Vx = 55.3 m/s

Vertical velocity, Vy = 36.5 m/s

The value of the resultant velocity is given by the vector sum of the two velocities which are acting at 90°.

$V=\sqrt{V_{x}^{2}+V_{y^{2}}}$

$V=\sqrt{55.3^{2}+36.5^{2}}}$

V = 66.26 m/s

Thus, the velocity of the vehicle is 66.26 m/s along its descent path.

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3 years ago

A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0

3 years ago

Find the voltage across the 15 Q resistor.<br> [?] V<br><br> No links please

Margaret [11]

Answer:

Explanation:

same idea as before Liam, first, find the parallel resistance in 35 || 20

(35*20) / (35+20) = 700 / 55 = 12.727272 ohms

now add the 12.727272 + 15 = 27.727272 ohms total resistance

V = IR

10 = I * 27.727272

10 / 27.727272 = I

0.360655 = I

V = IR (again, but across the 15 ohm resistor)

V = 0.360655 * 15

V = 5.4098

4 0

2 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

ArbitrLikvidat [17]

Answer:

The mass of the solid cylinder is $m = 1612.5 \ kg$

Explanation:

From the question we are told that

The radius of the grinding wheel is $R = 0.330 \ m$

The tangential force is $F_t = 250 \ N$

The angular acceleration is $\alpha = 0.940 \ rad/s^2$

The torque experienced by the wheel is mathematically represented as

$\tau = I * \alpha$

Where I is the moment of inertia

The torque experienced by the wheel can also be mathematically represented as

$\tau = F_t * r$

substituting values

$\tau = 250 * 0.330$

$\tau = 82.5 \ N\cdot m$

$82.5 = I * \alpha$

$82.5 = I * 0.940$

$I = 87.8 \ kg \cdot m^2$

This moment of inertia can be mathematically evaluated as

$I = \frac{1}{2} * m* r^2$

substituting values

$87.8 = \frac{1}{2} * m* (0.330)^2$

=> $m = 1612.5 \ kg$

5 0

3 years ago