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posledela
3 years ago
14

A player uses a hockey stick to push a puck at a constant velocity across the ice. The weight of the puck is 170 N. The

Physics
2 answers:
soldi70 [24.7K]3 years ago
8 0

Answer: The minimum force by which player should hit the puck to move it is 0.102 N.

Explanation:

Weight of the puck = 1.70 N

This weight of the puck is the force acting normal to the surface that is:

N = 1.70 N

The coefficient of friction =

Frictional force =

The minimum force by which player should hit the puck to move it is 0.102 N.

Read more on Brainly.com - brainly.com/question/9177221#readmore

Bingel [31]3 years ago
6 0

Answer:

F=10.2N : Force with which the player must push the disc

Explanation:

Conceptual analysis

We apply Newton's first law because the speed is constant in horizontal direction (x):

∑Fx = 0 Formula (1)

We apply Newton's first law because the speed is zero in vertical direction (y):

∑Fy= 0 Formula (2)

∑F: algebraic sum of forces in Newton (N)  

We calculate the friction force with the following formula:

Ff=f*N  Formula (3)

Ff= friction force in Newtons (N)

f = coefficient of friction

N= Normal force in Newtons (N)

Known data:

Wp = 170 N  : puck weight

f = 0.0600 :coefficient of friction

Problem development

We replace the data in formula (1) and (2), considering that the force is positive (+) if it goes in the direction of the movement and negative (-) if it opposes the movement:

∑Fy= 0

N-Wp=0

N= Wp=170 N

∑Fx =0

F - Ff=0    F :force with which the player must push the disc

F - 0.0600*170 =0

F=10.2N

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If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
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3 years ago
How does the input distance of a third-class lever compare to the output distance​
Alexandra [31]

Answer:

A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force

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2 years ago
A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km
ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance  is given by r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m

(b) Y component of distance is given by r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m

These are the x and y component of position vector

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3 years ago
Substitute the value 26.7 (or the exact value of 80/3) for x in the first of the original equation (x + y = 40) to find the valu
Tatiana [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

the value of  y = 13.3

Explanation:

From the question we are told that

       The equation is  (x + y = 40)

        The first value of x is  x_1 =  26.7

        The second equation is  (0.75x + 1.5y = 40)

So  substituting x_1

     26.7 + y  =  40

=>   y = 13.3

Now substituting y and  x_1 into second equation

       0.75(26.7) +  (1.5* 13.3) =  40

 =>  40 = 40

So  y = 13.3

         

4 0
3 years ago
A potential difference of 1.20 V will be applied to a 33.0 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (
Nostrana [21]

Answer:A) Current = 1.739A, B)current density, J = 2.147x10^6 A/m2

magnitude of electric field , E =  0.036 N/C

)rate of thermal energy, P  =2.086W

Explanation:

Resistance  = R =   ρL/A

But the cross-section area of the wire. is given as

Diameter / 2 = 0.04/2 =0.02in to m = 0.02 / 39.37= 0.000508

A = πr^2 = π x  0.000508^2 = 8.10 x 10^-7

since resistivity of copper,ρ= 17x10-9 ohm.m

so resistance is   R =   ρL/A

17x10-9  x 33 / 8.1x10-7

= 0.69 ohm.

A) Current =    I = Voltage /Resistance =1.20/0.69 =1.739A

B)current density, J = Current /Area

= 1.739/8.1x10-7

= 2.147x10^6 A/m2

c)magnitude of electric field , E =  Current density x resistivity =J ρ

E = 2.147 x 10^6  x 17  x 10^-9

E = 0.036 N/C

D)rate of thermal energy, P  = I² R =1.739² X 0.69

=2.086W

6 0
2 years ago
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