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cricket20 [7]
2 years ago
13

Please view attached images. The first one is #2 and the second one is #1.

Physics
1 answer:
MakcuM [25]2 years ago
8 0
The proton would be 2 and d a part of 1 then calculate that hope this helped
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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
2 years ago
Eddie tosses his phone up to his friends window, which is 4 m above the ground. what energy conversion takes place with respect
yan [13]
C the anwer is c hope i help
3 0
3 years ago
Read 2 more answers
Calculate the amount of energy transferred when a 40w light bulb is left on for 30 minutes.
Alinara [238K]

Answer:

E=72000J or 72kj

Explanation:

The formula is E=pt you need to convert your t from minutes to seconds before proceeding

3 0
3 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

4 0
3 years ago
Read 2 more answers
I'll give brainliest can someone help me with 2.4.1-2.4.3?with explanation. ​
KIM [24]

Answer:

220÷4=550 per hour

2.4.3 550×4=2200

hope it's right

8 0
2 years ago
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