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andreyandreev [35.5K]
4 years ago
7

Will give 15 points.

Physics
1 answer:
dem82 [27]4 years ago
5 0
Bar Graph, Line Graph, and Circle Graph.
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Suppose you have a car with a battery that applies 12.5 V to the starter.
Marat540 [252]

Answer:

R = 0.1 ohms

Explanation:

It is given that,

Voltage of the battery, V = 12.5 V

Current flowing in the car's starter, I = 125 A

We need to find the effective resistance of a car's starter. It can be calculated using Ohm's law. Let R is the resistance.

V=IR\\\\R=\dfrac{V}{I}\\\\R=\dfrac{12.5}{125}\\\\R=0.1\ \Omega

So, the resistance of the car's starter is 0.1 ohms.

7 0
3 years ago
A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.
expeople1 [14]

Answer:

Magnitude = 3.64 × 10^6  

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × 10^6    kilometers

turn = 70°

travel an additional = 2.7 × 10^6   kilometers

solution

we will consider here

Px = 1.7 × 10^6  

Py = 0

Qx =2.7 × 10^6  cos(70)

Qy= 2.7 × 10^6  sin(70)

so that

Hx = Px + Qx    ............1

Hx = 2.62 × 10^6  

and

Hy = Py + Qy      ..........2

Hy = 2.53 × 10^6  

so Magnitude = \sqrt{((2.62 \times 10^6  )^2+(2.53 \times 10^6)^2)}

Magnitude = 3.64 ×  

so direction  will be

tan စ = Hy ÷ Hx    ......................3

tan စ  = \frac{2.53}{2.62}

tan စ  = 0.9656

စ = 43.9°

3 0
3 years ago
A simple Atwood’s machine uses a massless
Lady bird [3.3K]
The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t). 
<span>Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) </span>
<span>then a = 6.7/1.9 = 3.526 m/s². </span>
<span>The Atwood's Machine also has the formula d= distance = 1/2a(t²) </span>
<span>distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², </span>
<span>a = 3.526 m/s² the same acceleration. </span>
<span>a= g(m1-m2) / m1+m2) </span>
<span>m1a + m2a = m1g - m2g </span>
<span>m1a - m1g = -m2g - m2a </span>
<span>3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 </span>
<span>-6.28 m1 = -13.34 m2 </span>
<span>0.47 m1= m2 </span>
<span>if 24J = 1/2mv² </span>
<span>then 24J = 1/2 m1 ( 6.7)² </span>
<span>48/ 44.89 = m1 </span>
<span>1.069 kg = m1 , then </span>
<span>0.47(1.069) = m2 </span>
<span>0.503 kg = m2</span>
6 0
4 years ago
Describe and apply Newton’s Three Laws
Maurinko [17]
Newton's<span> First </span>Law of Motion<span>: I. Every object in a state of uniform </span>motion<span> tends to remain in that state of </span>motion<span> unless an external force is applied to it. This we recognize as essentially Galileo's concept of inertia, and this is often termed simply the "</span>Law<span> of Inertia".</span>
5 0
3 years ago
An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed incre
yarga [219]

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

v^2=v_o^2+2ax         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}

Next, you use the second Newton law to calculate the force:

F=ma

m: mass of the electron = 9.11*10^-31kg

F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N

The quotient between the weight of the electron and the force F is:

\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}

The force F is 2.59*10^11 times the weight of the electron

8 0
3 years ago
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