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V125BC [204]
3 years ago
9

A truck travels at a constant speed of 45 kilometers per hour. How far does the truck travel in 20 minutes?

Physics
1 answer:
Mekhanik [1.2K]3 years ago
6 0
The answer is 15 kilometers in 20 minutes.
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ss7ja [257]

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im pretty sure it is 3.0 K

Explanation:

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How did Bourne believe that an object could be made to rise or sink at will?
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Bourne believed that an object would float or sink at will as long as he could <span>manipulate the effect's of buoyancy which control and object to sink or float. Hope this helps!

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Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of
Aleonysh [2.5K]

Answer:

r = 0.0548 m

Explanation:

Given that,

Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.

We need to find the radius of their circular path. The formula for the radius of path is given by :

r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}

m is mass of Singly charged uranium-238 ion, m=3.95\times 10^{-25}\ kg

q is charge

So,

r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m

So, the radius of their circular path is equal to 0.0548 m.

4 0
2 years ago
A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w
lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

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