All categories

2 years ago

Which of the following is a unit of speed?

Physics

2 answers:

exis [7]2 years ago

8 0

Answer:

D, km/s^2

Explanation:

This is a unit of speed because it is saying kilometers over seconds squared, and that is a unit of speed similar to miles per hour or mph or m/h.

dangina [55]2 years ago

7 0

Answer:

Option c that it means cm/year.

Explanation:

Nm means nanometer, 1x10^9m

Km/s^2 is acceleration

h.. it may be height

You might be interested in

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball m

ikadub [295]

(a) $-39.4^{\circ}$

Let's take the initial direction (before the collision) of the cue ball has positive x-direction.

Along the y-direction, the total initial momentum is zero:

$p_y =0$

Therefore, since the total momentum must be conserved, it must be zero also after the collision. So we write:

$0 = m v_1 sin \phi_1 + m v_2 sin \phi_2 \\0 = m(4.60) sin (28^{\circ}) + m(3.40) sin \phi_2$

where

m is the mass of each ball

$v_1= 4.60 m/s$ is the velocity of the cue ball after the collision

$v_2 = 3.40 m/s$ is the velocity of the second ball after the collision

$\phi_1=28.0^{\circ}$ is the angle of the cue ball with the x-axis

$\phi_2$ is the angle of the second ball

Solving for $\phi_2$ , we find the angle between the direction of motion of the second ball and the original direction of motion:

$sin \phi_2 = -\frac{4.60 sin 28}{3.40}=-0.635\\\phi_2 = -39.4^{\circ}$

(b) 6.69 m/s

To find the original speed of the cue ball, we analyze the situation along the horizontal direction.

First, we calculate the total momentum along the x-direction after the collision, which is:

$p_x = m v_1 cos \phi_1 + m v_2 cos \phi_2 \\0 = m(4.60) cos (28^{\circ}) + m(3.40) cos (-39.4^{\circ})=6.69 m$

The initial total momentum along the x-direction as

$p_x = m u$

where

m is the mass of the cue ball

$u$ is the initial velocity of the cue ball

The momentum along this direction must be conserved, so we can equate the two expressions and find the value of u:

$mu = 6.69 m\\u = 6.69 m/s$

7 0

3 years ago

A proton in a particle accelerator is traveling at a speed of 0.99c.(a) If you use the approximate nonrelativistic equation for

____ [38]

Answer:

a) p = 4.96 10⁻¹⁹ kg m / s , b) p = 35 .18 10⁻¹⁹ kg m / s ,

c) p_correst / p_approximate = 7.09

Explanation:

a) The moment is defined in classical mechanics as

p = m v

Let's calculate its value

p = 1.67 10⁻²⁷ 0.99 3. 10⁸

p = 4.96 10⁻¹⁹ kg m / s

b) in special relativity the moment is defined as

p = m v / √(1 –v² / c²)

Let's calculate

p = 1.67 10⁻²⁷ 0.99 10⁸/ √(1- 0.99²)

p = 4.96 10⁻¹⁹ / 0.141

p = 35 .18 10⁻¹⁹ kg m / s

c) the relationship between the two values is

p_correst / p_approximate = 35.18 / 4.96

p_correst / p_approximate = 7.09

4 0

3 years ago

Krishne wants to measure the mass and volume of a key. Which tools should she use? a balance and a beaker of water a balance and

Ratling [72]

Balance and beaker of water. The balance will measure mass and the beaker will measure the volume when you take the initial volume without the key subtracting that value from the value with the key in the beaker

8 0

2 years ago

Read 2 more answers

Through which material does the light travel the fastest?

eimsori [14]

The answer is A because it gives you a straight line which would make it easy for you go just go to end fast as a car .

6 0

2 years ago

calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame

kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

$\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2$

0 - p × Area for trachea =

$\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}$

$-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)$

⇒ $-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}$

$\frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s$

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0

2 years ago

Read 2 more answers