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irinina [24]
3 years ago
12

Two​ vehicles, a car and a​ truck, leave an intersection at the same time. the car heads east at an average speed of 50 miles pe

r​ hour, while the truck heads south at an average speed of 60 miles per hour. find an expression for their distance apart d​ (in miles) at the end of t hours.
Physics
1 answer:
olya-2409 [2.1K]3 years ago
5 0

The car heads east at an average speed of 50 miles per​ hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per​ hour from the intersection point towards South.

The distance of car from the intersection point after t hours is 50t.

The distance of truck from the intersection point after t hours is 60t.

Since these distances are perpendicular to each other, distance apart d​ (in miles) at the end of t hours is

d=\sqrt{(50t)^2+(60t)^2} \\ d=10\sqrt{61} t\\ d=78.1t

Thus the distance apart is d=78.1t \;miles

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Answer:

reactants --> products + thermal energy

Explanation:

Heat is absorbed in endothermic reaction.

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3 years ago
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).
qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

or

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

 

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

7 0
3 years ago
In the diagnostic radiology energy range (which includes mammography) from 23 to 150 kVp, which of the following tissues possess
Semmy [17]

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Bone

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The voltages used in diagnostic tubes range from roughly 20 kV to 150 kV and thus the highest energies of the X-ray photons range from roughly 20 keV to 150 keV.

The tests and equipment used sometimes involves low doses of radiation to create highly detailed images of an area.

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3 years ago
If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he
forsale [732]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
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If we re-arrange the formula, we get
C_s =  \frac{Q}{m \Delta T}
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
C_s =  \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C
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I uploaded the answer to^{} a file hosting. Here's link:

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