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madreJ [45]
3 years ago
15

heathers uncle has HIV/Aids .after visiting ,heathers uncle kisses her good bye on the cheek. Should heather and her uncle be co

ncerned about HIV transmission ?
Physics
1 answer:
Allushta [10]3 years ago
8 0

Answer:

no

Explanation:

HIV transmits only during fluid interaction, only if she Heather had an cut exposing the inner skin then HIV can be transmitted

You might be interested in
Ii.
notka56 [123]

Answer:

  • toaster -- 15 A, 8 Ω
  • fry pan -- 10.83 A, 11.08 Ω
  • lamp -- 0.83 A, 144 Ω
  • fuse will blow

Explanation:

  P = VI

  I = P/V = P/120

  R = V/I = V/(P/V) = V^2/P = 14400/P

<u>Toaster</u>: I = 1800/120 = 15 . . . amps

  R = 14400/1800 = 8 . . . ohms

<u>Fry pan</u>: I = 1300/120 = 10.833 . . . amps

  R = 14400/1300 = 11.08 . . . ohms

<u>Lamp</u>: I = 100/120 = 0.833 . . . amps

  R = 14400/100 = 144 . . . ohms

The total current exceeds 20 A, so will blow the fuse.

5 0
4 years ago
Una persona y su perro se encuentran inicialmente en reposo y paradas arriba de una caja que está apoyada sobre un piso rugoso.
alexandr1967 [171]

Answer:

La masa de la mascota es de 11,24 kg.

Explanation:

Dado que la masa total de = 80 kg

Tenemos

Fuerza de fricción = 80 kg × 9.81 × 0.3 = 235.44 N

La mascota salta horizontalmente con una velocidad, v, de 6.0 m / s para dar un movimiento de 1 m a la caja de masa y a la persona

Trabajo realizado por el salto de la mascota = (80 - m) × 9.81 × 0.3

La energía del movimiento del perro = 1/2 × m × v² = Trabajo realizado al saltar de la mascota

1/2 × m × 6² = (80 - m) × 9.81 × 0.3

20943 · m = 235440

m = 235440/20943 = 11,24 kg

5 0
4 years ago
An athlete produced 840 watts of power in 0.6 seconds. How much work did this individual perform during the exercise
Veseljchak [2.6K]

An athlete produced 840 watts of power in 0.6 seconds. And will produce 504 joule of work did this individual perform during the exercise.

<u>Solution:</u>

we know that power is the work done completed in a given interval of time.

Power = \frac{work}{time}

also, Work = Power × Time

Work = 840 × 0.6 = 504 joule

What is Power?

  • The quantity of work that can be accomplished in a certain amount of time is measured by power.
  • Power is expressed in joules per second (J/s) this is because work is denoted by the symbol J and time by the symbol s.
  • This is the watt, which is also known as the power unit in SI (W). One joule of labor per second is equal to one watt.
  • The number of watts that a device uses is indicated on labels for lightbulbs and other small appliances like microwaves.
  • A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction.
  • The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item. Work has no direction and only magnitude. Work is a scalar quantity as a result.

Know more about work power numerical brainly.com/question/181496

#SPJ4

7 0
2 years ago
A 5000 kg railcar hits a bumper (a spring) at 1 m/s, and the spring compresses 0.1 meters. Assume no damping. a) Find the spring
ludmilkaskok [199]

Answer:

k = 0.5 MN/m

Explanation:

Mass of the railcar, m = 5000 kg

Speed of the rail car, v = 1 m/s

The Kinetic energy(KE) of the railcar is given by the equation:

KE = 0.5 mv²

KE = 0.5 * 5000 * 1²

KE = 2500 J

The spring's compression, x = 0.1 m

The potential energy(PE) stored in the spring is given by the equation:

PE = 0.5kx²

PE = 0.5 * k * 0.1²

PE = 0.005k

According to the principle of energy conservation, Kinetic energy of the railcar equals the potential energy stored in the spring

KE = PE

2500 = 0.005k

k = 2500/0.005

k = 500000 N/m

k = 0.5 MN/m

8 0
4 years ago
A 3.0 kg cart moving due east at 4.0 m/s collides with a 6.0 kg cart moving due west. the carts stick together and come to rest
denpristay [2]
The problem about describes a perfectly inelastic collision. We are tasked to find the initial velocity of an object having a mass of 6 kg moving due west. It is given in the problem that after collision the cart sticks together and it stops. Thus, the final mass is the sum of the two cart and the final velocity is zero. For a perfectly inelastic collision,

m1v1-m2v2=vf(m1+m2)
By Substitution,

3(4)-6(v2)=0
6v2=12
v2=2

Therefor, the initial velocity if a 6 kg cart is 2 m/s
3 0
4 years ago
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