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2 years ago

6

A 5000kg freight car moving at 2 m/s East collides with a 10,000kg freight car at rest. Upon collision, they got stuck and moved

with same velocity. Find their final velocity after collision

1 answer:

mr_godi [17]2 years ago

6 0

Answer:

A

Explanation:

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A camera lens focuses on an object 75.0 cm from the lensThe image forms 3.50 cm behind the lens. What is the magnification of th

N76 [4]

Answer:

7/150

Explanation:

The following data were obtained from the question:

Object distance (u) = 75cm

Image distance (v) = 3.5cm

Magnification (M) =..?

Magnification is simply defined as:

Magnification (M) = Image distance (v)/ object distance (u)

M = v /u

With the above formula, we can obtain the magnification of the image as follow:

M = v/u

M = 3.5/75

M = 7/150

Therefore, the magnification of the image is 7/150.

6 0

3 years ago

16x^2y^2-25a^2b^2<br>factorize the expression

SIZIF [17.4K]

Answer:

(4xy+5ab)(4xy-5ab)

Explanation:

16 $x^{2}$ $y^{2}$ -25 $a^{2}$ $b^{2}$

4^2 is 16 and 5^2 is 25,

Also, (x-a)(x+a) = x^2-a^2

So, this factorized is:

(4xy+5ab)(4xy-5ab)

Hope this helps!

8 0

2 years ago

A cart loaded with bricks has a total mass of 22.2 kg and is pulled at constant speed by a rope. The rope is inclined at 27.5 ◦

musickatia [10]

Answer:

W = 1.432 KJ

Explanation:

given,

mass = 22.2 Kg

angle of the rope = 27.5°

distance on the ground = 24 m

kinetic friction= μ = 0.32

acceleration due to gravity, g = 9.8 m/s²

Work done = ?

W = F d cosθ

a = 0 because it is moving with constant speed

equating all the forces acting in x direction

F cosθ = F friction = μN

equating all the forces acting in y direction

F sinθ + N -mg =0

now,

N = mg - F sinθ

putting value of N

F cosθ = μ mg -μ F sinθ

F (cosθ + μsinθ ) = μ mg

$F = \dfrac{\mu mg}{cos\theta + \mu sin\theta}$

$F = \dfrac{0.32 \times 22.2 \times 9.8}{cos 27.5^0+0.32 \times sin27.5^0}$

F =67.28 N

now,

W=F d cosθ

W =67.28 x 24 x cos(27.5)

W =1432.27 J

W = 1.432 KJ

7 0

2 years ago

Read 2 more answers

Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Degger [83]

B. kinetic energy increases and potential energy decreases

5 0

3 years ago

Jupiter's semimajor axis is 7.78×1011 m. The mass of the Sun is 1.99×1030 kg. (a) What is the period of Jupiter's orbit in secon

dedylja [7]

Explanation:

It is given that,

Semi major axis of the Jupiter, $a=7.78\times 10^{11}\ m$

Mass of the sun, $M=1.99\times 10^{30}\ kg$

(a) Let T is the period of Jupiter's orbit. It is given by :

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{GM}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.99\times 10^{30}}\times (7.78\times 10^{11})^3$

$T=3.74\times 10^8\ s$

(b) We know that,

$1\ year=3.154\times 10^7\ s$

or

$1\ s=3.171\times 10^{-8}\ year$

$3.74\times 10^8\ s={3.171\times 10^{-8}}\times {3.74\times 10^8}$

T = 11.859 earth years

Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers