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olganol
1 day ago
11

A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.

The driver takes the turtle to a town 1.1 km to the north with an average speed of 12 m/s. What is the magnitude of the average velocity of the turtle for its entire journey?
Physics
1 answer:
kicyunya [2.2K]1 day ago
8 0
The average velocity is 3.59 m/s. To determine this, we first calculate the total displacement. The turtle initially walks 18m south and is then carried 1.1 km (or 1100m) north, resulting in a net displacement of 1082m north (1100m - 18m). The time taken includes the duration of walking and the time spent in the truck. The turtle's walking time amounts to 3.5 minutes, equating to 210 seconds. For the truck's travel time, using the formula Time = Distance / Speed, we find the truck time for the 1100m distance at a speed of 12m/s to be approximately 91.67s. Therefore, the total time taken sums to 301.67s. The average velocity, calculated as total displacement divided by total time, thus equals 3.59 m/s to the north.
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An ideally efficient heat pump delivers 1000 J of heat to room air at 300 K. If it extracted heat from 260 K outdoor air, how mu
Yuliya22 [2420]

Answer:

Wnet, in, = 133.33J

Explanation:

Provided that

Pump heat QH = 1000J

Hot temperature TH= 300K

Cold temperature TL= 260K

Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:

According to the first law of thermodynamics,

COP(HP, rev) = 1/(1-TL/TH)

COP(HP, rev) = 1/(1-260/300)

COP(HP, rev) = 1/(1-0.867)

COP(HP, rev) = 1/0.133

COP(HP, rev) = 7.5

The power necessary to operate the heat pump is given by

Wnet, in = QH/COP(HP, rev)

Wnet, in = 1000/7.5

Wnet, in = 133.333J. QED

Thus, the 133.33J represents the initial work input during the heat transfer process.

<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">

QL=QH-Wnet, in

QL=1000-133.333

QL=866.67J

</pbased></padditionally...>
5 0
21 day ago
Jo, Daniel and Helen are pulling a metal ring. Jo pulls with a force of 100N in one direction and Daniel with a force of 140N in
inna [2205]

Answer:

She exerts a force of 40 N.

Explanation:

The fact that the ring remains stationary indicates that the forces are in equilibrium.

Let’s denote Jo's force as x.

The equation to consider is

140 = x + 100

x = 40

5 0
14 days ago
When 999mm is added to 100m ______ is the result​
Keith_Richards [2256]

Answer:

The outcome of adding 999mm to 100m is 101m.

Explanation:

That's my belief.

6 0
26 days ago
Read 2 more answers
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation
Ostrovityanka [2204]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

6 0
20 days ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Sav [2226]

Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

The question indicates that

    The wire’s diameter is  d = 0.205cm = 0.00205 \ m

     The radius of the wire is  r = \frac{0.00205}{2} = 0.001025 \ m

     Aluminum's resistivity is 2.75*10^{-8} \ ohm-meters.

       The electric field variation is described as

         E (t) = 0.0004t^2 - 0.0001 +0.0004

     

The charge is effectively given by the equation

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area expressed as

       A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 Thus,

       \frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

By substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.

The question states that t =  5 seconds

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
1 month ago
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