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Elza [17]
3 years ago
15

A person stands 35.0 m from a flag pole. With a protractor at eye level, he finds that the angle to the top of the flag pole is

25.0 degrees. How high is the flag pole?
Physics
1 answer:
Phoenix [80]3 years ago
6 0

Answer:

The flag is 16.321 meters high.

Explanation:

The situation is represented in the attached figure

In the figure we have

tan(\theta )=\frac{H}{D}

Given that

θ = 25.0 degrees

D = 35.0 meters

Applying values in the above equation we have

tan(25)=\frac{H}{35.0}\\\\\therefore H=35.0\times tan(25^{o})=16.321meters

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forces F1 (east) and F2 are simultaneously applied to a 3.0kg mass. when F2 is east, a=5.0m/s² east and when F2 is west a=1.0m/s
zhenek [66]

Answer:

345453-5676

its the right answer

5 0
3 years ago
A bullet of mass m and speed v is fired at, hits and passes completely through a pendulum bob of mass M on the end of a stiff ro
dexar [7]

Required value of initial speed of the bullet be ( 4M/m)√(gL).

Given parameters:

Mass of the bullet =m.

Mass of the bob of the pendulum = M.

speed of the bullet before collision = v

Speed of the bullet after collision = v/2.

Length of the pendulum stiff rod = L.

Let speed transmitted to the pendulum be u.

Using principle of conservation of momentum:

mv = Mu + mv/2

⇒ Mu = mv/2

⇒ u = (m/M)v/2

We know that: to make the bob over the top of the trajectory without falling backward in its circular path, required speed be = √(4gL). [ where g = acceleration due to gravity]

To be minimum initial speed the bullet must have in order for the pendulum bob to just barely swing through a complete vertical circle:

u = √(4gL)

⇒  (m/M)v/2 = √(4gL)

⇒ v =( 4M/m)√(gL).

Hence, minimum required  speed of the bullet be ( 4M/m)√(gL).

Learn more about speed here:

brainly.com/question/28224010

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7 0
1 year ago
A ball rolls off a table and it traveling with a horizontal velocity of 2 m/s and 1 point
morpeh [17]

Answer:

The velocity when the ball hits the ground is obtained using v2. 2 = v1. 2 + 2 g Dy with v1=0 and Dy=h. Thus solving for v2 yields 17.1 m/s v2 = 2 g h =.

21 pages·330 KB

4 0
3 years ago
Read 2 more answers
A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup
ohaa [14]
Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

Because the molar mass of oxygen is 16, the number of moles of O₂ is
n = (240 g)/(2*16 g/mol) =  7.5 mol

As an ideal gas,
p*V = nRT
or
V = (nRT)/p
where R = 8.314 J/(mol-K)

When
p = 0.910 atm = (0.910 atm) * (101325Pa/atm) = 92205.75 Pa
T = 27 °C = (27 + 273) K = 300 K
then the volume is
V= \frac{(7.5 \, mol)*(8.314 \, J/(mol-K))*(300 \, K)}{(92205.75 \, Pa)} =0.2029 \, m^{3}

V = (0.2029 m³)*(10³ L/m³) = 202.9 L

Answer: 203 liters
3 0
3 years ago
Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di
dlinn [17]

Answer:

E=29\times 10^{-5}eV

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

E^{+}=E+\mu B,

E^{-}=E-\mu B,

E^{0}=E

Here, E is the energy in the absence of electric field.

And

E^{+} and E^{-} are the highest and the lowest energies.

The difference of these energies

\Delta E=2\mu B

\mu=9.3\times 10^{-24}J/T is known as Bohr's magneton.

B=2.5 T,

Therefore,

\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J

Now,

Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is E=29\times 10^{-5}eV

6 0
3 years ago
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