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3 years ago

Find the change in kinetic energy of a 1.0 kg fish leaping to the right at 12.0 m/s.

Physics

2 answers:

WARRIOR [948]3 years ago

8 0

Answer:

Hey yall, the answer is 72 J.

Explanation:

sp2606 [1]3 years ago

4 0

Answer:

Explanation:

Given parameters:

Mass of fish = 1kg

Velocity = 12m/s

Unknown:

Change in kinetic energy = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically given as:

K.E = $\frac{1}{2}$ m v²

Now, insert the parameters and solve;

K.E = $\frac{1}{2}$ x 1 x 12 = 6J

The change in kinetic energy is 6J

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A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

Where:

N = is the number of turns = 1

ΔΦ = ΔB*A

Δt = is the time = 0.3 s

A = is the loop of wire area = πr² = πd²/4

ΔB: is the magnetic field = (0 - 1.04) T

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

4 years ago

The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

AVprozaik [17]

Answer:

v=115 m/s

v=414 km/h

Explanation:

Given data

$A_{area}=0.140m^{2}\\ p_{air}=1.21 kg/m^{3}\\ m_{mass}=80kg$

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

$mg=1/2*C*p_{air}*v^{2}*A_{area}\\ v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s$

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

$v=(115*3.6)km/h\\v=414km/h$

5 0

3 years ago

What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?

goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0

3 years ago

block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally

Alexeev081 [22]

Answer:

$v_{2} =\frac{1}{2}$

Explanation:

From the second law of Newton movement laws, we have:

$F=m*a$ , and we know that a is the acceleration, which definition is:

$a=\frac{dv}{dt}$ , so:

$F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}$

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

$dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt$

$v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t)$ (This is the indefinite integral), the definite one is:

$v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}$

3 0

3 years ago

15 The group of elements called the noble gases

kogti [31]

Answer:

C. a full outer shell of valence electrons.

Explanation:

The noble gases has a full outer shell so they don't have to react with other elements to gain or loose electrons (to have a full outer shell and be stable).

8 0

3 years ago

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