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Nutka1998 [239]
3 years ago
13

The following objects have velocity and mass as given: Object A: m=5 kg, ⃗v=−11 ^j Object B: m=6 kg, ⃗v=5^ i +8.7 ^ j Object C:

m=10 kg, ⃗v=−10 ^i Find the x- and y-components of the net momentum of the system objects if we define the system to consist of (a) objects A and C (b) objects B and C (c) objects A, B, and C
Physics
1 answer:
Fofino [41]3 years ago
4 0

Answer:

For object A

m = 5 kg ,   v= -11 j

For B object

m = 6 kg  ,  v= 5 i +8.7 j

For object C

m = 10 kg   ,   v= -10 i

We know that

Linear momentum   P= m v   kg.m/s

a) A and C

Momentum in y direction

Py=- 5 x 11 j= - 55 j  kg.m/s

Momentum in x direction

Px=- 10 x 10 j= - 100 i    kg.m/s

b) B and C

Momentum in y direction

Py=6 x 8.7 j= 52.2 j    kg.m/s

Momentum in x direction

Px=( 6 x 5 - 10 x 10 ) i = - 70 i   kg.m/s

c) A ,B and C

By using data of a and b

Momentum in y direction

Py= 6 x 8.7 - 5 x 11 j= -2.8 j    kg.m/s

Momentum in x direction

Px=  6 x 5 -10 x 10 i = -70 i   kg.m/s

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A worker pushes a 200-N box (weight of box, you have to find mass) resting on a level floor with a force of 30 N. What is the ac
QveST [7]

For the problem we will apply the concepts related to Newton's second law. Recall that this is defined as the product between mass and acceleration, and that in special cases, when the acceleration is equivalent to the force of gravity, the force is equivalent to the weight of the person. From these relationships we will find the mass, and then the acceleration with the given force.

F_w = mg

Here,

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g = Gravity

Replacing and rearranging we have that the mass is,

m = \frac{F_w}{g} = \frac{200}{9.81}

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Now using the value of the force, but solving for the acceleration with the previous value of the mass we have,

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a = \frac{30}{20.387}

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8 0
3 years ago
A cart, which has a mass of m - 2.50 kg, is sitting at the top of an inclined plane which is 3.30 meters long and which meets th
SashulF [63]

Answer:

Part(i) the time taken for this cart to reach the bottom of the inclined plane is 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane is 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane is 25.663 J

Explanation:

Given;

mass of the cart = 2.5 kg

angle of inclination, β = 18.5⁰

length of inclined plane = 3.3m

Part(i) the time taken for this cart to reach the bottom of the inclined plane

s = ut + ¹/₂×at²

initial vertical velocity, u = 0

s = 3.3 m

s =  ¹/₂×at²

t = \sqrt{\frac{2s}{a} }

acceleration, of the cart, a = gsinβ

a = 9.8sin(18.5) = 3.11 m/s²

t = \sqrt{\frac{2X3.3}{3.11 }}= 1.457 s

Part(ii) the velocity of the cart when it reaches the bottom of the inclined plane

V = a×t

V = 3.11 × 1.457 = 4.531 m/s

Part(iii) the kinetic energy of the cart when it reaches the bottom of the inclined plane

KE = ¹/₂MV²

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tatuchka [14]

Answer:

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