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3 years ago

Sound is faster than light that is why we hear thunder before we see lightning

Physics

1 answer:

Umnica [9.8K]3 years ago

8 0

Answer:

false

Explanation:

sound travels slower than light. that is why we see lightning before we hear the thunder

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The bohr shift on the oxygen-hemoglobin dissociation curve is produced by changes in

Alisiya [41]

Answer:

THE BOHR SHIFT ON THE OXYGEN-HEMOGLOBIN DISSOCIATION CURVE IS PRODUCED BY CHANGES IN THE CONCENTRATION OF CARBON IV OXIDE.

Explanation:

The oxygen-hemoglobin dissociation curve shows the relationship between the saturated hemoglobin concentration and oxygen. It shows how the blood hold on to and releases oxygen. The Bohr shift can occur as a result of changes in concentration of carbon iv oxide and other factors such as acidity or pH, 2,3-bisphosphoglycerate, exercise, also temperature of the body. These factors contributes to the right or left shift on the curve. Carbon iv oxide prevents the binding of oxygen to the hemoglobin. The is because hemoglobin has the same binding site for both oxygen and carbon iv oxide. Carbon iv oxide increase also leads to a change in the pH of the blood through the formation of bicarbonate ion. Bicarbonate ion formation causes reduced acidity and therefore lead a shift in the dissociation curve for more of the carbon iv oxide to be excreted as hemoglobin's affinity for oxygen reduces. And when the concentration of carbon iv oxide is low in the plasma, acidity increases and this provides more affinity for oxygen by the hemoglobin.

7 0

3 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago

A 58 kg skier is going down a 35 degree slope. The areaof each

maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

$N = mgcos\theta$

Since the skier is standing on two skis, his weight will be divide by two

$N' = \frac{mgcos\theta}{2}$

Pressure is given as the force applied in a given area, that is

$P = \frac{F}{A}$

Replacing F with N'

$P = \frac{N'}{A}$

$P = \frac{\frac{mgcos\theta}{2}}{A}$

Our values are given as,

$m = 58kg$

$g = 9.8m/s^2$

$\theta = 35\°$

$A = 0.3m^2$

Replacing we have that

$P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}$

$P = 776.01Pa$

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0

3 years ago

If the escalator pulls you up a slope of 30.0o and moves with a velocity of 3.10m/s. What is the vertical component of your v

ioda

Answer:

$1.55\ \text{m/s}$

$2.68\ \text{m/s}$

Explanation:

v = Velocity of the elevator = 3.1 m/s

$\theta$ = Angle of the slope = $30^{\circ}$

Vertical component is given by

$v_y=v\sin\theta\\\Rightarrow v_y=3.1\sin30^{\circ}\\\Rightarrow v_y=1.55\ \text{m/s}$

The vertical component of the velocity is $1.55\ \text{m/s}$ .

Horizontal component is given by

$v_x=v\cos\theta\\\Rightarrow v_x=3.1\times \cos30^{\circ}\\\Rightarrow v_x=2.68\ \text{m/s}$

The horizontal component of the velocity is $2.68\ \text{m/s}$ .

8 0

3 years ago

Why can't we implement the center tapped full wave rectifier without center-tapped transformer

maksim [4K]

For a full wave bridge you don't want a center tap

6 0

3 years ago