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Julli [10]
3 years ago
11

(a) Suppose that a NASCAR race car is moving to the right with a constant velocity of +93 m/s. What is the average acceleration

of the car? (b) Twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed. What is the average acceleration of the car?
Physics
1 answer:
Doss [256]3 years ago
4 0
A. the average acceleration would be 0 since there is no change in velocity

b. Average acceleration = change in velocity/time

93- (-93)/12

= 186 /12

= 15.5 m

hope this helps

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A negative ion of charge -2e is located at the origin and a second negative ion of charge -3e is located nearby at x = 3.8 nm ,
Rus_ich [418]

Answer:

\vec{F}_{21}=-5.63\times 10^{-11}N\\\\\vec{F}_{21}=\\

Explanation:

Given that

Q_1 = -2e\, C\\\\Q_2=-3e\,C\\\\x= 3.8 \times 10^{-9}\,m\\\\y= 3.2 \times 10^{-9}\,m\\\\r=\sqrt{x^2+y^2}\\\\r= 4.96\times 10^{-9} m\\

As both charges are negative so there exist force of repulsion in direction as shown in figure.

F_{12}=\frac{kQ_1Q_2}{r^2}\\\\F_{12}= \frac{(9\times 10^9)(6)(1.602\times 10^{-19})^2}{(4.96\times 10^{-9})^2}\\\\F_{12}=5.63\times 10^{-11}N

Angle at which force F12 is acting is

\theta=tan^{-1}\frac{3.2}{3.8}\\\\\theta=tan^{-1}\frac{y}{x}\\\\\theta= 40.1^o

F_{x}=F_{12}cos\theta\\\\F_{x}=(5.63\times 10^{-11})cos(40.1)\\\\F_{x}=4.306\times 10^{-11}N\\\\F_{y}=F_{12}sin\theta\\\\F_{y}=(5.63\times 10^{-11})sin(40.1)\\\\F_{y}=3.62\times 10^{-11}N\\\\

\vec{F}_{12}=\vec{F}_{x}+\vec{F}_y\\\\\vec{F}_{12}=4.30\times 10^{-11}\,\hat{i} + 3.62\times 10^{-11}\,\hat{j}\\\\\vec{F}_{12}=

Force exerted on charge -2e is equal in magnitude to F12 but is in opposite direction

F_{21}=-5.63\times 10^{-11}N

\vec{F}_{21}=

7 0
3 years ago
If the energy of a photon is 1.32 × 10¯18 j, what is its wavelength in nm?
Katarina [22]
I think I remember hold on let me see if I can solve it
8 0
3 years ago
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Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction betw
jarptica [38.1K]

Answer:

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

         θ = 90         cos 90 = 0

        W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

3 0
2 years ago
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Answer:

W=-21,870,000\ J

Explanation:

<u>Work and Kinetic Energy </u>

The work an object does due to its motion is equal to the change of its kinetic energy. Being ko and k1 the initial and final kinetic energy respectively and m the mass of the object, then

W=\Delta k=k_1-k_0

Since

\displaystyle k=\frac{mv^2}{2}

We have

\displaystyle W=\frac{mv_1^2}{2}-\frac{mv_0^2}{2}

The truck has a mass of 60,000 kg and is moving at 27 m/s. The runaway truck ramp must stop the truck, so the final speed is 0. Thus

\displaystyle W=\frac{(60,000)0^2}{2}-\frac{(60,000)(27)^2}{2}

W=0-21870000\ J

\boxed{W=-21,870,000\ J}

3 0
3 years ago
Identify the following as
bazaltina [42]
3. Kinetic energy
4. Potential energy
5. Kinetic energy because it’s moving towards the waterfall otherwise there wouldn’t be a waterfall.
6. Kinetic energy
7. Kinetic energy
8. Potential energy
9. Potential energy
10. Kinetic energy
6 0
3 years ago
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