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denis23 [38]
2 years ago
12

A hockey puck is pushed by a stick with a force of 750 newtons. The puck travels 2.0 meters in 0.30 seconds. How powerful is the

push? (Ignore frictional effects.)
Physics
1 answer:
nekit [7.7K]2 years ago
5 0
The work done by the force pushing the puck is equal to the product between the force and the distance covered:
W=Fd=(750 N)(2.0 m)=1500 J

And the power is the work done per unit time:
P= \frac{W}{t}
and using t=0.30 s, we find the power:
P= \frac{1500 J}{0.30 s}=5000 W
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How are uniform circular motion maps the same as linear motion maps? Check all that apply.
Flauer [41]

Answer:

A: Time is represented by dots at 1-second intervals.

D: Velocity vectors change direction as the object’s direction changes.

Explanation:

3 0
2 years ago
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There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame
slavikrds [6]

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Smaller diameter paintball = 5 cm

d₂ = Larger diameter paintball = 9 cm

V₂ = Volume of larger diameter paintball

Volume of smaller diameter paintball

V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3

Similarly

V_2=\frac{4}{24}\pi d_2^3

Dividing the above two equations, we get

\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3

∴ The larger one hold 163.296 cm³ of paint

5 0
2 years ago
A group of students must conduct an experiment to determine how the location of an applied force on a classroom door affects the
schepotkina [342]

Answer:

the answer the correct one is the  d

Explanation:

In the gate rotation experiment several things are measured.

- the distance from the hinges to the applied force, which must be measured with a tape measure

- The value of the force that is devised with a dynamometer

- the rotated angle that is measured with a protractor

- the time it takes to turn an angle, which is measured with a stopwatch

When examining the answer the correct one is the  d

8 0
2 years ago
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of th
Anton [14]

Answer:

Option B, 93 cm

Explanation:

An diagram of the seed's motion is attached to this solution.

This is very close to a projectile motion question. And the quantity to be calculated, how far along the grant a seed released would travel is called the Range.

And this would be obtained from the equations of motion,

First of, the height of the plant is related to some quantities of the motion with this relation.

H = u(y) t + 0.5g(t^2)

U(y) = initial vertical component of velocity = 0 m/s, H = height at which motion began, = 20cm = 0.2 m

That means t = √(2H/g)

The horizontal distance covered, R,

R = u(x) t + 0.5g(t^2) = u(x) t (the second part of the equation goes to zero as the vertical component of the acceleration of this motion is 0)

(substituting the t = √(2H/g) derived from above

R = u(x) √(2H/g)

Where u(x) = the initial horizontal component of the bomb's velocity = maximum initial speed, that is, 4.6 m/s, H = vertical height at which the seed was released = 20 cm = 0.2 m, g = acceleration due to gravity = 9.8 m/s2

R = 4.6 √(2×0.2/9.8) = 0.929 m = 0.93 m = 93 cm. Option B.

QED!

6 0
2 years ago
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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is
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To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

F = AY\alpha \Delta T

Where

A = Cross-sectional Area

Y = Young's modulus

\alpha= Coefficient of linear expansion for steel

\Delta T= Temperature Raise

Our values are given as,

A = 4.45cm^2

T = 37K

\alpha = 1.17*10^{-5}K^{-1}

Y = 200*10^9Gpa

Replacing we have,

F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)

F = 38526.1N

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0
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