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2 years ago

Topic : Collisions and Conservation of Momentum

1 answer:

7 0

A 0.2 kg ball moves to the right with a speed of 3 m/s. It hits a 0.5 kg ball that at rest. After the collision, the second ball moves to the right with ...

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A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

pav-90 [236]

<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

Displacement, s = 0 m

Acceleration, a = -9.81 m/s²

Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

s = ut + 0.5 at²

0 = 128.89 x t + 0.5 x (-9.81) x t²

t² - 26.28 t = 0

t ( t- 26.28) = 0

t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

5 0

3 years ago

A train has an initial velocity of 44m/s and an accelaration of _4m/s calculate its velocity

Kobotan [32]

Complete question:

A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity after 10s ?

Answer:

the final velocity of the train is 4 m/s.

Explanation:

Given;

initial velocity of the train, u = 44 m/s

acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)

time of motion, t = 10 s

let the final velocity of the train = v

The final velocity of the train is calculated using the following kinematic equation;

v = u + at

v = 44 + (-4 x 10)

v = 44 - 40

v = 4 m/s

Therefore, the final velocity of the train is 4 m/s.

7 0

3 years ago

Can someone check my answers and tell me if their correct?

Otrada [13]

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi = 2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

<u>Formula</u>

a = (vf - vi)/t

<u>Solve</u>

-9.81 = (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0 m/s But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11

6 0

3 years ago

A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem

Shtirlitz [24]

Answer:

U = (ε0AV^2) / 2d

Explanation:

Where C= capacitance of the capacitor

ε0= permittivity of free space

A= cross sectional area of plates

d= distance between the plates

V= potential difference

First, the capacitance of a capacitor is obtained by:

C = ε0A/d.

Starting at the formula , U= (CV^2)/2. Formula for energy stored in a capacitor

Substitute in for C:

U = (ε0A/d) * V^2 / 2

Hence:

U = (ε0AV^2) / 2d

3 0

3 years ago

WILL GIVE BRAINLIEST TO CORRECT ANSWER PLEASE HELP ME

koban [17]

Answer:

The total distance is 381.5 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

$v_{f} =v_{o} +(a*t)$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 7 [s]

Vf = 0 + (2*7)

Vf = 14 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

14² = 0 + (2*7*x)

x = 196/(14)

x = 14 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

$v_{f}= v_{i} +(a*t)$

Vf = final velocity [m/s]

Vi = initial velocity = 14 [m/s]

a = desacceleration = 4 [m/s²]

t = time = 8 [s]

Vf = 24 + (4*8)

Vf = 56 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

56² = 14² + (2*4*x)

x = 2940/(8)

x = 367.5 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 14 + 367.5 = 381.5 [m].

4 0

3 years ago