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Aleksandr-060686 [28]
3 years ago
14

A tree is 257 ft high. To the nearest tenth of a meter, how tall is it in meters? There are 3.28 ft in 1 m.

Physics
1 answer:
olga2289 [7]3 years ago
7 0

Answer:

Height of tree = 78.35 meters.

Explanation:

We have

          1 meter = 3.28 feet

That is

          1 ft = \frac{1}{3.28}=0.3048m

Here height of tree = 257 ft

Height of tree = 257 x 0.3048 = 78.35 m

Height of tree = 78.35 meters.

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A centrifuge is a device that rotates an object to produce an acceleration many times that of gravity Pilots are trained in such
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Answer:

ω = 0.571 rad/s

Explanation:

given data

radius = 30 m

solution

we take here g = 9.8 m/s²

and g is express as

g = r × ω²     ....................1

put here value and we get

9.8 = 30 × ω²

solve it we get

ω = 0.571 rad/s

5 0
3 years ago
Examine the following equation.
alex41 [277]

Answer:

E)brain decay

Explanation:

Looking at the question causes it.

6 0
3 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr
xxMikexx [17]

Answer:

Radius, r=2.14\times 10^{-5}\ m

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, E=4.9\times 10^{-19}\ J

Kinetic energy is given by :

E=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{2E}{m}}

v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}            

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

qvB\ sin90=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }

r=2.14\times 10^{-5}\ m

So, the radius of the circular path is 2.14\times 10^{-5}\ m. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
Which of these is an advantage of parallel circuits over series circuits?
adell [148]

Answer:

d

Explanation:

study island said so

6 0
3 years ago
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