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erma4kov [3.2K]
2 years ago
9

The change in the momentum of an object is represented by the following formula:

Physics
1 answer:
Ede4ka [16]2 years ago
6 0

Hi there!

Recall that:

Change in momentum = mass × change in velocity

Or:

Δp = mΔv = m(vf - vi)

Plug in the given values. We can assign east to be positive and west to be negative in this instance (Velocity is a vector with direction).

Thus:

Δp = (1)(-21 - 10) = -31 kgm/s OR 31 kgm/s WEST.

The correct answer is B.

Change in momentum is EQUIVALENT to the quantity of IMPULSE.

The correct answer is H.

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block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
3 years ago
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, fla
lara31 [8.8K]

Answer:

Explanation:

Given

Speed of ball u=3.2\ m/s

Plane is inclined at an angle 20^{\circ}

To win the Game we need to hit the target at x=2.4\ m away

Launch angle of ball \theta

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is g\sin 20

Range of Projectile is given by

R=\frac{u^2\sin 2\theta }{g}

for R=2.4\ m

2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}

\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}

\sin 2\theta =0.7855

2\theta =51.77

\theta =25.88^{\circ}

so ball must be launched at an angle of 25.88^{\circ}

4 0
2 years ago
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma
ahrayia [7]

Incomplete question as we have not told to find what.So the complete question is here

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-60 m/s)i and a second piece, also of mass m, moves with velocity (-60 m/s)j. The third piece has mass 3m.Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?

Answer:

V_{3}=(20i+20j)m/s

Explanation:

Given data

The vessel at rest

Piece one,of mass m,moves with velocity=(-60 m/s)i

Piece two,of mass m,moves with velocity=(-60 m/s)j

Piece three,of mass 3m

As the linear momentum is conserved in this system,Because the system is closed and no external  force acting on it

So momentum is given as

p_{initial}=p_{final}

As the vessel at rest so the initial momentum is zero

So

m_{1}V_{1}+m_{2}V_{2}+m_{3}V_{3}=0\\m_{3}V_{3}=-m_{1}V_{1}-m_{2}V_{2}\\V_{3}=\frac{-m_{1}V_{1}-m_{2}V_{2}}{m_{3}} \\V_{3}=\frac{-m_{1}(-60m/s)i-m_{2}(-60m/s)j}{3m}\\V_{3}=(20i+20j)m/s

 

5 0
3 years ago
By what angle should the second polarized sheet be rotated relative to the first to reduce the transmitted intensity to one-half
algol [13]

Answer:

   θ  = 45º

Explanation:

The light that falls on the second polarized is polarized, therefore it is governed by the law of Maluz

              I = I₀ cos² θ

in the problem they ask us

            I = ½ I₀

let's look for the angles

             ½ I₀ = I₀ cos² θ

             cos θ  = √ ½ = 0.707

            θ  = cos 0.707

           θ  = 45º

6 0
3 years ago
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