Answer:
The minimum distance in which the car will stop is
x=167.38m
Explanation:
∑F=m*a
∑F=u*m*g
The force of friction is the same value but in different direction of the force moving the car so it can stop so
<span>The magnetic field does not continually spread outward from the wire.</span>
Answer:
1.45 K
I had the same question and i got it right.
Answer:
a) P = 1240 lb/ft^2
b) P = 1040 lb/ft^2
c) P = 1270 lb/ft^2
Explanation:
Given:
- P_a = 2216.2 lb/ft^2
- β = 0.00357 R/ft
- g = 32.174 ft/s^2
- T_a = 518.7 R
- R = 1716 ft-lb / slug-R
- γ = 0.07647 lb/ft^3
- h = 14,110 ft
Find:
(a) Determine the pressure at this elevation using the standard atmosphere equation.
(b) Determine the pressure assuming the air has a constant specific weight of 0.07647 lb/ft3.
(c) Determine the pressure if the air is assumed to have a constant temperature of 59 oF.
Solution:
- The standard atmospheric equation is expressed as:
P = P_a* ( 1 - βh/T_a)^(g / R*β)
(g / R*β) = 32.174 / 1716*0.0035 = 5.252
P = 2116.2*(1 - 0.0035*14,110/518.7)^5.252
P = 1240 lb/ft^2
- The air density method which is expressed as:
P = P_a - γ*h
P = 2116.2 - 0.07647*14,110
P = 1040 lb/ft^2
- Using constant temperature ideal gas approximation:
P = P_a* e^ ( -g*h / R*T_a )
P = 2116.2* e^ ( -32.174*14110 / 1716*518.7 )
P = 1270 lb/ft^2
Options:
universal appeal
flattery
association
bandwagon
Answer: Association
Explanation: Advertisement is a marketing technique through which business organisations utilize the opportunities provided by both the print, electronic and other channels of communication to market a product to the target audience.
Association Advertising is a type of Advertising where certain attributes which have been known to be associated with good and quality products are used to market the products to the target audience.
