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3 years ago

The bell rings and a physics student heads to class. They stop to talk to a few friends. They slow

Physics

1 answer:

grin007 [14]3 years ago

3 0

Answer:

420m

Explanation:

Given parameters:

Time = 5minutes

Average speed = 1.4m/s

Unknown:

Distance covered = ?

Solution:

Speed is the rate of change of distance with time.

Mathematically;

Speed = $\frac{distance}{time}$

Distance = speed x time

Insert the parameters and solve;

Convert the time to seconds;

1 minute = 60s

5 minute = 5 x 60 = 300s

So,

Insert the parameters and find the distance;

Distance = 300 x 1.4 = 420m

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A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

3 0

3 years ago

An object attached to an ideal spring oscillates with an angular frequency of 2.81 rad/s. the object has a maximum displacement

NikAS [45]

Ω = 2.81
A = 0.232
k = 29.8

x = A cos(ωt + Ф)

at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0

at t = 1.42, with Ф = 0:
x = A cos(ωt)

U = 1/2 k x² = 1/2 k [A cos(ωt)]²

4 0

3 years ago

A 5cm object is located 12 cm from a convex mirror with a focal length of 14cm. Calculate the

Tema [17]

Answer:

-86.415485655

Explanation:

3 0

3 years ago

Can there be situation when velocity is constant but speed is not

klasskru [66]

Answer:

no I don’t think there can be so my answer is No.

Okay then yes sorry that I must have gotten it wrong before.

Explanation:

4 0

2 years ago

Read 2 more answers

a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?

devlian [24]

For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s² So the average acceleration is 1.75 m/s²

5 0

4 years ago