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Serjik [45]
3 years ago
5

A runner circles the track exactly 2 times for distance of 800m. it takes 4.0min. what is her average speed in m/s? what is her

average velocity?
Physics
1 answer:
Sindrei [870]3 years ago
5 0

If the runner ends where she started, then her total displacement is \Delta x=0\,\mathrm m, which means her average velocity will be

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{0\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=0\,\dfrac{\mathrm m}{\mathrm s}

The total distance she traversed, however, was \Delta d=800\,\mathrm m, which gives her an average speed of

\bar s=\dfrac{\Delta d}{\Delta t}=\dfrac{800\,\mathrm m}{4.0\,\mathrm{min}\cdot\frac{60\,\mathrm s}{1\,\mathrm{min}}=3.3\,\dfrac{\mathrm m}{\mathrm s}

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An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

\phi = 1.89 \times 10^{-7}t

(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0
2 years ago
A skater is using very low friction rollerblades. A friend throws a Frisbee at her, on the straight line along which she is coas
kupik [55]

Answer:

a)  perfectly inelastic,  b)  collision is inelastic,  c)   elastic  

Explanation:

In this exercise, it is asked to identify what type of shock occurs between the skater and the frisbee, for this we must define a system formed by the skater and the fribee, so that the forces during the crash have been internal and the amount of movement is preserved

Initial instant. Before the skater touches the frisbee

    p₀ = M v₁ + m v₂

where M and m are the masses of the skater and frisbee, respectively

for the final moment they give us several possibilities, in all case the moment is conserved

       p₀ = p_{f}

case a)

Final instant. grabs the frisbee and holds it

    p_{f} = (M + m) v '

     p₀ = p_{f}

We can see that this shock is perfectly inelastic, it holds the fressbee

case b)

final instant.

This case is similar to the previous one, but the final speed of fresbee is zero, therefore this collision is inelastic and the kinetic energy is not conserved.

case c)

final instant. Grab the fressbee and resend it

      p_{f} = M v_{1f} + m v_{2f}

this is an elastic Shock since the equivalent of a rebound of the fressbee, the kinetic energy is conserved.

5 0
2 years ago
A 15 g turntable is covered with a uniform layer of dry ice that has a mass of 9.0 g. The angular speed of the turntable and dry
pychu [463]

Answer:

1.2 rad/s

Explanation:

m1 = 15 g, m2 = 9 g, ω1 = 0.75 rad/s

Let the new angular speed is ω2 and the radius of the table be r.

The angular momentum is conserved when no external torque is applied.

I1 ω1 = I2 ω2

(m1 + m2)x r^2 x 0.75 = m1 x r^2 x ω2

(15 + 9) x 0.75 = 15 x ω2

ω2 = 1.2 rad/s

3 0
2 years ago
A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density ρ is not un
baherus [9]

A negative charge and a positive charge

5 0
3 years ago
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