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slamgirl [31]
3 years ago
6

4. Jed drops a 10 kg box off of the Eiffel Tower. After 2.6 seconds, how fast is the box moving? (Neglect air resistance.)

Physics
1 answer:
Gennadij [26K]3 years ago
5 0

Answer:29.4m/s

Explanation: Vf=Vi +at= 0 initial velocity +(9.8)(3)= 29.4 m/s

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How many cubic feet are in a 55-gallon drum?
san4es73 [151]
1 gallon = 231 cubic inches
1 cubic foot = 1728 cubic inches

                       (55 gal) x (231 in³/gal) x (1 ft³/1728 in³)

                   =    (55 x 231 / 1728)  ft³

                   =         7.352 cubic feet       (rounded)  
5 0
3 years ago
Solve for the BMI weight 58kg Height 1.61​
vaieri [72.5K]

Answer:

Explanation:

BMI= weight/(height × height)          ; weight in kilogram and height in metter

     = 58kg / (1.61m  × 1.61m )

     = (58/ 2.5921) kg/m^{2}

     = 22.375  kg/m^{2}

     ≈ 22.4 kg/m^{2}

7 0
2 years ago
An object of mass 'm' is on an inclined plane with an
Levart [38]

Answer:

F_{\text{par}} = F_{\text{frict}}

F_{\text{norm}} = F_{\text{perp}}

Explanation:

See attachment for complete work.

Download pdf
7 0
2 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force e
WINSTONCH [101]

Answer: N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

Explanation:

See attached for a sketch.

From the attachment.

.

N = normal reaction force on block

W = weight of the block

theta = angle of the inclined plane to the horizontal

From the sketch, we can see that

N is equal in magnitude but opposite direction to Wy

N = Wy

And

Wy = Wcos(theta)

Wx = Wsin(theta)

Then,

N = Wy = Wcos(theta)

But W = mass × acceleration due to gravity = mg

N = Mgcos(theta)

Therefore, the Normal reaction force is equal to Mgcos(theta)

7 0
3 years ago
Read 2 more answers
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