Ask question

Ask question

All categories

3 years ago

13

Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, i

s going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65∘ south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car A going just before the collision?

1 answer:

devlian [24]3 years ago

6 0

Answer:

The answer of the part (a) is v2 = 7.09 m/s

and the answer of the part (b) is vA1 = 5.25 m/s

Explanation:

Explanation of the both parts of answer is in the following attachments

You might be interested in

Can anyone help me with this

Murrr4er [49]

Answer:

event s

Explanation:

the cow gets oxygen from the tree and proceeds to respirate.

8 0

3 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Given that,

Kinetic energy = 6.2 MeV

Radius = 0.500 m

We need to calculate the acceleration

Using formula of acceleration

$a=\dfrac{v^2}{r}$

Put the value into the formula

$a=\dfrac{\dfrac{1}{2}mv^2}{\dfrac{1}{2}mr}$

Put the value into the formula

$a=\dfrac{6.2\times10^{6}\times1.6\times10^{-19}}{\dfrac{1}{2}\times1.67\times10^{-27}\times0.51}$

$a=2.32\times10^{15}\ m/s^2$

We need to calculate the rate at which it emits energy because of its acceleration is

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$

Put the value into the formula

$\dfrac{dE}{dt}=\dfrac{(1.6\times10^{-19})^2\times(2.3\times10^{15})^2}{6\pi\times8.85\times10^{-12}\times(3\times10^{8})^3}$

$\dfrac{dE}{dt}=3.00\times10^{-23}\ J/s$

The energy in ev/s

$\dfrac{dE}{dt}=\dfrac{3.00\times10^{-23}}{1.6\times10^{-19}}\ J/s$

$\dfrac{dE}{dt}=1.875\times10^{-4}\ ev/s$

We need to calculate the fraction of its energy that it radiates every second

$\dfrac{\dfrac{dE}{dt}}{E}=\dfrac{1.875\times10^{-4}}{6.2\times10^{6}}$

$\dfrac{\dfrac{dE}{dt}}{E}=3.02\times10^{-11}$

Hence, The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

5 0

3 years ago

Pls help answer embed <br>

Savatey [412]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

Weight of the object = drag force

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

5 0

3 years ago

4. Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 7.3 times tha

HACTEHA [7]

Answer:

Explanation:

Acceleration of particle A is 7.3 times the acceleration of particle B.

Let the acceleration of particle B is a, then the acceleration of particle A is

7.3 a.

Let the period of particle A is T and the period of particle B is 2.5 T.

Let the radius of particle A is RA and the radius of particle B is RB.

Use the formula for the centripetal force

$a=r\omega ^{2}=r\times \frac{4\pi^{2}}{T^{2}}$

So, $r = a\frac{T^{2}}{4\pi^{2}}$

The ratio of radius of A to the radius of B is given by

$\frac{R_{A}}{R_{B}}=\frac{a_{A}\times T_{A}^{2}}{a_{B}\times T_{B}^{2}}$

$\frac{R_{A}}{R_{B}}=\frac{7.3 a\times T^{2}}{a\times 6.25T^{2}}$

RA : RB = 1.17

3 0

3 years ago

Read 2 more answers

Straight-in spaces can leave you a safer out if

kiruha [24]

Answer:

Explanation:

O

4 0

4 years ago

Read 2 more answers