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3 years ago

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the fabric of the hot air balloon can stretch, but it does not allow air to pass through it. think about the change in air densi

ty you expect as the pilot adds hear to the air in the balloon. what change would you expect to see in the fabric of the hot air balloon. explain

1 answer:

Karolina [17]3 years ago

3 0

It would definitely stretch and wouldn't do anything as long as there is no holes in it.

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A bird has a mass of 0.8 kg and flies at a speed of 11.2 m/s. How much kinetic energy does the bird have?

riadik2000 [5.3K]

The kinetic energy of an object is directly proportional to its mass and the square of its velocity

KE = 1/2 (mv²)

KE = Kinetic Energy
m = mass in kg
v = velocity in m/s

Given:

m = .8 kg
v = 11.2 m/s

Substitute:

KE = 1/2 (.8)(11.2²)
KE = 50.18 J

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3 years ago

Read 2 more answers

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

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3 years ago

A 3.8kw elective motor powers an inclined conveyer belt. It is designed to lift heavy boxes from the warehouse floor to loading

zhannawk [14.2K]

Answer:

See the answers below

Explanation:

To solve this problem we must use the definition of power and work in physics.

a)

The function of the conveyor belt is to carry the boxes from an initial point that is at low altitude to an end point that is at high altitude. In this way the conveyor belt prints a speed to the box to be able to raise it to the required vertical distance.

Since we have a velocity at the beginning and then we place the box at a high position, where then the box remains at rest, we can say that it converts kinetic energy to potential energy.

b)

Power is defined as the relationship of work over time. Therefore we have:

$P=W/t$

where:

P = power = 3.8 [kW] = 3800 [W]

W = work [J]

t = time = 14 [s]

$W=P*t\\W=3800*14\\W= 53200[J] = 53.2[kJ]$

c)

Since the given time is equal to the given time at Point b, we can use the same work calculated.

We know that work is defined as the product of force by the distance traveled.

$W =F*d$

So, the force is equal to:

$F=W/d\\F=53200/5.3\\F=10037.73[N]$

Now we know that force is defined as the product of mass by gravitation acceleration.

$F =m*g$

where:

F = force or weight = 10037.73 [N]

g = gravity acceleration = 9.81 [m/s²]

m = mass [kg]

$m=F/g\\m = 10037.73/9.81\\m = 1023.2 [kg]$

d)

This part can be solved by means of the energy conservation theorem, where the potential energy is transformed into kinetic energy or vice versa.

$E_{pot}=m*g*h = E_{kin}=0.5*m*v^{2}$

where:

h = elevation = 4.7 [m]

v = velocity [m/s]

$m*g*h=0.5*m*v^{2}\\g*h=0.5*v^{2} \\v=\sqrt{\frac{g*h}{0.5} } \\v=\sqrt{\frac{9.81*4.7}{0.5} } \\v = 9.6 [m/s]$

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3 years ago

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and han

sleet_krkn [62]

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

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3 years ago

A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

8 0

3 years ago