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3 years ago

Which statement describes an unanticipated impact of a technological

1 answer:

6 0

Answer: the development of antibiotics allowed people to be treated for bacterial infections, but their overuse had led to strains of bacteria that cannot be killed by available antibiotics

Explanation: I just took the test

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How mesuremerment of gravity using simple pendulum <br>

ale4655 [162]

Answer:

See below

Explanation:

Set up your pendulum

measure its length and time the period ( you could time 100 of them and divide the time result by 100 to get the period, T)

then use

T = 2 pi sqrt (L/g) T = period L = length g = gravity

6 0

2 years ago

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

4 0

3 years ago

A Michelson interferometer uses light with a wavelength of 452.446 nm. Mirror M2 is slowly moved a distance x, causing exactly 2

I am Lyosha [343]

Answer:

Option b. Dark spot

Explanation:

Michelson interferometer gives the path difference of the light as path difference, d as twice the distance moved or covered by the mirror $M_{2}$ , x and is given as:

d= 2x

Since, its an even multiple, we obtain a bright fringe

now, when the move half the distance, i.e., $\frac{x}{2}$

Therefore, the path difference for half the distance $\frac{x}{2}$ is:

d = x

As it is clear that its an odd multiple which correspond to dark spot as the final image

3 0

4 years ago

A student produces a power of p = 0.87 kw while pushing a block of mass m = 75 kg on an inclined surface making an angle of Î¸ =

Paul [167]

When block is pushed upwards along the inclined plane

the net force applied on the block will be given as

$F_{net} = mg sin\theta + \mu_k mg cos\theta$

here we know that

m = 75 kg

$\theta = 8.5 degree$

$\mu_k = 0.16$

now plug in all values into this

$F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5$

$F = 225 N$

now for finding the power is given as

$P = Fv$

$0.87 \times 10^3 = 225 \time v$

$v = \frac{870}{225} = 3.87 m/s$

6 0

3 years ago

Cathode rays were shown to be a stream of _____.

guapka [62]

Cathode rays were shown to be a stream of "electrons".

Cathode rays (likewise called an electron beam) are streams of electrons saw in vacuum tubes. In the event that a cleared glass tube is outfitted with two anodes and a voltage is connected, the glass inverse the negative terminal is seen to sparkle from electrons radiated from the cathode. Electrons were first found as the constituents of cathode beams. The picture in an exemplary TV is made by centered light emission redirected by electric or magnetic fields in cathode ray tubes (CRTs).

7 0

4 years ago

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