A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s3)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j .a) What are ax(t) and ay(t), the x- and y- components of cars acceleration as a function of time?
b) What are the magnitude and direction of the velocity of the car at t= 8 sec?
c) What is the magnitude and direction of cars acceleration at t=8 sec

Answers

Answer 1

Answer: Maybe try A for your answer

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What is a light year

Answers

Answer:

A light-year is the distance light travels in one year.

Answer:

Explanation:

a unit of astronomical distance equivalent to the distance that light travels in one year, which is 9.4607 × 1012 km (nearly 6 million million miles).

An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball

Answers

Answer:

v = 3.951 m/s

Explanation:

Given that,

Mass of a ball, m = 6.5 kg

Radius of the circle, r = 0.9 m

Angular speed of the ball, $\omega=0.7\ rev/s=4.39\ rad/s$

Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :

$v=r\omega\n\nv=0.9* 4.39\n\nv=3.951\ m/s$

So, the tangential speed of the ball is 3.951 m/s.

A string is attached to the rear-view mirror of a car. A ball is hanging at the other end of the string. The car is driving around in a circle, at a constant speed. Which of the following lists gives all of the forces directly acting on the ball?a. tension
b. tension and gravity
c. tension, gravity, and the centripetal force
d. tension, gravity, the centripetal force, and friction

Answers

Final answer:

The forces directly acting on the ball hanging from a rear-view mirror while a car drives in a circle are tension, gravity, and the centripetal force.

Explanation:

The correct answer is c. tension, gravity, and the centripetal force.

When the car is driving in a circle, the ball experiences both tension and gravity. The tension in the string is what keeps the ball from falling, while gravity pulls the ball downward.

In addition to tension and gravity, the ball also experiences the centripetal force. This force is directed towards the center of the circular motion and keeps the ball moving in a circular path.

Learn more about Forces acting on a ball in circular motion here:

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7. When the block has fallen a distance d = 4.2 m, find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block. (Note : Take the downward direction positive)

Answers

Answer:

a) W₁ = - 127 J, b) W₂ = 148.18 J, c) $v_(f)$ = 3.43 m/s and d) $v_(f)$ = 3.43 m / s

Explanation:

The work is given the equation

W = F. d

Where the bold indicates vectors, we can also write this expression take the module of each element and the angle between them

W = F d cos θ

They give us displacement, let's use Newton's second law to find strength, like the block has an equal acceleration (a = g / 7). We take a positive sign down as indicated

W-T = m a

T = W -m a

T = mg -mg/7

T = mg 6/7

T = 3.6 9.8 6/7

T = 30.24 N

Now we can apply the work equation to our problem

a) the force of the cord is directed upwards, the displacement is downwards, so there is a 180º angle between the two

W₁ = F d cos θ

W₁ = 30.24 4.2 cos 180

W₁ = - 127 J

b) the force of gravity is directed downwards and the displacement is directed downwards, the angle between the two is zero (T = 0º)

W₂ = (mg) d cos 0º

W₂ = 3.6 9.8 4.2

W₂ = 148.18 J

c) kinetic energy

K = ½ m v²

Let's calculate speed with kinematics

$v_(f)$ ² = vo² + 2 a y

v₀ = 0

a = g / 7

$v_(f)$ ² = 2g / 7 y

$v_(f)$ = √ (2 9.8 4.2 / 7)

$v_(f)$ = 3.43 m/s

We calculate

K = ½ 3.6 3.43²

K = 21.18 J

d) the speed of the block and we calculate it in the previous part

$v_(f)$ = 3.43 m / s

What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?

Answers

Answer:

q = 8.85 x 10⁻¹¹ C

Explanation:

given,

Electric field, E = 1.18 N/C

distance, r = 0.822 m

Charge magnitude = ?

using formula of electric field.

$E = (kq)/(r^2)$

k is the coulomb constant

$q= (Er^2)/(k)$

$q= (1.18* 0.822^2)/(9* 10^9)$

q = 8.85 x 10⁻¹¹ C

The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?