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jeka57 [31]
3 years ago
11

HELP ME! One airline limits the size of carry-on luggage to a volume of 40,000 cm^3. A passenger has a carry-on that has an area

of 1,960 cm^2 and is 23 cm high. Is the passenger's luggage OK to carry onto the airplane? show your work.
Physics
2 answers:
EleoNora [17]3 years ago
8 0
You must times the area by the volume, look at it as if the area is just one of 23 layers that makes up the volume.
1960x23=45080
so no it cannot be carried as it is 5080cm^3 over the limit
Delvig [45]3 years ago
7 0

Answer:

yes it is ok

Explanation:

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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

v = 30\cdot t (2)

Free fall (v_{o} = 900\,\frac{m}{s}, a = -9.807\,\frac{m}{s}, t_{o} = 30\,s, 30\,s \le t \le 120\,s)

v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

5 0
3 years ago
The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center
jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0
3 years ago
A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi
Ahat [919]
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

6 0
3 years ago
A line from the Brackett series of hydrogen has a wavelength of 1945nm (or 1.0979x10^7m). From which state did the electron orig
Zarrin [17]
We use the Rydberg Equation for this which is expressed as:

<span>1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
</span>
where lambda is the wavelength, where n represents the final and initial states. Brackett series means that the initial orbit that electron was there is 4 and R is equal to 1.0979x10^7m<span>. Thus,
</span>
1/ lambda = R [ 1/(n2)^2 - 1/(n1)^2]
1/1.0979x10^7m = 1.0979x10^7m [ 1/(n2)^2 - 1/(4)^2]

Solving for n2, we obtain n=1.
5 0
3 years ago
Why does the evaporation of sweat cool down your skin?
Ne4ueva [31]
So sweat<span> helps </span>cool<span> you </span>down<span> two ways. First, it makes </span>your skin<span> feel cooler when it's wet. And when it </span>evaporates<span> it removes some heat. But </span>sweat<span> will only </span>evaporato<span>in an environment where there isn't much water in the air.</span>
6 0
3 years ago
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