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True [87]
2 years ago
14

A car travels 6 m north for 2 seconds and 8 m south for 5 seconds. What is the car's

Physics
2 answers:
Degger [83]2 years ago
8 0
2 m/s. 6+8/2+5 = 14/7 = 2
Zina [86]2 years ago
5 0

Answer:

  (b)  0.29 m/s

Explanation:

The average speed is found by dividing the net displacement by the amount of time it took to achieve that displacement.

__

The net displacement is ...

  6 m -8 m = -2 m . . . . . where north is positive

The total time is ...

  2 s + 5 s = 7 s

Then the distance per unit time (speed) is ...

  (2 m)/(7 s) = 2/7 m/s ≈ 0.29 m/s

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . The velo
Goshia [24]

Answer:

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 = 83-44.3=38.7^{\circ}

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, d^2 representing the distance between the planes, we see that:

d^2=2117^2 + 1711^2 -2\times (2117)\times (1711)cos(38.7)=1755480.2482

d = 1324.9453 m

4 0
3 years ago
Distinguish between basic and derived units
alex41 [277]

Answer:

Base units are defined units based on specific objects or events in the physical world. Derived units are defined by combining base units.

Base units are defined by a particular process of measuring a base quantity whereas derived units are defined as algebraic combinations of base units. For example, length is a base quantity in both SI and the English system, but the meter is a base unit in the SI system only.

8 0
2 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
If you have two uncertainties, and they are from two different sources and contribute to the uncertainty of a measurement, what
Darya [45]

The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.

                           Δm = ∑  | \frac{dm}{dx_i} | \ \Delta x_i

Physical quantities are precise values ​​of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the  cases, all the errors add up.

If m is the calculated quantity, x_i the measured values ​​and Δx_i the uncertainty of each value, the total uncertainty is

                      Δm = ∑  | \frac{dm}{dx_i } | \ \Delta x_i    | dm / dx_i | Dx_i

               

for instance:

If the magnitude is  a average of two magnitudes measured each with a different error

                     m = \frac{m_1+m_2}{2}

                     Δm = | \frac{dm}{dx_1} |  Δx₁ + | \frac{dm}{dx_2} | Δx₂

                     \frac{dm}{dx_1} = ½

                     \frac{dm}{dx_2} = ½

                     Δm = \frac{1}{2} Δx₁ + ½ Δx₂

                     Δm = Δx₁ + Δx₂

In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.

Learn more about propagation errors here:

brainly.com/question/17175455

6 0
3 years ago
A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential
BARSIC [14]

Answer:

1.88\cdot 10^{-5} A

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{\epsilon_0 A}{d} (1)

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

Q=CV (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

J=\frac{Q}{t} (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}

where we have

A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2

d = 0.58 mm = 5.8\cdot 10^{-4} m

\frac{V}{t}=500,000 V/s

Substituting into the equation, we find

J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A

6 0
3 years ago
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