Which statement best describes the polarization of light

The four-firm concentration ratio for newstands is 17 and for electric lamp makers it is 89. The HHI for newstands is 105 and f

TEA [102]

Answer:

is; is not

Explanation:

The four-firm concentration ratio for newstands is 17 and for electric lamp makers it is 89. The HHI for newstands is 105 and for electric lamp makers it is 2 comma 850. The newstand market is an example of monopolistic competition. The electric lamp market is not an example of monopolistic competition

4 0

3 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

3 years ago

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

7 0

3 years ago

Can someone help me with this real quick?

ch4aika [34]

Answer:

7,546 J

Explanation:

recall that Potential energy is given by

P.E = mgΔh

where m = 70kg (given)

g = 9.8 m/s² (acceleration due to gravity)

Δh = change in height

= distance from top of building to top of car

= height of building - height of car

= (5+8) - 2

= 11m

substituting all these into the equation:

P.E = mgΔh

= 70 x 9.8 x 11

= 7,546 J

4 0

3 years ago

A net force of 6.8 N accelerates a 31 kg scooter across a level parking lot. What is the magnitude of the scooter’s acceleration

mrs_skeptik [129]

Answer:

a)0.22 m/s².

Explanation:

Given that

Net force ,F= 6.8 N

mass ,m = 31 kg

From the second law of Newton's

F = m a ---------------1

Where

F=Net force ,m=mass

a=Acceleration

Now by putting the values in the equation 1

F = m a

6.8 = 31 x a

$a=\dfrac{6.8}{31}\ m/s^2$

$a = 0.219\ m/s^2$

$a = 0.22\ m/s^2$

Therefore the acceleration of the scooter will be 0.22 m/s².

The answer will be "a".

a)0.22 m/s².

3 0

3 years ago