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natta225 [31]
2 years ago
11

Two bricks are resting on the edge of the lab table. Shirley Sheshort stands on her toes and spots the two bricks. She acquires

an intense desire to know which of the two bricks are most massive. Since Shirley is vertically challenged, she is unable to reach high enough and lift the bricks; she can however reach high enough to give the bricks a push. Discuss how the process of pushing the bricks will allow Shirley to determine which of the two bricks is most massive. What difference will Shirley observe and how can this observation lead to the necessary conclusion?​
Physics
1 answer:
PSYCHO15rus [73]2 years ago
4 0

Explanation:

  • since force =mass×acceleration and acceleration being g which is = 10m/s^2
  • the brick with the larger mass will reach the ground first since it has a greater force to resist resistance forces such as air resistance

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A 4kg bird has 8 joules of kinetic energy, how fast is it flying?
Igoryamba
I believe the answer is 2m/s
7 0
2 years ago
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An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th
pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Q=mc\Delta T

Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

5 0
2 years ago
Use the formula h = −16t2 + v0t. (if an answer does not exist, enter dne.) a ball is thrown straight upward at an initial speed
makkiz [27]
Using the given formula with v0=56 ft/s and h=40 ft 
h = -16t2 + v0t  
40 = -16t2 + 56t 
16t2 - 56t + 40 = 0  
Solving the quadratic equation:  
t= (-b+/-(b^2-4ac)^1/2)/2a = (56+/-((-56)^2-4*16*40)^1/2)/2*16 = (56 +/- 24) / 32 
 We have two possible solutions  
t1 = (56+24)/32 = 2.5 
t2 = (56-24)/32 = 1  
So initially the ball reach a height of 40 ft in 1 second.
3 0
2 years ago
A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i
enyata [817]

Answer:

Emitted power will be equal to 7.85\times 10^{-5}watt

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing I=10^{-12}W/m^2

We have to find the emitted power

Emitted power is equal to P=I\times A

P=I\times 4\pi R^2

P=10^{-12}\times 4\times 3.14\times  2500^2=7.85\times 10^{-5}watt

So emitted power will be equal to 7.85\times 10^{-5}watt

4 0
2 years ago
The Slowing Earth The Earth's rate of rotation is constantly decreasing, causing the day to increase in duration. In the year 20
NNADVOKAT [17]

Answer:

The average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

Explanation:

Given data,

The period of 365 rotation of Earth in 2006, T₁ = 365 days, 0.840 sec

                                                                                  = 3.1536 x 10⁷ +0.840

                                                                                 = 31536000.84 s

The period of 365 rotation of Earth in 2006, T₀ = 365 days

                                                                               = 31536000 s

Therefore, time period of one rotation on 2006, Tₐ = 31536000.84/365

                                                                                   = 86400.0023 s

The time period of rotation is given by the formula,

                                <em>Tₐ = 2π /ωₐ</em>

                                 ωₐ = 2π / Tₐ

Substituting the values,

                                  ωₐ = 2π /  365.046306        

                                      = 7.272205023 x 10⁻⁵ rad/s

Therefore, the time period of one rotation on 1906, Tₓ = 31536000/365

                                                                                    = 86400 s

Time period of rotation,

                                   Tₓ = 2π /ωₓ

                                    ωₓ = 2π / T

                                           =  2π /86400

                                          = 7.272205217  x 10⁻⁵ rad/s

The average angular acceleration

                                   α  = (ωₓ -   ωₐ) /  T₁

             = (7.272205217  x 10⁻⁵ - 7.272205023 x 10⁻⁵) / 31536000.84

                                    α  = 6.152 X 10⁻²⁰ rad/s²

Hence the average angular acceleration of the Earth, α  = 6.152 X 10⁻²⁰ rad/s²

3 0
2 years ago
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