All categories

4 years ago

Along the frictionless path you have chosen, the main force(s) acting on the puck after receiving the kick is (are):

Physics

1 answer:

Vaselesa [24]4 years ago

5 0

Gravity, and Normal. Check the comments for why Applied isn't one.

You might be interested in

The most likely models of the planet Mercury indicate that more than half the planet may be composed of________.

Oxana [17]

Answer:

iron

Explanation:

4 0

3 years ago

When you must add three vectors together, what is not true this process? You must only give a magnitude of the resultant vector

vovangra [49]

The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.

The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.

4 0

3 years ago

PLEASE ANSWER ILL GIVE YOU BRAIN

steposvetlana [31]

1:4
2:44 minutes
3:11

4 0

3 years ago

If a weightlifter lifts a 286 kg mass 0.25 meters above his head, how much PEg does the mass have?

vazorg [7]

Answer:

Change in potential energy is 700.7 J.

Explanation:

Given:

Mass of the object is, $m=286$ kg

Height to which the object is raised above his head, $h=0.25$ m

Acceleration due to gravity is, $g=9.8$ m/s²

We know that for a mass $m$ raised to a height $h$ , the change in potential energy is given as:

$\Delta PE=mgh$ , where, $\Delta PE\rightarrow \textrm{Change in Potential Energy}$

Now, plug in the given values and solve.

$\Delta PE=286\times 9.8\times 0.25=700.7\textrm{ J}$

Therefore, the change in the potential energy is 700.7 J.

3 0

4 years ago

Read 2 more answers

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving in a circular simulation. Suppose that the ten

leva [86]

Answer:

Tangential speed, v = 2.64 m/s

Explanation:

Given that,

Mass of the puck, m = 0.5 kg

Tension acting in the string, T = 3.5 N

Radius of the circular path, r = 1 m

To find,

The tangential speed of the puck.

Solution,

The centripetal force acting in the string is balanced by the tangential speed of the puck. The expression for the centripetal force is given by :

$F=\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{Fr}{m}}$

$v=\sqrt{\dfrac{3.5\ N\times 1\ m}{0.5\ kg}}$

v = 2.64 m/s

Therefore, the tangential speed of the puck is 2.64 m/s.

3 0

3 years ago