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lina2011 [118]
3 years ago
14

A cyclist rides at 6.20 m/s through a intersection. A stationary car begins to

Physics
1 answer:
Xelga [282]3 years ago
8 0

Answer:

The width of the intersection is 20 meters

Explanation:

The speed with which the cyclist is riding, v₁ = 6.20 m/s

The rate at which the car starts to accelerate, a = 3.844 m/s²

The initial velocity of the car = The car is stationary at the start = 0 m/s

The time at which the cyclist and the car reach the other side of the intersection = The same time;

Let 't' represent the time at which the cyclist and the car both reach the other side of the intersection, we have;

The distance travelled by the cyclist = The distance traveled by the car

∴ v₁ × t = 1/2 × a × t²

Plugging in the values for 'v₁', and 'a' in the above equation, we get;

6.20 × t = 1/2 × 3.844 × t²

∴ 1.922·t² - 6.20·t = 0

∴ t·(1.922·t - 6.20) = 0

t = 0, or t = 6.20/1.922 = 100/31

The time at which the cyclist and the car both reach the other side of the intersection, t = 100/31 seconds

The with of the intersection, w = v₁ × t

∴ w = 6.20 × 100/31 = 100/5 = 20

The width of the intersection, w = 20 meters.

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A shotputter throws the shot with an initial speed of 15.6 m/s at a 30.0 degree angle to the horizontal. Calculate the horizonta
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An object of mass 0.50 kg is transported to the surface of Planet X where the object’s weight is measured to be 20 N. The radius
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Answer:

g'=40\ m.s^{-2}

Explanation:

Given:

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<u>Now free fall acceleration on planet X:</u>

W=m.g'

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Two particles are fixed on an x axis. Particle 1 of charge 44.9 μC is cated at х- 24.5 cm; particle 2 of charge Q is located at
Kryger [21]

The force that is being exerted on particle 3 by particle 1 is equal to:

F_{13}=\frac{K*Q_1*Q_3}{r13^2}

r_{13} = \sqrt{(0.245m)^2 + (0.245m)^2}=0.3465m^2

F_{13}=\frac{9*10^9Nm^2/C^2*44.9*10^-6C*38.9*10^-6C}{0.3465m^2}=45.37N

As both particles has positive charges, the particles will repel each other and the resultant force will have the direction of the vector r_{13}. Therefore, F_{13} will have x and y components equal to:

F{13x}=F{13}*\frac{r13x}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N

F{13y}=F{13}*\frac{r13y}{|r13|}=45.37N*\frac{0.245m}{0.3465m}=32.08 N

In order to calculate Force between particles 2 and 3, we first assume Q2 to be possitive (if it's negative the result will have a negative value, so this doesn't matter). We follow the same line of reasoning we used to calculate F13, just that Q2 will be unknown.

F_{23}=\frac{K*Q_2*Q_3}{r13^2}

r_{23} = \sqrt{(-0.0753m)^2 + (0.245m)^2}=0.2563m^2

F_{23}=\frac{9*10^9Nm^2/C^2*Q_2*38.9*10^-6C}{0.2563m^2}=1.36*10^6Q_2

F{23x}=F{23}*\frac{r23x}{|r23|}=1.36*10^6Q_2*\frac{-0.0753m}{0.2563m}= -401286.15Q_2

F{23y}=F{23}*\frac{r23y}{|r23|}=1.36*10^6Q_2*\frac{0.245m}{0.2563m}=1.305*10^6 Q_2

a) For incise a, F13y + F23y has to be equal to 0:

F{13y}+F{23y}=0

32.08 N+1.305*10^6 Q_2=0

Q_2=\frac{-32.08}{1.305*10^6}=-24.6*10^{-6}C =-24.6uC

b) For incise b, F13x + F23x has to be equal to 0:

F{13x}+F{23x}=0

32.08 N - 401286.15 Q_2=0

Q_2=\frac{32.08}{401286.15}=80*10^{-6}C =80uC

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