Ask question

Ask question

All categories

Karo-lina-s [1.5K]

3 years ago

8

6.If a 250. gram cart moving to the right with a velocity of .31 m/s collides inelastically with a 500. gram cart traveling to t

he left with a velocity of -0.22 m/s, what is the total momentum of the system before the collision

2 answers:

spin [16.1K]3 years ago

5 0

Answer:

The total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

Explanation:

Given that,

Mass of the cart, m = 250 g = 0.25 kg

Initial velocity of the cart, u = 0.31 m/s (due right)

Mass of another cart, m' = 500 g = 0.5 kg

Initial velocity of the another cart u' = -0.22 m/s (due left)

Let p is the total momentum of the system before the collision. It is given by :

$p=mu+m'u'\\\\p=0.25\times 0.31+0.5\times (-0.22)\\\\p=-0.0325\ kg-m/s$

So, the total momentum of the system before the collision is 0.0325 kg-m/s due east direction.

miv72 [106K]3 years ago

3 0

Answer:

Explanation:

mass of I cart, m = 250 g = 0.25 kg

initial velocity of I cart, u = 0.31 m/s

mass of II cart, m' = 500 g =0.5 kg

initial velocity of II cart, u' = - 0.22 m/s

Total momentum before collision = m x u + m' x u'

= 0.25 x 0.31 - 0.5 x 0.22

= 0.0775 - 0.11

= - 0.0325 kg m/s

You might be interested in

Which accurately explains concave and convex lenses?

barxatty [35]

Answer:

its C! I just finished the test on edg :)

3 0

3 years ago

Read 2 more answers

How well do you know your friend.

VladimirAG [237]

Explanation:

goood and mark this answer as brainliest

5 0

3 years ago

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring

azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin $\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

Put the values from equation (1) & (2) we get

16.45 = 120 $x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0

3 years ago

Calculate the wavelength of the electromagentic waves with the given frequencies, and determine the type of electromagnetic radi

ololo11 [35]

To develop this problem we require the concepts related to wavelength and its expression to calculate it.

The wavelength is given by

$\lambda =\frac{c}{f}$

Where,

$c=3*10^8 m/s$ light velocity

f = frequency.

Our values are given by,

$f_1=4.38*10^{14}Hz$

$f_2= 4.14*10^{20}Hz$

$f_3 = 3.24*10^{12}Hz$

Then,

$\lambda_1=\frac{3*10^8}{4.38*10^{14}}= 6.8493*10^{-7}m$ Visible

$\lambda_2=\frac{3*10^8}{4.14*10^{20}}= 7.24*10^{-13}m$ Gamma Ray

$\lambda_3=\frac{3*10^8}{3.24*10^{12}}= 9.259*10^{-5}m$ Infrared

<em>*Note the designation on the type of rays that are, can be found in consulted via On-line or in the optical books referring to the electromagnetic spectrum table with their respective ranges.</em>

5 0

3 years ago

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago