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3 years ago

A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec

1 answer:

5 0

2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.

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Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot

Gelneren [198K]

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = $\frac{kQq}{r^{2} }$

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

$N_{x}$ = N cosine 60° + (-N cosine 60°) = 0

for the y-component

$N_{y}$ = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( $N_{x}$ = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

5 0

4 years ago

The IMA of the pulley show is

Lera25 [3.4K]

The IMA of the pulley shown is 2.

4 0

3 years ago

Read 2 more answers

How many atoms are represented by one formula unit of aluminum sulfite, al2(so3)3?

ratelena [41]

Formula unit is the empirical formula of an ionic or a covalent compound. It is the chemical formula that gives the proportions of the elements present in a certain compound however it does not give the actual numbers or the arrangement of the atoms. For this case, there are 14 atoms of different elements that is involved in the compound.

4 0

3 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$