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3 years ago

Oops didnt mean to ask a question

Physics

2 answers:

Marizza181 [45]3 years ago

7 0

Answer:

it fine lol..i give hug soooo [hug] dere you go now i must go back to school

Explanation:

LekaFEV [45]3 years ago

6 0

Answer:

Its fine. You can ask me if you need help, and I can answer it.

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Which ray diagram demonstrates the phenomenon of absorption? tum

lions [1.4K]

Answer:

hjbyvtf ghbj

Explanation:

uhgbnm,likjh

4 1

3 years ago

How many square meters are in a rectangular piece of carpet which measures 12.0 feet by 22.0 feet? 1 m = 39.37 in., 1 ft = 12 in

gavmur [86]

Answer:

Area of the rectangle will be $24.520m^2$

Explanation:

We have given sides of rectangle is 12 feet by 22 feet

We know that 1 feet = 0.3048 m

So 12 feet will be equal to 12×0.3048 = 3.657 m

And 22 feet will be equal to 22×0.3048 = 6.705 m

We have to find the area of the rectangle in square meter

Area of rectangle is equal to multiplication of sides of the rectangle

So area A $=3.657\times 6.705=24.520m^2$

6 0

3 years ago

In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nu

nikitadnepr [17]

Answer:

Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)

Kinetic energy at the beginning: KEi = $mv^{2}$ /2

Potential energy at the end: Uf =(Qq/r)*k

Kinetic energy at the beginning: KEf = 0

From energy conservation:

r = 2*Qk/ $v^{2}$ *q/m

Ranking of the closest approach to the gold nucleous comes from q/m:

- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest

- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking

- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4

Explanation:

Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)

Kinetic energy at the beginning: KEi = $mv^{2}$ /2

Potential energy at the end: Uf =(Qq/r)*k

Kinetic energy at the beginning: KEf = 0

From energy conservation:

r = 2*Qk/ $v^{2}$ *q/m

Ranking of the closest approach to the gold nucleous comes from q/m:

- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest

- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking

- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4

7 0

3 years ago

Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,

Nitella [24]

Answer: The amount of heat needed is 29000 Cal.

Explanation:

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 200 g

$T_2$ = final temperature = $0^oC$

$T_1$ = initial temperature = $-30^oC$

Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

For process 2:

$q_2=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 200 g

$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

For process 3:

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3$

$Q=3000+16000+10000=29000Cal$

Hence, the amount of heat needed is 29000 Cal.

7 0

3 years ago

Listening to the radio, you can hear two stations at once. Describe this wave interaction

Sedaia [141]

Actually says Pantazis, since their frequencies are so wildly different, brain waves don’t interfere with radio waves. Even if that was the case, brain waves are so weak, they are hardly measurable at all. For comparison, says Pantazis, “the magnetic field of the earth is just strong enough to move the needle of a compass. Signals from the brain are a billionth of that strength.”

8 0

4 years ago

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