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2 years ago

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If a hockey puck is made to slide along a straight line on a completely frictionless surface, which statement about the motion o

f the hockey puck is true?
The hockey puck will come to a halt immediately.

The hockey puck will continue moving with constant velocity along a straight line.

The hockey puck will accelerate and continue moving in straight line.

The hockey puck will continue moving in random direction.

2 answers:

77julia77 [94]2 years ago

8 0

C. the hockey puck will accelerate and continue moving in a straight line

Llana [10]2 years ago

7 0

Answer:

The hockey puck will continue moving with constant velocity along a straight line.

Explanation:

As per Newton's first law we know that when there is no net force on an object then the velocity of object will always remains constant

Or we can say that the acceleration of an object will be zero when the force applied on the object is zero.

As per Newton's law we know

$F = ma$

so we will have

a = 0

so object will continue in straight line with same uniform speed without any change in magnitude as well as any change in direction

so correct answer will be

The hockey puck will continue moving with constant velocity along a straight line.

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A camera has a single converging lens with a fixed focal length f. (a) How far should the lens be from the film (or in a present

Varvara68 [4.7K]

Answer:

a) Due to the characteristic that a converging lens focuses light rays from infinity and parallel to its main axis. Therefore, the lens should be placed at a distance "f" from the film, in this way it will form the image of the object placed at infinity in said film.

b) Since the converging lens produces an image of an object placed at a distance of 2f, the lens must be placed at the same distance (2f), so that this object that is placed at a distance of 2f is focused.

Explanation:

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2 years ago

A uniformly charged, straight filament 5.10 m in length has a total positive charge of 2.00 µC. An uncharged cardboard cylinder

Ratling [72]

Answer:

70509.8039216 N/C

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

q = Charge = 2.00 µC

l = Length of filament = 5.1 m

r = Radius of cylinder = 10 cm

$\lambda=\dfrac{q}{l}$

Electric field is given by

$E=\dfrac{2k\lambda}{r}\\\Rightarrow E=\dfrac{2\times 8.99\times 10^9\times \dfrac{2\times 10^{-6}}{5.1}}{10\times 10^{-2}}\\\Rightarrow E=70509.8039216\ N/C$

The electric field at the surface of the cylinder is 70509.8039216 N/C

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3 years ago

Read 2 more answers

In a pickup game of dorm shuffleboard,students crazed by final exams use a broom to propel a calculus book along the dorm hallwa

Yakvenalex [24]

Answer:

Coefficient of friction between the book and floor is 0.582.

Explanation:

Using the velocity formula;

v^2 = 2as

a = v^2/(2s)

a = 1.6^2/(2*0.9)

a = 2.56/1.8

a = 1.42 m/s^2

the force necessary to give the book the acceleration is

F = ma = 3.5*1.42 (m is mass of the book i.e. 3.5 kg)

F = 4.98 N

The difference in the force is the friction force, which is

Ff = 25 - 4.98 = 20 N

Ff = mgμ

where μ is coefficient of friction and g is acceleration due to gravity that is 9.8 m/s^2

μ = Ff/mg

μ = 20/(3.5*9.81)

μ = 0.582

Coefficient of friction between the book and floor is 0.582.

6 0

2 years ago

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine th

Luba_88 [7]

Here is the complete question

The internal shear force V at a certain section of a steel beam is 80 kN, and the moment of inertia is 64,900,000 . Determine the horizontal shear stress at point H, which is located L = 20 mm below the centriod

The missing image which is the remaining part of this question is attached in the image below.

Answer:

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

Explanation:

Given that :

The internal shear force V = 80 kN = 80 × 10³ N

The moment of inertia = 64,900,000

The length = 20 mm below the centriod

The horizontal shear stress $\tau$ can be calculated by using the equation:

$\tau = \dfrac{VQ}{Ib}$

where;

Q = moment of area above or below the point H

b = thickness of the beam = 10 mm

From the centroid ;

Q = $Q_1 + Q_{2}$

Q = $A_1y_1 + A_{2}y_{2}$

Q = ( ( 70 × 10) × (55) + ( 210 × 15) (90 + 15/2) ) mm³

Q = ( ( 700) × (55) + ( 3150 ) ( 97.5) ) mm³

Q = ( 38500 + 307125 ) mm³

Q = 345625 mm³

$\tau_H = \dfrac{VQ}{Ib}$

$\tau_H = \dfrac{80*10^3 * 345625}{64900000*10 }$

$\tau_H = \dfrac{2.765*10^{10}}{649000000 }$

$\tau_H = 42.60400616 \ N/mm^2$

$\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

The horizontal shear stress at point H is $\mathbf{\tau_H \approx 42.604 \ N/mm^2}$

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3 years ago

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n200080 [17]

Explanation:

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