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3 years ago

Will mark as BRAINLIEST.......

Physics

2 answers:

Alborosie3 years ago

7 0

This is the written answers to find the velocity and acceleration of r

lubasha [3.4K]3 years ago

6 0

Answer:

We have the position vector given in terms of time t. r(t) = t^3*i + t^2*j

To find the velocity vector we have to differentiate r(t) with respect to time.

r'(t) = 3t^2*i + 2t*j

The vector representing acceleration is the derivative of the position vector

r''(t) = 6t*i + 2*j

When time t = 2.

The velocity vector is 3*2^2*i + 2*2*j

=> 12*i + 4*j

The speed is the absolute value of the velocity vector or sqrt(12^2 + 4^2) = sqrt (144 + 16) = sqrt 160

The acceleration vector is 6*2*i + 2*j

=> 12*i + 2*j

The required acceleration at t=2 is 12*i + 2*j and the speed is sqrt 160.

Explanation:

Can I have thx and brainliest?

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Which type of energy is produced

Andreyy89

Answer:

C. sound energy

Explanation:

hope this helps

5 0

3 years ago

Which unit is equivalent to j/s

mrs_skeptik [129]

To be honest, there's no sure way to answer that, because you haven't defined your terms and we can't be sure of what j or s might be.

Tell you what I'll do:. I'll assume definitions for j and s, and then I'll answer the question that I invented.

Assume that j stands for Joule, the unit of energy. And assume that s stands for 'second', the unit of time.

Then j/s is the rate of transferring energy or doing work.

Its unit is the Watt, equivalent to 1 Joule per second.

In your system of notation, it would be 'w' .

4 0

3 years ago

What is the magnitude of fs on an object lying on a flat surface without moving, on

Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

magnitude of force on the object, F = 10 N

angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

7 0

2 years ago

A Martian rover is descending a hill sloped at 30° with the horizontal. It travels with a constant velocity of 2 meters/second.

kodGreya [7K]

Vx = 2*cos30 = 1.73m's
Other words:
D = Vo*t = 1.4 * 21 = 31.5m. @ 30 Deg.
Dy = 31.5*sin30 = 15.75 m.

8 0

3 years ago

Read 2 more answers

A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai

shutvik [7]

Answer:

d) None of the above

Explanation:

$v_{o}$ = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m

t = time of travel

using the equation

$y=v_{oy} t+(0.5)a_{y} t^{2}$

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

$v_{ox}$ = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

$a_{x}$ = acceleration along the horizontal direction = 0 m/s²

t = time of travel = 2.1 s

Using the kinematics equation

$x =v_{ox} t+(0.5)a_{x} t^{2}$

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

$v_{oy}$ = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

$a_{y}$ = acceleration along the vertical direction = - 9.8 m/s²

$y_{o}$ =initial vertical position at the time of launch = 20 m

$y$ = vertical position at the maximum height = 20 m

$v_{fy}$ = final velocity along vertical direction at highest point = 0 m/s

using the equation

${v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})$

$0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)$

$y$ = 20.02 m

h = height above the starting height

h = $y$ - $y_{o}$

h = 20.02 - 20

h = 0.02 m

7 0

3 years ago