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weqwewe [10]
2 years ago
9

Dave's sister gets very stressed out during final exam time; however, Dave isn't affected by the stress. What does this illustra

te?
A.
Fight or flight response

B.
Personal stress

C.
Relaxation response

D.
Subjective stress
Physics
1 answer:
dmitriy555 [2]2 years ago
6 0
The answer is b personal stress
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A 5-turn square loop (10 cm along aside, resistance = 4.0 ) is placed in a magnetic field that makes an angle of30o with the pla
andreev551 [17]

Answer:

Explanation:

Area A of the coil = .1 x .1 = .01 m²

no of turns n = 5

magnetic field B = .5 t²

Flux  Φ perpendicular to plane passing through it.= nBA sin30

rate of change of flux

dΦ/dt = nAdBsin30 / dt

= nA d/dt (.5t²x .5 )

= nA x 2 x .25 x t

At t = 4s

dΦ/dt = nA x 2

= 5x .01 x 2

= .1

current = induced emf / resistance

= .1 / 4

= .025 A

= 25 mA.

4 0
3 years ago
The more active a cell is, the more of these structures it has.
Aleks [24]

Answer:pretty sure its mitochondria

Explanation:

6 0
3 years ago
A car tire is filled with air at pressure 325000 pa at 303 K. When the tire cools,its pressure reduces to 299000 Pa. What is the
max2010maxim [7]

Answer:

new temperature of the tire will be 278.76 K

Explanation:

when the temperature increases, the particles will have greater kinetic energy and also the pressure will be increase for the gas particles.

so when the temperature increases, pressure  will also increase  and vice versa

 T is directly proportional to P

T1  = initial temperature= 303 k

P1= Initial pressure =  325000 pa

T2= Final temperature=  ?

P2= Final pressure =   299000 pa

mathematically

P1/T1= P2/T2

T2= P2 x T1/P1

T2 = 299000 x 303/ 325000=  278.76 k

6 0
3 years ago
Read 2 more answers
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe
spin [16.1K]

Answer:

1.19 m/s²

Explanation:

The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so

f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀    (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

g = 2π√(71.753/200 Hz)/3.16

g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

6 0
3 years ago
The vacuum inside the thermos bottle shown in Figure 16-1 stops which type of thermal energy transfer to keep the liquid hot
BARSIC [14]
The answer is D- both a and b
7 0
3 years ago
Read 2 more answers
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