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2 months ago

One of the Lady Spartans was falling to the ground after

Physics

1 answer:

ValentinkaMS [3.4K]2 months ago

0 0

Answer:

Explanation:

Considering that,

A woman falls and reaches a final velocity of 4m/s

v = 4m/s

The mass of the woman is 60kg.

m = 60kg

Employing energy conservation, the potential energy from the point of the woman's fall is transformed into her final kinetic energy.

Thus,

P.E = K.E(final) = ½mv²

P.E = ½ × 60 × 4²

P.E = 480 J.

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The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Yuliya22 [3333]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

It is known that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v are constant on both sides. Therefore we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let the final pressure be P and the temperature $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide equation (2) by equation (3)

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, the left side temperature equals 1.48 times the right side temperature

6 0

1 month ago

A trained sea lion slides from rest down a long

Keith_Richards [3271]

Answer:

1.5 m/s²

Explanation:

Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.

Considering the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

This gives us N = mg cos θ

Next, examining the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substituting for N yields:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

hence a = g (sin θ − μ cos θ)

If we set θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

this results in a = 1.48

When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².

5 0

29 days ago

A resistor with resistance R and an air-gap capacitor of capacitance C are connected in series to a battery (whose strength is "

kicyunya [3294]

Answer:

a) Q = C*emf

b) Decrease in electric field strength and electric potential

c) Initial current through the resistor = emf/R

d) The final charge = K*C*emf

Explanation:

a) The resistors and capacitors are linked in series with the battery

According to Kirchoff's voltage law, the total voltage in the circuit must equal zero

Let $V_{R}$ represent the Voltage across the Resistor

$V_{c}$ and

represent the Voltage across the capacitor

Implementing KVL;

$emf - V_{R} - V_{c} = 0\\$

.........................(1)

Since this is a series connection, the same current traverses through the circuit

$V_{R} = IR\\Q = CV_{c} \\V_{c} = Q/C$

Integrating $V_{c}$ and $V_{R}$ into equation (1)

$emf - IR - Q/C = 0$

Initially, as the capacitor reaches full charge, the current will drop to zero because of equilibrium

$I = 0A\\emf = Q/C\\Q = C* emf$

b) When the plastic sheet is inserted between the plates, current begins to flow because both the electric field intensity and electric potential decrease. As a result, charge diminishes, leading to current flow

c) The current through the resistor equates to the total current within the circuit (given the series connection)

$I = I_{o} \exp(\frac{-t}{RC} )\\At time the initial time, t\\t = 0\\ I_{o} = \frac{emf}{R} \\$

Substituting the values of t and I₀ into the aforementioned formula for I

$I = \frac{emf}{R} \exp(0)\\I = \frac{emf}{R}$

d) Note: The initial charge on the capacitor equals C * emf

Following the insertion of the plastic, the new charge will be:

$Q = K* Q_{initial} \\Q_{initial} = C *emf\\Q_{final} = KCemf$

4 0

1 month ago

If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f

serg [3582]

Response:

E = ρ ( R1²) / 2 ∈o R

Clarification:

Provided information

Two cylinders are aligned parallel

Distance = d

Radial distance = R

d < (R2−R1)

To determine

Express the response using the variables ρE, R1, R2, R3, d, R, and constants

Solution

We have two parallel cylinders

therefore, area equals 2 $\pi$ R × l

And we apply Gauss's Law

EA = Q(enclosed) / ∈o......1

Initially, we calculate Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( $\pi$ R1² × l )

Thus, inserting all values into equation 1

produces

EA = Q(enclosed) / ∈o

E(2 $\pi$ R × l) = ρ ( $\pi$ R1² × l ) / ∈o

This simplifies to

E = ρ ( R1²) / 2 ∈o R

6 0

1 month ago

A baseball is hit high into the upper bleachers of left field. Over its entire flight the work done by gravity and the work done

serg [3582]

Answer:

Explanation:

As the baseball ascends, gravitational forces as well as air resistance act downward, whereas the displacement is moving upward which results in an angle of 180° between the force and displacement. Therefore, the work done by both the gravitational force and air resistance is negative, confirming option (d) as accurate.

6 0

1 month ago