All categories

3 years ago

Urgent!!!! Timed quiz. Calculate the velocity of the pink car.

Physics

1 answer:

dem82 [27]3 years ago

3 0

Answer:

Option A

20 m/s

Explanation:

From the law of conservation of linear momentum, the sum of momentum before and after collision equals zero hence for this case also, the sum of momentum of the two cars before collision should be equal to momentum after collision.

Momentum is given as the product of mass and velocity, where velocity considers the direction. Momentum, p=mv where m is the mass of the object and v is the velocity.

Momentum before collision

Since the velocities of both cars before collision are zero hence their momentum are zero ie 0.5*0+(1*0)=0

Momentum after collision

For 0.5 Kg car, its momentum will be 0.5*-40=-20 Kg.m/s

For the 1 kg car its momentum will be 1*v=1v

v-20=0

v=20 m/s

You might be interested in

What is the

Len [333]

Answer:

$d=5\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

$d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3$

So, the density of the object is equal to $5\ g/cm^3$ .

6 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago

ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly

Igoryamba

The period T of a pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$ . Using this data, we can solve the previous formula to find L:
$L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm$

4 0

3 years ago

1. Indicate whether these objects or atoms are positively,<br> negatively or neutrally charged.

vovangra [49]

Answer:

Neutrally charged!!!!!!!!!!!!!!!!!!!!!

Explanation:

6 0

3 years ago

Please help soon!!!!

anyanavicka [17]

The Answer: Parallel