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Sauron [17]
3 years ago
14

At a certain point on a heated metal plate, the greatest rate of temperature increase, 3 degrees Celsius per meter, is toward th

e northeast. If an object at this point moves directly north, at what rate is the temperature increasing?
Physics
2 answers:
Katena32 [7]3 years ago
3 0

Explanation:

Formula to determine the rate of temperature increasing will be calculated as follows.

              D_{u}F = \Delta F.\vec{u}

                       = |\Delta F|.|j| cos \theta

                       = 3.|1|.cos(\frac{\pi}{4})

                       = 3 \times \frac{1}{\sqrt{2}}

                       = \frac{3 \sqrt{2}}{2}

                       = 2.121

Thus, we can conclude that the rate at which the temperature increasing is 2.121.

LekaFEV [45]3 years ago
3 0

Answer:

Explanation:

The rate of temperature increase towards north east is 3 °C/m

So, the component of increase in temperature in north is the component of the increase in temperature towards north east

Desired component = 3 cos 45° = 2.12 °C/m

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Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

r_1 = Distance between center of mass of person and pole = 0.45 m

r_2 = New distance between center of mass of person and pole = 0.46 m

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Angular speed is given by

\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s

The new angular speed is 1.70344 rad/s

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A girl throws a ball of mass 0.8 kg against a wall. The ball strikes the wall horizontally with a speed of 11 m/s, and it bounce
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