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3 years ago

15

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]3 years ago

8 0

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

Llana [10]3 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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Anika [276]

B. Heat goes from the drink to the ice

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3 years ago

Two slits separated by a distance of d = 0.190 mm are located at a distance of D = 1.91 m from a screen. The screen is oriented

Svetlanka [38]

Answer:

$\theta = 0.195^0$

Explanation:

wavelength $\lambda = 648 nm \ = 648*10^{-9}m$

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

$dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})$

$\theta = 0.195^0$

The first maximum will appear at an angle $\theta = 0.195^0$ from the beam axis

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3 years ago

4. Lead has a density of 11.5g/cm3. A rectangular block of lead measures 7cm ×5cm×2cm.

MArishka [77]

Answer:

a.) volume = 70cm^3

b.) mass = 805 grams

Explanation:

see the explanation in attachment.

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3 years ago

A) The current theory of the structure of the Earth, called plate tectonics, tells us that the continents are in constant motion

suter [353]

A) The mass of the continent is $2.5\cdot 10^{21} kg$

B) The kinetic energy is 2016 J

C) The speed of the jogger should be 7.1 m/s

Explanation:

A)

The mass of the continent can be calculated as

$m = \rho V$

where

$\rho = 2800 kg/m^3$ is its density

V is its volume

We have to calculate its volume. We know that the continent is represented as a slab of side 5900 km (so its surface is 5900 x 5900, assuming it is a square) and depth of 26 km, so its volume is:

$V=(5900 km)^2 (26 km)=9.05\cdot 10^8 km^3 =9.05 \cdot 10^8 \cdot (10^9 m^3/k^3)=9.05\cdot 10^7 m^3$

So, the mass of the continent is

$m=\rho V = (2800)(9.05\cdot 10^{17})=2.5\cdot 10^{21} kg$

B)

The kinetic energy of a body is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the continent, we have:

$m=2.5\cdot 10^{21} kg$ is the mass

$v=4 cm/year$ is the speed

We have to convert the speed into SI units. we have:

1 cm = 0.01 m

$1 year = (365)(24)(60)(60) s = 3.15\cdot 10^7 s$

So, the speed is

$v=4 cm/year = 0.04 m/year \cdot \frac{1}{3.15\cdot 10^7}=1.27\cdot 10^{-9} m/s$

Therefore, the kinetic energy is

$K=\frac{1}{2}(2.5\cdot 10^{21} kg)(1.27\cdot 10^{-9} m/s)^2=2016 J$

C)

Again, the kinetic energy of an object is

$K=\frac{1}{2}mv^2$

For the jogger in this problem, his mass is

m = 80 kg

And we want its kinetic energy to be equal to that of the continent, so

K = 2016 J

Re-arranging the equation for v, we find what speed the jogger needs to have this kinetic energy:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2016)}{80}}=7.1 m/s$

Learn more about kinetic energy here:

brainly.com/question/6536722

#LearnwithBrainly

8 0

3 years ago

A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 0.9 m from

posledela

Answer:

<em>a) the final angular speed is 0.738 rad/s</em>

<em>b) the change in kinetic energy = 0.3 J</em>

Explanation:

the two 1 kg objects have a total mass of 2 x 1 = 2 kg

radius of rotation of the objects = 0.9 m

moment of inertial of the student and the chair = 6 kg-m^2

initial angular speed of rotation of the sitting student and object system ω1 = 0.61 rad/s

final angular speed of rotation of the sitting student and object system ω2 = ?

moment of inertia of the rotating object is

$I = mr^{2}$ = 2 x $0.9^{2}$ = 1.62 kg-m^2

total moment of inertia of sitting student and object system will be

==> 6 + 1.62 = 7.62 kg-m^2

The initial angular momentum of the sitting student and object system will be calculated from

==> Iω1 = 7.62 x 0.61 = 4.65 kg-rad/s-m^2

if the radius of rotation of the object is reduced to 0.39 m,

new moment of inertia of the rotating object will be

$I = mr^{2}$ = 2 x $0.39^{2}$ = 0.304 kg-m^2

new total moment of inertia of the sitting student and object system will be

==> 6 + 0.304 = 6.304 kg-m^2

The final momentum of the sitting student and object system will be calculated from

==> Iω2 = 6.304 x ω2 = 6.304ω2

<em>According to conservation of angular momentum, initial momentum of the system must be equal to the final momentum of the system</em>. Therefore,

4.65 = 6.304ω2

ω2 = 4.65/6.30 =<em> 0.738 rad/s</em>

b) Rotational kinetic energy of the system = $\frac{1}{2} Iw^{2}$

for the initial conditions, kinetic energy is

==> $\frac{1}{2} Iw1^{2}$ = $\frac{1}{2}* 7.62*0.61^{2}$ = 1.417 J

for the final conditions, kinetic energy is

==> $\frac{1}{2} Iw1^{2}$ = $\frac{1}{2}*6.304*0.738^{2}$ = 1.717 J

change in kinetic energy = final KE - initial KE

==> 1.717 - 1.417 = <em>0.3 J</em>

6 0

3 years ago