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3 years ago

List of places in our bodies where HIV is present

Physics

1 answer:

d1i1m1o1n [39]3 years ago

4 0

Answer: The brain

Explanation:

The brain is an ideal reservoir for our bodies. That is the main one.

You might be interested in

Which of the following is constant during uniform circular motion?

tatuchka [14]

An object undergoing <span>uniform circular motion </span>is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction. So I'm thinking velocity

8 0

3 years ago

A stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. What is the spe

KIM [24]

Answer: V = 15 m/s

Explanation:

As stationary speed gun emits a microwave beam at 2.10*10^10Hz. It reflects off a car and returns 1030 Hz higher. The observed frequency the car will be experiencing will be addition of the two frequency. That is,

F = 2.1 × 10^10 + 1030 = 2.100000103×10^10Hz

Using doppler effect formula

F = C/ ( C - V) × f

Where

F = observed frequency

f = source frequency

C = speed of light = 3×10^8

V = speed of the car

Substitute all the parameters into the formula

2.100000103×10^10 = 3×10^8/(3×10^8 -V) × 2.1×10^10

2.100000103×10^10/2.1×10^10 = 3×108/(3×10^8 - V)

1.000000049 = 3×10^8/(3×10^8 - V)

Cross multiply

300000014.7 - 1.000000049V = 3×10^8

Collect the like terms

1.000000049V = 14.71429

Make V the subject of formula

V = 14.71429/1.000000049

V = 14.7 m/s

The speed of the car is 15 m/s approximately.

8 0

3 years ago

The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

6 0

3 years ago

Three small balls of the same size but different masses are hung side-by-side in parallel on the strings of same length. They to

andrey2020 [161]

Answer:

m1/6 ( c )

Explanation:

since all the balls starts having the same momentum after the two collisions we will apply the principal of conservation of energy

After first collision

m1v = m1v1 + m2v2 --- ( 1 )

After second collision

m2v2 = m2v2 + m3v3 ---- ( 2 )

combining equations 1 and 2

m1v = m1v1 + m2v2 + m3v3 ----- ( 3 )

All balls moving at the same momentum ( p ) = m1v1 = m2v2 = m3v3

note ; 3p = m1v ∴ m3 = $\frac{m1v}{3v3}$ ----- ( 4 )

applying conservation of energy

3v = v1 + v2 + v3 ------- ( 5 )

also 3m1v1 = m1v = v1 = v/3 =

v2 + v3 = 8/3 v ----- ( 6 )

next eliminate V3 for equation 6 by applying conservation of energy and momentum

m1 = 2m2 ------ ( 7 )

now using p1 = p2 = m1v1 = 1/2 m1v1 hence v2 = 2v1 where v1 = 1/3 v

hence ; v2 = 2/3 v ------- ( 8 )

solving with equation 6 and 8

v3 = 2v ------ ( 9 ) ∴ v/v3 = 1/2 ---- ( 10 )

solving with equation 9 and 10

m3 = m1/3 * 1/2 = m1/6

8 0

3 years ago

An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N, find the coefficient of kinetic

Stolb23 [73]

Answer:

The coefficient of kinetic friction = 0.026

Explanation:

An 56 kg sled is being pulled across the snow, at constant speed,by a horizontal force of 15 N.

Here we have to note that the weight is pulled at a constant speed . This means that the net force acting on the weight is zero.

The external force acting on the body is in the forward direction and the friction acts in the backward direction.

Friction increases as the mass of the body increases.

Friction = $u_{k}\times m \times g$

We now equate this to the external force of 15 N.

15 = $u_{k} \times 56 \times 10$

$u_{k}$ = $\frac{15}{560}$

$u_{k}$ = 0.026

The coefficient of kinetic friction = 0.026

8 0

3 years ago