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2 years ago

14

A 497−g piece of copper tubing is heated to 89.5°C and placed in an insulated vessel containing 159 g of water at 22.8°C. Assumi

ng no loss of water and a heat capacity for the vessel of 10.0 J/°C, what is the final temperature of the system (c of copper = 0.387 J/g·°C)?

1 answer:

algol [13]2 years ago

6 0

Answer:

36.6 degree centigre

Explanation:

Check the attached file below

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Two radio antennas A and B radiate in phase. Antenna B is a distance of 120 m to the right of antenna A. Consider point Q along

LuckyWell [14K]

Answer:

240 m

120 m

Explanation:

d = Path difference = 120 m

For destructive interference

Path difference

$d=\dfrac{\lambda}{2}\\\Rightarrow \lambda=2d\\\Rightarrow \lambda=2\times 120\\\Rightarrow \lambda=240\ m$

The longest wavelength is 240 m

For constructive interference

$d=\lambda\\\Rightarrow 120\ m=\lambda$

The longest wavelength is 120 m

4 0

3 years ago

A bullet fired horizontally hits the ground in 0.5 sec. If it had been fired with a much higher speed in the same direction, and

stira [4]

Answer:

3. 0.5 sec.

Explanation:

A bullet fired horizontally follows a projectile motion, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with constant acceleration, g = 9.8 m/s^2, towards the ground

The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

$y(t) = h + v_{0y} t - \frac{1}{2}gt^2$

where y is the vertical position at time t, h is the initial height, and $v_{0y}$ is the initial vertical velocity of the bullet.

Since the bullet is fired horizontally, $v_{0y}=0$ . So the equation becomes

$y(t) = h - \frac{1}{2}gt^2$

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

$t=\sqrt{\frac{2h}{g}}$

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.

4 0

2 years ago

A basketball player can leap upward 0.42 m. What is the magnitude of his initial velocity at the start of the leap? Use g=9.81 m

Tatiana [17]

Farmula 2gh=vf²-vi²
2×0.42×9.81=0²-vi²
8.2404=vi²
Vi=✓8.2404
Vi=2.87m/sec

8 0

2 years ago

The materials that made up the solar nebula can be categorized into the four general types as follows. Rank these materials from

san4es73 [151]

Answer:

Hydrogen and Helium
Rock
Metals
Hydrogen compounds

Ranked from left to right based on their abundance in the solar nebula:

Hydrogen and Helium-Hydrogen compounds-Rocks-Metals

I hope you find this information useful and interesting! Good luck!

5 0

2 years ago

A cylindrical rod 120 mm long and having a diameter of 15.0 mm is to be deformed using a tensile load of 35,000 N. It must not e

astraxan [27]

Answer: Magnesium Alloy

Explanation:

Cross sectional area A = (pi/4)d sqr

A = pi / 4 x 0.015 sqr

A = 1.76 x 10^-4 m2

Stressed induced in rod S = P/A = 35000/1.76x 10^-4

S= 198.06 M Pa

Reduction in diameter of the titanium alloy = -v (alpha/E) x d

= - 0.33 (( 198.06 X 10^6) / 70 x 10^9 )x 0.015 = - 1.41 x 10^-2 mm

For steel alloy

= - 0.36 (( 198.06 X 10^6) / 150 x 10^9 )x 0.015 = - 1.02 x 10^-2 mm

For magnesium

= 0.27 (( 198.06 X 10^6) / 205 x 10^9 )x 0.015 = 0.391 x 10^-2 mm

The titanium alloy and the steel alloy does not satisfy the second criteria since the reduction in diameter exceeds the allowable limit of 1.2 x 10^2mm Considering the yield strength of the material, we find the aluminium alloy is not suitable and hence no need to check for the second criteria. By considering the both the given conditions, we find the magnesium alloy is the suitable material.

3 0

2 years ago