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3 years ago

15

What Grade will you give my One Pager so Far?

2 answers:

AnnZ [28]3 years ago

7 0

Answer:

98

Explanation:

Trava [24]3 years ago

7 0

Answer:

98.5

Explanation:

the .5 is because of the colors

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A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center?

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Answer:

A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center? [0.297 m/s] 6. A merry-go-round a.k.a “the spinny thing” is rotating at 15 RPM, and has a radius of 1.75 m A.

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A spring attached to a mass is at rest in the initial

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E=(1/2)kx^2

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An insulated Thermos contains 140 cm3 of hot coffee at 85.0°C. You put in a 15.0 g ice cube at its melting point to cool the cof

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Answer:

$T = 69^o C$

Explanation:

Here at thermal equilibrium we can say that thermal energy given by Hot coffee is equal to the thermal energy absorbed by ice cubes

So here we have

$Q_{ice} = Q_{coffee}$

now since ice cubes are added into coffee when it is at melting temperature

So here we can say that final temperature of coffee is T degree C

Now we have

$m_1L + m_1c_1\Delta T_1 = m_2c_2\Delta T_2$

here we have

$m_1 = 15 gram$

L = 333 kJ/kg = 333 J/g[/tex]

$c_1 = c_2 = 4186 J/kg C = 4.186 J/g C$

$\Delta T_1 = T - 0$

$\Delta T_2 = 85 - T$

now we have

$15(333) + 15(4.186)(T - 0) = 140(4.186)(85 - T)$

$4995 + 62.79T = 49813.4 - 586.04T$

$648.83 T = 44818.4$

$T = 69^o C$

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3 years ago

A 600kg car is moving at 5 m/s to the right and elastically collided with a stationary 900 kg car. What is the velocity of the 9

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Answer:

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

Explanation:

It says “Momentum before the collision is equal to momentum after the collision.” Elastic Collision formula is applied to calculate the mass or velocity of the elastic bodies.

$m_{1} v_{1}=m_{2} v_{2}$

$\mathrm{m}_{1} \text { and } \mathrm{m}_{2} \text { are masses of the object }$

$\mathrm{v}_{1} \text { velocity before the collision }$

$\mathrm{v}_{2} \text { velocity after the collision }$

$\mathrm{m}_{1}=600 \mathrm{kg}$

$\mathrm{m}_{2}=900 \mathrm{kg}$

$\text { Velocity before the collision } v_{1}=5 \mathrm{m} / \mathrm{s}$

$600 \times 5=900 \times v_{2}$

$3000=900 \times v_{2}$

$\mathrm{v}_{2}=\frac{3000}{900}$

$\mathrm{v}_{2}=3.3 \mathrm{m} / \mathrm{s}$

$\mathrm{v}_{2} \text { velocity after the collision is } 3.3 \mathrm{m} / \mathrm{s}$

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3 years ago

Its Acceleration during the upward Journey ?

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Acceleration will be 9.81 if it goes downwards. If it accelerates upwards it will be -9.81m/s^2

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