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timama [110]
1 year ago
7

Refer to the attached image!!!​

Physics
1 answer:
dimaraw [331]1 year ago
6 0

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

  • T is time of motion
  • g is acceleration due to gravity
  • θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

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I need an answer for this plzz!!<br>number 2 <br>anybody can help ??
Anuta_ua [19.1K]
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
8 0
3 years ago
If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, t
muminat

Answer:

75.5g

Explanation:

From the ionic equation, we can write

CU^{2+}+SO^{2-}_{4}\\

next we find the number of charge

Note Q=it

for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs

hence

Q=8.5*13500\\Q=114750C

Since one faraday represent one mole of electron which equal 96500C

Hence the number of mole produced by 114750C is

114750/96500=1.2mol

The mass of copper produced is

mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g

Hence the amount of copper produced is 75.5g

7 0
3 years ago
Read 2 more answers
How would you describe brass since it ia used in weapons, pipes, intruments and ect? A compound, Alloy, Element l, Or molecule?
Alik [6]
<span>Brass is an <u>alloy</u>. An alloy is a mixture of elements to form a unique material. Brass is a mixture of copper and zinc and the percentage of each element depends on the desired material. It has a higher malleability than bronze or zinc. Meaning that it can be bend easily into it desired form.</span>
5 0
2 years ago
A geologist is trying to determine what time period a layer of rock was formed.
vodka [1.7K]

Answer:

B

Explanation:

You always want to test as many samples as possible

8 0
2 years ago
An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get
Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

mV_i = (m-m_O)V_f - m_OV_O

where,

m=Total mass

m_O = Mass of Object

V_i = Velocity before throwing

V_f = Final Velocity

V_O = Velocity of Object

Our values are:

m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg

Solving to find the final speed, after throwing the object we have

V_f=\frac{mV_0+m_TV_O}{m-m_O}

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) 5.3Kg\rightarrow 15m/s

V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}

V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}

V_{f1}=0.8511m/s

B) 7.9Kg\rightarrow 11.2m/s

V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}

V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}

V_{f2} = 2.0173m/s

C) 10.5Kg\rightarrow 7m/s

V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}

V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}

V_{f3} = 3.63478m/s

Therefore the final velocity of astronaut is 3.63m/s

7 0
3 years ago
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