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abruzzese [7]
3 years ago
7

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

conservation of energy to show that the speed of the banana just before hitting the water is v
Physics
1 answer:
Mariana [72]3 years ago
6 0

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

              Total Energy before fall = Total Energy after fall

                                E_{i}  = E_{f}

Here, total energy is the sum of kinetic energy and potential energy

K.E_{i} + P.E_{i} = K.E_{f} + P.E_{f}       (a)

When banana is at height h, it has

                 K.E_{i} = 0    and    P.E_{i} = mgh          

and when it reaches the river, it has

       K.E_{f}  = 1/2mv^{2}    and   P.E_{f}  = 0

Putting the values in equation (a)

                              0 + mgh = 1/2mv^{2} + 0

                                      mgh = 1/2mv^{2}

<em>cutting 'm' from both sides</em>

<em>                                           </em>gh = 1/2v^{2}

                                          v = \sqrt{2gh}

Hence, the velocity of banana before hitting the water is

                                          v = \sqrt{2gh}

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Elan Coil [88]

Answer:

I think its C sorry if it's wrong

7 0
3 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
2 years ago
A piece of gym equipment states that the maximum load it can hold is 300 kg. Why do you think it is important not to go over thi
Crazy boy [7]

Answer:

The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.

5 0
2 years ago
A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the tau
slega [8]

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

V=\frac{8*4.89}{0.617}\\ V=63.4m/s

Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}

Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\  T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N

5 0
2 years ago
Sperm whales, just like bats, use echolocation to find prey. A sperm whale’s vocal system creates a single sharp click, but the
Tamiku [17]

Answer: The size of the whale can be estimated by using echo Sounder or Sonar

Explanation:The researchers and physicst uses the knowledge of reflection of sound in water or echo to solve a lot of puzzle under water , including calclulating the distance and size of objects like fish,sunk ship,etc

The sonar is able to receive the signals from the whales and the succesive echoes that follows when placed in a strategic position in the water in the path of the whale .

The sonar has a traducers which has both a transmitter and a receiver,so it receives pulses from the whale and transmit back same echo back to the whale and to the bottom of the sea.

Using the speed of sound underwater and time taken for succesive echoes of pulses ,it can calculate with the precision the location of the whale while the image of the whale is also capture by the traducers to determine the size.

3 0
3 years ago
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