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2 years ago

2 QUESTIONS!! PLEASE HELP QUICK

Physics

1 answer:

Zolol [24]2 years ago

4 0

Answer:

1. True 2.0

Explanation:

1. When balanced forces are acting on an object, they have a constant velocity, meaning that the velocity remains the same.

2. When an object begins at a starting point and returns back to that same starting point, its displacement would be 0.

I'm pretty confident in these answers, however, you must have studied this more than me, so let me know if you have any questions or concerns at all, I'll be glad to help. :)

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Need an answer pls!!! will give brainliest

solmaris [256]

Answer:

i think your answer is C

Explanation:

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2 years ago

Read 2 more answers

QUESTION 1

Molodets [167]

Question 1: C Question 2: B, Hope this Helps!

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2 years ago

A student travels 4.0m East, 8.0m South, 4.0m west, and finally 8.0m north. What is her displacement? What distance did she trav

erica [24]

no cap bro bro i thinks its poop Explanation:

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2 years ago

What is the mass of an object that has a weight of 110N ?

fgiga [73]

Weight (W) = 110 N
Acceleration due to gravity (g) = 9.8 m/s^2
Let the mass of the object be m.
By using the formula, W = mg, we get,
110 N = 9.8 m/s^2 × m
or, m = 110 N ÷ 9.8 m/s^2
or, m = 11.2 Kg

Answer:

The mass of the object is 11.2 Kg.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0

2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago