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3 years ago

Please help With the question! Ill mark brainliest for whoever answers First.

Physics

2 answers:

IgorC [24]3 years ago

8 0

Answer: Asteroids

Explanation: space rubble

IntroductionAsteroids, sometimes called minor planets, are rocky remnants left over from the early formation of our solar system about 4.6 billion years ago. Most of this ancient space rubble can be found orbiting the sun between Mars and Jupiter within the main asteroid belt.

kodGreya [7K]3 years ago

6 0

I think it’s Asteroid

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Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>

Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

$v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}$

6 0

3 years ago

A museum curator moves artifacts into place on various different display surfaces. If the curator moves a 145 kg aluminum sculpt

NikAS [45]

We are given the mass of an <span>aluminum sculpture which is 145 kg and a horizontal force equal to 668 Newtons. The coefficient of friction can be determined by dividing the horizontal force by the weight of the object. In this case, 668 N / 145 * 9.8 equal to coeff of friction of 0.47</span>

8 0

3 years ago

Read 2 more answers

Two identical strings, of identical lengths of 2.00 m and linear mass density of μ=0.0065kg/m, are fixed on both ends. String A

kolezko [41]

Answer:

beat frequency = 13.87 Hz

Explanation:

given data

lengths l = 2.00 m

linear mass density μ = 0.0065 kg/m

String A is under a tension T1 = 120.00 N

String B is under a tension T2 = 130.00 N

n = 10 mode

to find out

beat frequency

solution

we know here that length L is

L = n × $\frac{ \lambda }{2}$ ........1

so λ = $\frac{2L}{10}$

and velocity is express as

V = $\sqrt{\frac{T}{\mu } }$ .................2

frequency for string A = f1 = $\frac{V1}{\lambda}$

f1 = $\frac{\sqrt{\frac{T}{\mu } }}{\frac{2L}{10}}$

f1 = $\frac{10}{2L} \sqrt{\frac{T1}{\mu } }$

and

f2 = $\frac{10}{2L} \sqrt{\frac{T2}{\mu } }$

beat frequency is = f2 - f1

put here value

beat frequency = $\frac{10}{2*2} \sqrt{\frac{130}{0.0065}}$ - $\frac{10}{2*2} \sqrt{\frac{120}{0.0065} }$

beat frequency = 13.87 Hz

6 0

3 years ago

Which figure represents the electric field lines of two negative point charges?<br> Help

Yuliya22 [10]

Answer:

The question is incomplete because the options are not given. We'll solve this question with the options.

There are two properties of negative charges which we have to consider to find the figure of its electrical field.

Firstly, for a negative charge, the electrical field lines are always directed radially outwards and they don't intersect.

Secondly, we know that similar charges repel each other, so there will be no electrical field present directly between these two negative charges. Electrical field line will be present between two charges only when there is a force of attraction.

Taking both of these facts into consideration, the electrical field line between two negatives is represented by the figure below.