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user100 [1]
3 years ago
9

A hockey player uses a hockey stick to hit a puck such that the stick provides an applied force on the puck The puck travels for

distance of 0.85 m while experiencing the force from the stick
The puck leaves the stick with a speed of 8.0 m/s and travels at constant speed in a straight line along the horizontalice for a distance of 12 m. The frictional force between the puck and the ice
surface is negligible
Which of the following statements best describes the force or forces exerted on the puck after it is hit by the stick and slides on the ice?
(A) The applied force from the stick is still exerted on the puck during this time.
Since the puck's acceleration is zero, there are no forces exerted on the puck during this time.
Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck
There is no frictional force exerted on the puck and the gravitational force and normal force are exerted on the puck and are equal in magnitude
Physics
1 answer:
Norma-Jean [14]3 years ago
4 0

Answer:

Explanation:

Let's analyze the situation presented in order to know which answer is correct.

When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.

Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.

The correct answer is:

C)           Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck

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Carbon-14 has a half-life of 5,700 years. How long will it take for 6.25% of the Carbon-14 to be remaining?
IRISSAK [1]

Answer:

22,800 years

Explanation:

Half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.0625 = (½)^(t / 5700)

log 0.0625 = (t / 5700) log 0.5

4 = t / 5700

t = 22,800

It takes 22,800 years.

4 0
3 years ago
approximation to the average velocity in that time interval, what should be the sequence of calculations?Update the (vector) pos
Klio2033 [76]

Answer:

The steps are outlined in the explanation below.

Explanation:

The average velocity is derived midpoint from the initial to the final velocity. Here is the proof:

Find the total displacement:

let the displacement be given by the letter s

Then since the average velocity is defined as:  v_{av}  = \frac{x - x_{0} }{t - t_{0} }

where t = final time

           t₀ = initial time

           v = final speed

           v₀ = initial time

where x denotes the position, then

v_{ave} = \frac{v+v_{0} }{2}

where v = \frac{dx}{dt} and dx = change in distance with respect to time.

6 0
3 years ago
Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N
nadezda [96]
First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say F_{1}, and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
F_{1} = 7i, where i = Unit vector in the x-direction.

2) Take the second force, say F_{2}, and assume that it is making an angle \alpha with the first force F_{1}.

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

F_{2} = (15*cos \alpha)i + (15*sin \alpha )j

Now from simple vector addition, we know that,
F_{R} = F_{1} + F_{2} --- (A)

Where F_{R} = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find F_{1}+F_{2} first!
F_{1}+F_{2} =  7i+(15*cos \alpha)i + (15*sin \alpha )j

=> F_{1}+F_{2} =  (7+15*cos \alpha)i + (15*sin \alpha )j

Now the magnitude of F_{1}+F_{2} is,
| F_{1}+F_{2}| = \sqrt{ (7+ 15*cos \alpha)^{2} +  (15*sin \alpha )^{2}}

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}

Since (sin \alpha)^{2} + (cos \alpha)^{2} = 1, therefore,

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Since  | F_{1}+F_{2}| = |F_{R}|, and the magnitude of the resultant force is 20N, therefore,

 |F_{R}| = | F_{1}+F_{2}|
20 = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Take square on both sides,
400 = 49 + 225 + 210*(cos \alpha)
(cos \alpha) =  \frac{3}{5}

\alpha = 53.13^{o}

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0
3 years ago
a care starting from rest has an acceleration 0.3 m/s square, calculate the velocity and distance travelled by this car after 2
Anna [14]

Answer:

Final velocity (v) = 36 m/s

Distance traveled (s) = 2,160 m

Explanation:

Given:

Initial velocity (u) = 0

Acceleration (a) = 0.3 m/s

Time travel (t) = 2 minutes = 120 seconds

Find:

Final velocity (v) = ?

Distance traveled (s) = ?

Computation:

v = u + at

v = 0 + 0.3(120)

v = 0.3(120)

v = 36 m/s

Final velocity (v) = 36 m/s

Distance traveled (s) = ut + (1/2)at²

Distance traveled (s) = (0.5)(0.3 × 120 × 120)

Distance traveled (s) = 2,160 m

3 0
3 years ago
A fan that is rotating at 960 rev/s is turned off. It makes 1500 revolutions before it comes to a stop. a) What was its angular
Evgesh-ka [11]

Answer:

α = 1930.2 rad/s²

Explanation:

The angular acceleration can be found by using the third equation of motion:

2\alpha \theta=\omega_f^2-\omega_i^2

where,

α = angular acceleration = ?

θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad

ωf = final angular speed = 0 rad/s

ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s

Therefore,

2\alpha(9424.78\ rad) = (0\ rad/s)^2-(6031.87\ rad/s)^2\\\\\alpha = -\frac{(6031.87\ rad/s)^2}{(2)(9424.78\ rad)}

<u>α = - 1930.2 rad/s²</u>

<u>negative sign shows deceleration</u>

5 0
3 years ago
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