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klio [65]
3 years ago
14

A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floo

r and that its period is 27.5 s. (a) How tall is the tower
Physics
1 answer:
wariber [46]3 years ago
4 0

Answer: 187.68 m

Explanation:

Given

Time period T=27.5\ s

The time period of a pendulum is given by

\Rightarrow T=2\pi \sqrt{\dfrac{L}{g}}

where L=height of tower

Put values

\Rightarrow 27.5=2\pi \sqrt{\dfrac{L}{9.8}}\\\\\Rightarrow L=\dfrac{27.5^2\times 9.8}{(2\pi )^2}=187.68\ m

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A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle
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Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

6 0
3 years ago
The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo
hjlf

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m= h^2/(2*m*λ^2)

replacing we obtained the above values.

6 0
3 years ago
If the primary coil of a transformer has 200 turns and is supplied with 120 v ac power, how many turns must the secondary coil h
Oduvanchick [21]
The ratio of the turns to the voltage should be equal
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5 0
3 years ago
A charge of q= +15 uC moves in a Northeast direction with a speed 5 m/s, 25 degrees East of a magnetic field, pointing North, wi
grin007 [14]

Explanation:

It is given that,

Magnitude of charge, q=15\ \mu C=15\times 10^{-6}\ C

It moves in northeast direction with a speed of 5 m/s, 25 degrees East of a magnetic field.

Magnetic field, B=0.08\ j

Velocity, v=(5\ cos25)i+(5\ sin25)j

v=[(4.53)i+(2.11)j]\ m/s

We need to find the magnitude of force on the charge. Magnetic force is given by :

F=q(v\times B)

F=15\times 10^{-6}[(4.53i+2.11j)\times 0.08\ j]

<em>Since</em>, i\times j=k\ and\ j\times j=0

F=15\times 10^{-6}[(4.53i)\times (0.08)\ j]

F=0.00000543\ kN

F=5.43\times 10^{-6}\ kN

So, the force acting on the charge is 5.43\times 10^{-6}\ kN and is moving in positive z axis. Hence, this is the required solution.

6 0
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