When the forces applied to a body are balanced, the body is said to be in __________________________________

Can someone help me understand this assignment and give me some hints and tips on what I should do, please?

katen-ka-za [31]

First you have to note down you horoscope for 2-3 days.

Daily observe what is written in your horoscope is true or not. You can observe your entire day movements.

Now you have to conclude whether your horoscope was true or not and you have to write an essay explaining what had happened to you in past observation and conclude whether you believe in astronomy or not depending on your observations

6 0

3 years ago

A 121-cm-long, 4.00 g string oscillates in its m = 3 mode with a frequency of 180 Hz and a maximum amplitude of 5.00 mm. What ar

Alex777 [14]

Answer:

0.8067 m

69.696 N

Explanation:

m = Mode = 3

M = Mass of string = 4 g

f = Frequency = 180 Hz

l = Length of string = 121 cm

Length of the string is given by

$l=m\dfrac{\lambda}{2}\\\Rightarrow \lambda=2\dfrac{l}{m}\\\Rightarrow \lambda=2\times \dfrac{1.21}{3}\\\Rightarrow \lambda=0.8067\ m$

The wavelength is 0.8067 m

Linear density is given by

$\mu=\dfrac{M}{l}\\\Rightarrow \mu=\dfrac{4\times 10^{-3}}{1.21}$

Speed of the wave

$v=f\lambda\\\Rightarrow v=180\times 2\times \dfrac{1.21}{3}\\\Rightarrow v=145.2\ m/s$

Speed of wave is given by

$v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow T=\mu v^2\\\Rightarrow T=\dfrac{4\times 10^{-3}}{1.21}\times 145.2^2\\\Rightarrow T=69.696\ N$

Tension is given by 69.696 N

8 0

3 years ago

I need help for a assignment NOT A QUIZ

Debora [2.8K]

Answer:

what

ITS BLANK FOR ME I WISH I COULD HELP YOU

3 0

3 years ago

Read 2 more answers

A steep slope on a displacement vs. time graph indicates a larger velocity

Anni [7]

As we know that the velocity at any instant is defined as slope of displacement time graph at that instant

so we will have

$v = \frac{ds}{dt}$

now we will say

slope of the graph is also defined as

$slope = tan\theta$

now on increasing the value of the slope we will have more steep graph

so more steep graph will have more value of theta

so for steep graph

$\theta = large$

$tan\theta = large$

so value of the velocity will be large at steep graph

6 0

3 years ago

Read 2 more answers

A 4.00-kg object traveling 20.0 m/s west collides with a 6.00-kg mass object traveling 12.0 m/s east. The collision is perfectly

nlexa [21]

Answer:

The velocity of the 4.00 kg object after the collision is 12 m/s.

Explanation:

Given that,

Mass of object $m_{1} = 4.00\ kg$

Velocity of object $v_{1} = 20.0\ m/s$

Mass of another object $m_{2} = 6\ kg$

Velocity of another object $v_{2}= 12.0\ m/s$

We need to calculate the relative velocity

$v_{r}=v_{1}-v_{2}$

$v_{r}=20-12=8\ m/s$

The relative velocity is 8 m/s in west before collision.

We know that,

In one dimensional elastic collision, the relative velocity before collision equals after collision but with opposite sign.

So, The relative velocity after collision must be 8 m/s in east.

So, The object of 6.00 kg is going 20 m/s and the object of 4.00 kg is slows down to 12 m/s.

Hence, The velocity of the 4.00 kg object after the collision is 12 m/s.

4 0

3 years ago

When the forces applied to a body are balanced, the body is said to be in ___________________________________