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3 years ago

The downlink frequency is lower than the uplink frequency. True or False

Physics

2 answers:

labwork [276]3 years ago

5 0

The answer is true not false

miv72 [106K]3 years ago

4 0

Answer:

I think it is true

Explanation:

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The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the insta

BartSMP [9]

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm. Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

$F=kx$

where

$F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N$ is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

$k=\frac{F}{x}=\frac{14.7}{0.02}=735 N/m$

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

$U=\frac{1}{2}kx^2=\frac{1}{2}(735)(0.03)^2=0.33 J$

This energy is entirely converted into kinetic energy of the cart, which is:

$U=K=\frac{1}{2}mv^2$

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.33)}{0.5}}=1.15 m/s$

4 0

3 years ago

A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260

ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

$\tau=F\times r$

$I\times\alpha=F\times r$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r$

$F=\dfrac{M}{2}a$ ...(I)

Here, F = tension

The force is

$F=m(g-a)$ ...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

$\dfrac{M}{2}a=m(g-a)$

$a=\dfrac{g}{1+\dfrac{M}{2m}}$

Put the value into the formula

$a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}$

$a=6.84\ m/s^2$

We need to calculate the tension in the rope

Using equation (I)

$F=\dfrac{M}{2}a$

Put the value into the formula

$F=\dfrac{12.1}{2}\times6.84$

$F=41.38\ N$

Hence, The tension in the rope is 41.38 N.

6 0

2 years ago

Please help<br> question #70

Keith_Richards [23]

can i get the question so that i can answer your question

8 0

3 years ago

Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3

Ahat [919]

Answer:

Explanation:

Some of the mass

8 0

2 years ago

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positively charged ebonite rod is brought close to a small ball. The rod does not touch the ball, which is made from a conductin

xxMikexx [17]

Answer:

The rod attracts the ball

Explanation:

5 0

3 years ago