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2 years ago

If we could easily create machines or systems with no friction, what could we use them for?

Physics

1 answer:

Alexus [3.1K]2 years ago

3 0

Answer:

We could make a frictionless generator.

Explanation:

This would change life by making the cost for energy next to zero

You might be interested in

Why cant your rocket could never reach the speed of light?

Oksanka [162]

Answer:

The length of the object would shrink to zero which is not possible.

Explanation:

A rocket or any body cannot reach the speed of light because according to theory of relativity the and the Lorentz factor the length of the object would shrink to zero and the time dilation for that body would be infinite.

The Lorentz factor is given as:

$\gamma=\frac{1}{\sqrt{\frac{v^2}{c^2} } }$

where:

v = speed of the moving object

c = speed of light

4 0

3 years ago

What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?

lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

$v = H_o D\\\\H_o = \frac{v}{D}$

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

$H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}$

3 0

2 years ago

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at

Stolb23 [73]

Answer:

a. Fx = -8.089 N b. Fy = 3.525 N c. 8.824 N d. 336.45°

Explanation:

Since F₁ = 9.2 N and acts at 57° above the negative axis in the second quadrant, its x-component is -F₁cos57° and its y- component is F₁sin57°

Since F₁ = 5.2 N and acts at 53.7° below the negative axis in the third quadrant, its x-component is -F₂cos53.7° and its y- component is -F₂sin53.7°

Part A

What is the x component Fx of the resultant force?

The x component of the resultant force Fx = -F₁cos57° + -F₂cos53.7° = -9.2cos57° + (-5.2cos53.7°) = (-5.011 - 3.078) N = -8.089 N

Part B

What is the y component Fy of the resultant force?

The y component Fy of the resultant force = F₁sin57° + -(F₂sin53.7°) = 9.2sin57° - 5.2sin53.7° = (7.716 - 4.191) N = 3.525 N

Part C

What is the magnitude F of the resultant force?

The magnitude F of the resultant force = √(Fx² + Fy²)

F = √(-8.089² N + 3.525² N) = √65.432 + 12.426 = √77.858 = 8.824 N

Part D

What is the angle ? that the resultant force forms with the negative x axis?

The angle the resultant force makes with the negative x axis is given by

θ = tan⁻¹(Fy/Fx) = tan⁻¹(3.525/-8.089) = tan⁻¹-0.4358 = -23.55°.

To measure it from the negative x axis, we add 360. So, our angle = 360 -23.55 = 336.45°

7 0

3 years ago

1. A kangaroo hops 84 m to the east in 7 seconds.

DENIUS [597]

Explanation:

Given parameters:

Distance hopped = 84m

Displacement = 84m due east

Time = 7s

Unknown:

Speed of kangaroo = ?

Velocity of kangaroo = ?

Solution:

To solve this problem,

Speed = $\frac{distance}{time }$ = $\frac{84}{7}$ = 12m/s

Velocity = $\frac{displacement}{time}$ = $\frac{84}{7}$ = 12m/s due east

3 0

3 years ago

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4

dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0

3 years ago