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ira [324]
3 years ago
8

Not yet answered Points out of 2.00 Not flaggedFlag question Question text Two cars collide head-on on a level friction-free roa

d. The collision was completely inelastic and both cars quickly came to rest during the collision. What is true about the velocity of this system's center of mass?
Physics
1 answer:
Natali5045456 [20]3 years ago
5 0

Answer:

Velocity of center of mass must be ZERO

Explanation:

As we know that two cars are moving on friction-less surface

So here we can say that on the system of cars there is no external force along x direction

So we can say that

F_{net} = \frac{dP_{cm}}{dt}

so here if net force is zero

\frac{dP_{cm}}{dt} = 0

P_{cm} = constant

since finally the two cars comes to rest so here final velocity of center of mass of two cars will be zero and hence it will be always zero as there is no force on the system of two cars

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A motorcyclist drove 7 km at 57km/h and then another 7 km at 81 km/h. What was the average speed? ​
alex41 [277]

<u>Answer:</u>

<em>The average speed of the car is 66.9 km/h</em>

<u>Explanation:</u>

Here distance covered with the speed <em>57 km/h=7 km  </em>

distance covered with the speed of <em>81 km/h=7 km</em>

<em>Average speed is equal to the ratio of total distance to the total time. </em>

<em>total distance= 7 + 7= 14 km  </em>

<em>time= \frac{distance}{speed} </em>

<em>time taken to cover the first 7 km= 7/57 h  </em>

<em>time taken to cover the second part of the journey = 7/81 h </em>

<em>average speed =  14/(7/57+7/81)=(14 \times 57 \times 81)/945=66.9 km/h</em>

<u><em>Shortcut: </em></u>

<em>When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases. </em>

7 0
3 years ago
An infinite slab of charge of thickness 2z0 lies in the xy-plane between z=−z0 and z=+z0. The volume charge density rho(C/m3) is
Oksi-84 [34.3K]

Answer:

please read the answer below

Explanation:

To find the electric field you can consider the Gaussian law for a cylindrical surface inside the slab.

\int E dA=EA_{G}=\frac{Q_{int}}{\epsilon_o}

Q_{int}=\rho V_{G}

where Qint is the charge inside the Gaussian surface, AG is the area of the surface and rho is the charge density of the slab.

By using the formula for the volume of a cylinder you obtain:

V_{G}=\pi r^2h

where h is the height. If you assume that the slab is in the interval (-zo<z<z0) you can write VG:

V_{G}=\pi r^2 z

Finally, by replacing in the expression for E you get:

E=\frac{Q_{int}}{\epsilon_o A_G}=\frac{Q_{int}}{\epsilon_o \pi r^2}\frac{z}{z}=\frac{\rho z}{\epsilon_o}

E=\rho z/\epsilon_o

hence, for z>0 you obtain E=pz/eo > 0

for z<0 -> E=pz/eo < 0

7 0
3 years ago
A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The
IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

8 0
4 years ago
Calculate the speed for a car that went a distance of 125 kilometers in 2 hours time.
laila [671]

The speed of the car was exactly 125 km per 2 hours.

In a more familiar unit,  125km/2hr = 62.5 km/hr.

5 0
3 years ago
A man dropped a dime in a wishing well , he heard it 5 sec later ..Find distance traveled if moving at 10 m/s 2
Alenkasestr [34]

Answer:

The distance traveled is 109.58 m

Explanation:

The Speed of sound in air = 344 m/s

Let the time in which the dime dropped by the man reach an impact in the well = t₁

Let the time in which the sound travel from the well to the man = t₂

Then

1/2× 10 × t₁² = 344 × t₂ which gives;

5 × t₁² = 344 × t₂.........................(1)

Also the total time before the man heard the dime = t₁ + t₂ = 5

Therefore;

 t₂ = 5 - t₁

Substituting the value of t₂ in equation (1), we have;

5 × t₁² = 344 × (5 - t₁)

5·t₁² + 344·t₁ - 1720 = 0

Using the quadratic formula, we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

Which gives;

t_1 = \dfrac{-344\pm \sqrt{344^{2}-4\times 5\times (-1720)}}{2\times 5}

t₁ = 4.68 s or -73.48 s

Therefore, with the positive value for t₁ = 4.68 s, we have

The distance = 1/2× 10 × 4.68² = 109.58 m

The distance traveled = 109.58 m.

8 0
4 years ago
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