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3 years ago

Which of the following explains why doctors practice antiseptic medicine?

Physics

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

D.) to prevent doctors from introducing microbes that could infect the patient

Explanation:

An antiseptic is a substance that stops or slows down the growth of microorganisms. They're frequently used in hospitals and other medical settings to reduce the risk of infection during surgery and other procedures.

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A motor does 20. joules of work on a block, accelerating the block vertically upward. Neglecting friction, if the gravitational

Eva8 [605]

The motor does 20 J of work on the block, it means that the total mechanical energy of the block has increased by 20 J. But the increase in total mechanical energy is equal to the sum of the increases in potential energy and kinetic energy:
$\Delta E = \Delta U + \Delta K$
So, if the gravitational potential energy has increased by 15 J, the kinetic energy has increased by
$\Delta K = \Delta E - \Delta U = 20 J - 15 J= 5 J$

4 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim

miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
We can find this distance simply applying the definition of average velocity, as follows:

$\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)$

The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

$v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)$

where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
Solving for Δx, we get:

$\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)$

So, the total distance traveled was the sum of (1) and (3):
Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

$\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)$

Δx₂, is the distance traveled while decelerating, and can be obtained using (2):

$v_{omax} ^{2} = 2* a* \Delta x_{2} (7)$

Solving for Δx₂, we get:

$\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)$

Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
Taking the positive root in the quadratic formula, we get the following value for vomax:
v₀max = 24.3 m/s.

6 0

3 years ago

What happens when a roller coaster car moves down from the top of a hill?

VARVARA [1.3K]

-- It accelerates.
-- Its speed increases.
-- It gains momentum.
-- It loses altitude.
-- It loses potential energy.
-- It gains kinetic energy.
-- Its wheels make a lot of noise.
-- Everybody screams.

6 0

4 years ago

Read 2 more answers

A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the

olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0

4 years ago

A wave has angular velocity 12 rad/sec and maximum displacement X cm (X= last two digit [1] of your student’s ID). Calculate the

Naily [24]

Answer:

$a=1440\ m/s^2$

Explanation:

Given that,

The angular velocity of a wave, $\omega=12\ rad/s$

The maximum displacement of the wave, A = 10 cm (let)

The maximum acceleration of the wave is given by :

$a=-A\omega^2$

Put all the values,

$a=10\times 12^2\\\\a=1440\ m/s^2$

So, the maximum acceleration of the wave is equal to $1440\ m/s^2$ .

4 0

3 years ago