Help!! <br> A. 0.5 s <br> B. 1.0 s<br> C. 1.5 s<br> D. 2.0 s

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t

kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

$W=Fd cos \theta$ (1)

where

F is the magnitude of the force

d is the distance covered by the object

$\theta$ is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

$\theta=0$

And the distance covered is equal to the circumference of the circle, which is:

$d=2\pi r=2\pi (2.5 m)=15.7 m$

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

$W=(3.0)(15.7)(cos 0)=47.1 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

2 years ago

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

7 0

2 years ago

a 15 kg tv sots on a shelf at a height of 0.3 m. how much gravitational potential energy is added to the television when it is l

nata0808 [166]

Gravitational potential energy can be described as m*g*h (mass times gravity times height).

Originally,
15kg * 9.8m/s^2 *0.3 m = 44.1 kg*m^2/s^2 = 44.1 Joules.

After it is moved to a 1m shelf:
15kg * 9.8m/s * 1 = 147 kg*m^2/s^2= 147 Joules.

To find how much energy was added, we subtract final energy from initial energy:

147 J - 44.1 J = 102.9 Joules.

6 0

2 years ago

Please help

Hitman42 [59]

Answer:

3200 m/s

Explanation:

The speed of Sound through Argon is 319.

The speed of sound in any chemical element in the fluid phase has one temperature-dependent value. In the solid phase, different types of sound wave may be propagated, each with its own speed: among these types of wave are longitudinal (as in fluids), transversal, and (along a surface or plate) extensional.

If argon could exist as solid, then 3200 m/s is the best speed.

7 0

3 years ago

Read 2 more answers

The block slides on a horizontal frictionless surface. The block has a mass of 1.0kg and is pushed 5.0N at 45°. What is the magn

Firlakuza [10]

Answer:

$3.5\:\mathrm{m/s^2}$