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Nuetrik [128]
2 years ago
12

Only Wednesdays Pls help

Physics
2 answers:
Sindrei [870]2 years ago
4 0
No, because it’s will be bad for peoples health
madreJ [45]2 years ago
3 0
No, it should not be permitted. It can seriously damage one of the people participatings mind. It would overall not be good
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Order the steps to describe how information is sent. A signal is produced. A signal is seen, heard, or used. Radio waves are mod
nordsb [41]

Answer:

A signal is produced.

Radio waves are modulated and amplified.

Radio waves are amplified and demodulated.

A signal is seen, heard, or used.

Explanation:

Communication is the method of transmitting information between two or more participants. A radio communication network is a set of static and wireless radio equipment developed to facilitate an entity by enabling particular modes of communication such as one-to-many and one-to-one communication. Radio waves are being used to transmit information spatially from the sender to receiver, by modulating the radio signal in the transmitter. The above steps explains the loop of the radio communication system.

6 0
3 years ago
How to find yield strength of a load vs deflection?
liraira [26]
Σ/ε
σ = F/A
ε = ΔL/L
F = force
A = area
L = lenght
ΔL = |old lenght - new lenght|
6 0
3 years ago
Th answer is "electric attraction is a force that can act at a distance."
Tasya [4]
Electric forces is not action-by-distance. Charged particle emits a electric field radially outwards. It corresponds by the inverse-square, meaning it is 1/r^2.
7 0
3 years ago
A 20 m high filled water tank develops a 0.50 cm hole in the vertical wall near the base. With what speed does the water shoot o
Rina8888 [55]

Answer:

The speed of the water shoot out of the hole is 20 m/s.

(d) is correct option.

Explanation:

Given that,

Height = 20 m

We need to calculate the velocity

Using formula Bernoulli equation

\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}

Where,

v₁= initial velocity

v₂=final velocity

h₁=total height

h₂=height of the hole from the base

Put the value into the formula

v_{1}^2=2g(h_{2}-h_{1})

v_{1}=\sqrt{2g(h_{2}-h_{1})}

v_{1}=\sqrt{2\times9.8\times(20-0.005)}

v_{1}=19.7\ m/s= approximate\ 20\ m/s

Hence, The speed of the water shoot out of the hole is 20 m/s.

7 0
3 years ago
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
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