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Ksivusya [100]
2 years ago
10

Yasmin determines the velocity of a car to be 15 m/s. The accepted value for the velocity of the car is 15.6 m/s. What is Yasmin

's percent error? 0.04% 3.3% 3.8% 60%
Physics
2 answers:
snow_lady [41]2 years ago
6 0

Answer:

3.8%

Explanation:

edge

Flauer [41]2 years ago
3 0

answer above is correct

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Betelgeuse is the bright red star representing the left shoulder of the constellation Orion. All the following statements about
Sav [38]

Answer:

Option A; ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

Explanation:

A red supergiant star is a larger and brighter type of red giant star. Red supergiants are often variable stars and are between 200 to 2,000 times bigger than the Sun. Example is Betelgeuse.

Betelgeuse is one of the largest known stars, it has a diameter of about 700 times the size of the Sun or 600 million miles, it emits almost 7,500 times as much energy as the Sun, it has a rather low surface temperature (6000F compared to the Sun's 10,000F); this means that it has a more cooler surface than the Sun's surface.

This low temperature also means that the star will appear orange-red in color, and the combination of size and temperature makes it a kind of star called a red super giant.

Although, all the statements above are correct, the only one that can be inferred from the red color of Betelgeuse is that ITS SURFACE IS COOLER THAN THE SURFACE OF THE SUN.

3 0
2 years ago
Round your answers to one decimal place. This parallel circuit has two resistors at 15 and 40 ohms. What is the total resistance
goldenfox [79]
-- With two resistors in parallel, the total effective resistance is
the reciprocal of  (1/R₁ + 1/R₂).

1/R₁ + 1/R₂  = 1/15 + 1/40

= 8/120 + 3/120

= 11/120

So the total effective resistance is 120/11 = 10.9 ohms .

Current = (voltage) / (resistance)

= 12 / (120/11)

= (12 · 11) / 120

= 132/120  =  1.1 Amperes  
4 0
2 years ago
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Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b
kolbaska11 [484]
To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength 
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV 

= 1.240x10^-10 m 

= 1.240x10^-1 nm
3 0
2 years ago
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Three point charges are arranged on a line. Charge q3 = +5.00 nC and is at the origin. Charge q2 = -2.00 nC and is at x = 5.00 c
tatuchka [14]

Answer:

q₁= +0.5nC

Explanation:

Theory of electrical forces

Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

To solve this problem we apply Coulomb's law:

Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters

Data:

Equivalences

1nC= 10⁻⁹ C

1cm= 10⁻² m

Data

q₃=+5.00 nC =+5* 10⁻⁹ C

q₂= -2.00 nC =-2* 10⁻⁹ C

d₂= 5.00 cm= 5*10⁻² m

d₁= 2.50 cm=  2.5*10⁻² m

k = 8.99*10⁹ N*m²/C²

Calculation of magnitude and sign of q1

Fn₃=0 : net force on q3 equals zero

F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.

F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.

We propose the algebraic sum of the forces on q₃

F₂₃ - F₁₃=0

\frac{k*q_{2} *q_{3} }{d_{2}^{2}  } -\frac{k*q_{1} *q_{3} }{d_{1}^{2}  }=0

We eliminate k*q₃ of the equation

\frac{q_{1} }{d_{1}^{2}  } = \frac{q_{2} }{d_{2}^{2}  }

q_{1} =\frac{q_{2} *d_{1} ^{2} }{d_{2}^{2}  }

q_{1} =\frac{2*10^{-9}*2.5^{2}*10^{-4}   }{5^{2}*10^{-4}  }

q₁= +0.5*10⁻⁹ C

q₁= +0.5nC

4 0
2 years ago
A 100-meter sprint is a race using only the straight side of a racetrack. A 400-meter sprint is a race that makes one complete l
Effectus [21]
Velocity is a vector quantity and depends on both speed and direction.
In 100m you only travel straight in one direction.
But in 400m you have to turn corners and then go back the way you came, and then turn another corner, you're changing direction - hence changing velocity, even if the speed is the same.
5 0
2 years ago
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