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kirill [66]
3 years ago
12

An absent-minded Penn State professor drives from State College to Tyrone, 20 miles away, at 60 mph (miles per hour), before rea

lizing that he forgot his wallet. He immediately returns to State College at 60 mph, picks up his wallet and drives back to Tyrone at 60 mph. Assuming all the travel was done more or less on a straight line and neglecting the time taken to turn around or collect the professor's wallet, what is the magnitude of the professor's average velocity for the entire trip?
Physics
1 answer:
BaLLatris [955]3 years ago
8 0

Answer:20 mph

Explanation:

Given

distance between college to Tyrone=20 miles

Given professor drives with a velocity of 60 mph

he returns to college after reaching tyrone and then again drive to tyrone.

so his net displacement is 20 miles

time taken to cover 20 miles is \frac{1}{3} hr

average velocity=\frac{Displacement}{Time\ taken}

V_{avg}=\frac{20}{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}

V_{avg}=\frac{20}{1}=20 mph

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Two car collide in an intersection. The speed limit in that zone is 30 mph. The car (mass of 1250 kg) was going 17.4 m/s (38.9).
Musya8 [376]

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

(1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}

<u>u₂ = 3.7 m/s</u>

5 0
2 years ago
There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12
Licemer1 [7]

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = \frac{Q}{\Delta V}

        C = ε₀ \frac{A}{d}

we solve for the charge (Q)

        \frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = \epsilon_o \  \frac{A \ \Delta V_1 }{d_1}

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          \frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}

          ΔV₂ = \frac{d_2}{d_1 } \ \Delta V_1

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  \frac{0.005}{0.003}  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

3 0
2 years ago
I NEED HELP ASAP WHAT IS THE DISTANCE MOVED IN 15 SECONDS
Oduvanchick [21]

Answer:

5 I think

Explanation:

8 0
2 years ago
Read 2 more answers
If the merry-go-round makes one revolution in 10 seconds, what is the child’s linear speed?
Anton [14]
The child's linear speed is
              
    <em> (pi / 5) x (the child's distance from the center of the ride, in feet)</em>

                                                                                        feet per second.
5 0
3 years ago
Calculate the molarity of a 10. 0% cacl₂ solution. The density of the solution is 1. 0835 g/cm³.
brilliants [131]

The molarity of 10% CaCl2 is 0.9%

concentration of the given salt CaCl₂ = 10%

Density of a solution = 1.0835 g/cm³

Volume = m / d

= 100 / 1.0835

= 92.29 litres

Density = mass / volume

1.0835 × 92.29 = mass

mass = 99.99 gram

Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100

= 0.9008/ 92.29 X 100 %

= 0.009 X 100 %

= 0.9 %

The molarity of 10% CaCl2 is 0.9%

To know more about density and molarity you may visit the link which is mentioned below:

brainly.com/question/10710093

#SPJ4

5 0
1 year ago
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