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NISA [10]
4 years ago
15

Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.

Physics
1 answer:
miss Akunina [59]4 years ago
8 0

Answer:

Δt =  \frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt  Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 +  Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw} = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

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marissa [1.9K]

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container =  120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, M_{cu} = 200 g

initial temperature of the copper cube, T_c_u = 85°C

initial temperature of the aluminum chunk T_A_l = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}

where;

C_{AL} is specific heat capacity of aluminum

\delta T_{Al} is change in temperature of aluminum

C_c_u is the specific heat capacity of copper

\delta T_c_u is the change in temperature of copper

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg

Therefore, the mass of the aluminum chunk is 258 g

7 0
3 years ago
Is it possible to have a net torque when all of the forces sum to zero? Explain.
nexus9112 [7]

Answer:

Yes it is possible

Explanation:

When two equal magnitude forces are acting on the rod in opposite direction

Then the net force on the system is always zero in that case

so we will have

F - F = 0

now for the system net torque due to these forces is given by

\tau = F(r_1) + F(r_2)

here we know that

r_1, r_2 = distance of the forces from reference about which torque is measured

so here we can say that net force is zero on the system while torque is not zero

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4 years ago
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8 0
3 years ago
A closed-end organ pipe is used to produce a mixture of sounds. The third and fifth harmonics in the mixture have frequencies of
nexus9112 [7]

Answer:

F_1=366.67Hz

Explanation:

From the question we are told that:

Frequency of 3rd harmonics F_3=1100

Frequency of 5th harmonics F_3=1833

Generally the equation for Wavelength at 3rd Harmonics is mathematically given by

 \lambda_3=\frac{4}{3}l

Therefore

 F_3=\frac{3v}{4l}

Generally the equation for Wavelength at 1st Harmonics is mathematically given by

 \lambda_1=\frac{4}{1}l

Therefore

 F_1=\frac{v}{4l}

Generally the equation for the frequency of the first harmonic is mathematically given by

 F_1=\frac{F_3}{3}

 F_1=\frac{1100}{3}

 F_1=366.67Hz

7 0
3 years ago
Calculate the magnitude of the electric field 2.00 m from a point charge of 5.00 mC (such as found on the terminal of a Van de G
MakcuM [25]

Answer:

The magnitude of the electric field is 1.124 X 10⁷ N/C

Explanation:

Magnitude of electric field is given as;

E = \frac{kq}{r^2} , N/C

where;

E is the magnitude of the electric field, N/C

q is the point charge, C

k is coulomb's constant, Nm²/C²

r is the distance of the point charge, m

Given;

q = 5mC = 5×10⁻³ C

r = 2m

k = 8.99 × 10⁹ Nm²/C²

Substitute these values and solve for magnitude of electric field E

E = \frac{kq}{r^2}  = \frac{(8.99 X10^9)(5X10^{-3})}{2^2}  = 1.124 X10^7 N/C

Therefore, the magnitude of the electric field is 1.124 X 10⁷ N/C

7 0
3 years ago
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