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gulaghasi [49]
3 years ago
14

The energy transfer diagram represents the energy of a person diving into a pool.

Physics
2 answers:
NNADVOKAT [17]3 years ago
8 0

Answer:

A) 1000 J

Explanation:

HACTEHA [7]3 years ago
5 0
The correct answer is: 
<span>A) 1000 J

In fact, energy cannot be destroyed nor created: this means that the total energy of the person must remain constant.

The initial energy of the person is U=8000 J, in form of gravitational potential energy. When he jumps and dives into the pool, this energy is converted into kinetic energy (K, 7000 J) and partially into thermal energy, Et. The final energy of the person, sum of kinetic energy and thermal energy, must be equal to the initial energy:
</span>U=K+E_t
<span>from which we can find the thermal energy:
</span>E_t = U-K=8000 J - 7000 J = 1000 J<span>
</span>
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raketka [301]
It would be D I believe! Depending on the angle of the mirror and distance positioned!
8 0
3 years ago
Consider the following statement:
vovangra [49]

Answer:

The statement "The magnetic field of a magnet comes out of the north pole and goes into the south pole" is imprecise

Explanation:

This is because the zero divergence equation (∇ · B = 0 ) is valid for any magnetic field, even if it is time dependent rather than static. Physically, it means that there are no magnetic charges otherwise we would have ∇ · B ∝ ρmag instead of ∇ · B = 0. Consequently, the magnetic field lines never begin or end anywhere in space; instead they form closed loops or run from infinity to infinity.

3 0
4 years ago
Dante is leading a parade across the main street in front of city hall. Starting at city hall, he marches the parade 4 blocks ea
Ipatiy [6.2K]

Answer:

The correct option is A)

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

Given that Dante is leading a parade across the main street in front of city hall.

Let, Initial location of parade is 0i+0j

One block of city is one units on the XY- graph

Statement 1: Parade marches the parade 4 blocks east, then 3 blocks south

New location of parade is 4i-3j

Statement 2: The parade marches 1 block west and 9 blocks north and finally stops.

Final location of parade is (4i-3j)+(-1i+9j)=3i+6j

Displacement is given by

Displacement = (Final destination)-(Initial destination)

Displacement = (3i+6j)-(0i+0j)=3i+6j

Thus,

Magnitude of displacement = \sqrt{3^{2}+6^{2}}

                                              = 6.71 m

Direction of displacement =  tan^{-1}(\frac{Y}{X} )

                                           =  tan^{-1}(\frac{6}{3} )

                                           = 63.43 NE

Therefore, the correct option is A) Displacement: 6.71 m, Direction: 63.4 degrees north of east

5 0
3 years ago
Jessica is running a 10K. She alternates between running and walking each kilometer. She runs at a rate of 1 kilometer every 5 m
saveliy_v [14]

Answer:

c.100 minutes

Explanation:

Total distance = 10 km

Runs for 1 km every 5 minutes

walks 1 km every 15 min

She alternates between walking and running  so,  Jessica will walk 5 km and run 5 Km

Time taken by Jessica for walking : 5 km

Time taken to walk 1 km=5 minutes

Time taken to walk 5 km

=> 5 X 5

=>25 minutes

Time taken by Jessica for Running : 5km

Time taken to run 1 km = 15 minutes

 

=> 5 X 15

=>75 minutes

Total time taken   = Time taken by Jessica for walking + Time taken by Jessica for Running

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8 0
3 years ago
Read 2 more answers
A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

emf=NBA\omega sin(\omega t)

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}

(c) emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V

(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

3 0
3 years ago
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