Physics question help

The path of a stunt car that is driven off a cliff follows a curve typical of a projectile launched horizontally

hodyreva [135]

Answer:

option B: The same (see picture)

6 0

2 years ago

A 5400 kg elephant sees a mouse and turns the opposite direction with 8000N of force. What is the elephant's acceleration?

OlgaM077 [116]

Answer:

acceleration= force / mass

a= 1.48 m/sec^2

3 0

4 years ago

Why do we see Moon phases?

wlad13 [49]

Answer:

We see moon phases cause the moon goes around the earth and the sun makes a white part on it

Explanation:

6 0

3 years ago

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The spectral type of this star is listed as K5V. (V means it is a main sequence star.) Based on the table below, how does the su

notka56 [123]

It turns out that the star in question belongs to the K5V star class. Thus, the surface temperature of this must be between 5200 K and 3700 K. It has a surface temperature of 6200 K to 5200 K and is a G2V class star, commonly known as Bernard's star. In other words, the star in question has a colder surface than HIP 87937. This is further explained below.

<h3>What is surface temperature?</h3>

Generally, The temperature at a surface is referred to as the surface temperature. In particular, it may refer to the surface air temperature, which is the temperature of the air that is relatively close to the earth's surface.

In conclusion, The star in question is of the K5V type. Therefore, the temperature of the surface of this must be somewhere between 5200 and 3700 kelvin. The star with the designation HIP 87937, often known as Bernard's star, is of the G2V type. Stars of this type have a surface temperature that falls anywhere between 6200 and 5200 degrees Celsius. Therefore, it should come as no surprise that the surface temperature of the given star is lower than that of HIP 87937.

Read more about surface temperature

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3 0

2 years ago

A cast-iron flywheel has a rim whose OD is 1 m and whose ID is 0.8 m. The flywheel weight is to be such that an energy fluctuati

grin007 [14]

Answer:

A.Coefficient of speed fluctuation of the flywheel = 0.222

B. The width <em>(thickness)</em> of the rim should be 0.131 m (131 mm)

Explanation:

A.

Coefficient of speed fluctuation $(C_{s})$ $= \frac{N_{2}-N_{1}}{N}$

$N_{1}$ = minimum speed = 200 rpm

$N_{2}$ = maximum speed = 250 rpm

$N$ = average speed $= \frac{N_{2}+N_{1}}{2} = \frac{250+200}{2} = 225 rpm$

∴ $Cs = \frac{250-200}{225}=0.222$

Hence the coefficient of speed fluctuation of the flywheel = 0.222

B.

The moment of Inertia , $I=\frac{E_{2}-E_{1}}{C_{s}\times\omega^{2}}$

Where

$E_{2}-E_{1}=$ energy fluctuation of flywheel = 6.75 J

$\omega^{2}=$ angular velocity of flywheel = $\frac{2\pi N}{60} = \frac{2\pi \times 225}{60}= 23.56 rad/sec$

$C_{s}=$ coefficient of speed fluctuation of the flywheel = 0.222

Hence,

$I=\frac{6.75\times10^{3}}{0.222\times(23.56)^{2}}=54.78 Nms^{2}$

Similarly,

I = $\frac{m}{8}\times(d_{o}^{2}- d_{i}^{2})$

From the moment of Inertia, we can get the weight of the flywheel as

$m=\frac{8I}{(d_{o}^{2}+ d_{i}^{2})}= \frac{8\times 54.78}{(1^{2}+0.8^{2})}=267.22kg$

From this weight, we will be able to calculate the volume of the flywheel and hence, estimate the thickness. But to do this, we need to know its density first. this can be got from standard tables.

Specific weight of cast iron = $70.6KN/m^{3}$ ( from standard material property table)

density of cast iron , $\rho =\frac{70.6\times 10^{3}}{9.81}= 7,197kg/m^{3}$

Volume of cast iron flywheel = $\frac{m}{\rho}= \frac{267.22}{7197}= 0.03713 m^{3}$

similarly, the volume of the flywheel can also be obtained through the formula :

$V= \frac{\pi t(d_{o}^{2}-d_{i}^{2})}{4}$

we can easily estimate the thickness of the flywheel from here by solving for t as shown below

$0.03713=\frac{\pi t(1^{2}-0.8^{2})}{4}$

$0.03713=0.2827t$

$t=\frac{0.03713}{0.2827}$

$\therefore t= 0.131m \approx 131mm$

The width of the rim = 131 mm

8 0

3 years ago