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larisa [96]
3 years ago
12

ListenA bicycle and its rider have a combined mass of 80. kilograms and a speed of 6.0 meters per second. What is the magnitude

of the average force needed to bring the bicycle and its rider to a stop in 4.0 seconds?
a.1.2 × 102 N
b.3.2 × 102 N
c.4.8 × 102 N
d.1.9 × 103 N
Physics
1 answer:
Setler [38]3 years ago
3 0

Answer:

a) 1.2\times 10^2\ N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-6}{4}\\\Rightarrow a=-1.5\ m/s^2

The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

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Newton's 2nd law of motion: 

                             Force = (mass) x (acceleration)

Divide each side by (mass):     

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Define linear expansivity
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Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
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8.829 m/s²

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g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

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Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

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Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
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The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

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  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
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To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

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Two point charges, a +45nC charge X and a +12nC charge Y are separated by a distance of 0.5m.
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A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

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C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

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Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

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Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

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Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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