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3 years ago

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Two bowling balls collide. A red 5 kg bowling ball is traveling with a velocity of 1 m/s to the left. A blue 4 kg bowling ball i

s traveling with a velocity of 2 m/s to the right. The 5 kg ball moves 0.8 m/s to the right after the collision. What is the final velocity of the 4kg ball?

1 answer:

lana66690 [7]3 years ago

6 0

Answer:

idk use google

Explanation:

google

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3 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

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To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$

$m_1u_1 +m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

$m_2u_2 = (m_1+m_2)v_f$

Rearranging to find the final velocity

$v_f = \frac{m_2u_2}{ (m_1+m_2)}$

$v_f = \frac{ 12500*7.8}{ 12500+7430}$

$v_f = 4.8921ft/s$

The expression for the impulse received by the first car is

$I = m_1 (v-u)$

$I = \frac{W}{g} (v-u)$

Replacing,

$I = \frac{12500}{32.2}(4.89-7.8)$

$I = -1129.65lb\cdot s$

The negative sign show the opposite direction.

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3 years ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

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3 years ago