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charle [14.2K]
2 years ago
15

Ken received a 66 on his first math exam, which counted for 20% of his final grade; he now believes that he won't be able to pas

s the class. His conclusion best illustrates a pessimistic outlook. the fight-or-flight response. problem-focused coping. relative deprivation.
Physics
1 answer:
pogonyaev2 years ago
6 0

Answer:

His conclusion best illustrates a pessimistic outlook.

Explanation:

As seen in the question above, Ken got 20% of his final grade in the first test he did for this class, that is, there will be other tests that can provide him to reach the grade needed to pass the class. However, even if there are possibilities, he believes that he will not pass the class, he does not have a positive and optimistic view of his future in this class and is sure that he will fail. This negative view of the future is an example of a pessimistic outlook.

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The curved section of a horizontal highway is a circular unbanked arc of radius 740m. If the coefficient of static friction betw
fiasKO [112]
Coefficient of static friction = tan(a) = 0.4
r = 740 m
g = 9.8 m/s²
v \:  =  \:  \sqrt{gr \tan( \alpha ) }
v = √(9.8 × 740 × 0.4) m/s
v ≈ 53.85908 m/s
6 0
2 years ago
Read 2 more answers
A monoatomic ideal gas undergoes an isothermal expansion at 300 K, as the volume increased from 0.010 m^3 to 0.040 m^3. The fina
Natali [406]

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

\Delta U = \frac{3}{2}nR\Delta T

now for isothermal process since temperature change is zero

so change in internal energy must be ZERO

4 0
2 years ago
In coordinates with the origin at the barn door, the cow walks from x 0 to x 6.9 m as you apply a force with x component Fx 320.
Stella [2.4K]

Answer:

-209.42J

Explanation:

Here is the complete question.

A balky cow is leaving the barn as you try harder and harder to push her back in. In coordinates with the origin at the barn door, the cow walks from x = 0 to x = 6.9 m as you apply a force with x-component Fx=−[20.0N+(3.0N/m)x]. How much work does the force you apply do on the cow during this displacement?

Solution

The work done by a force W = ∫Fdx since our force is variable.

Since the cow moves from x₁ = 0 m to x₂ = 6.9 m and F = Fx =−[20.0N+(3.0N/m)x] the force applied on the cow.

So, the workdone by the force on the cow is  

W = ∫₀⁶°⁹Fx dx = ∫₀⁶°⁹−[20.0N+(3.0N/m)x] dx

= ∫₀⁶°⁹−[20.0Ndx - ∫₀⁶°⁹(3.0N/m)x] dx

= −[20.0x]₀⁶°⁹ - [3.0x²/2]₀⁶°⁹

= -[20 × 6.9 - 20 × 0] - [3.0 × 6.9²/2 - 3.0 × 0²/2]

= -[138 - 0] - [71.415 - 0] J = (-138 - 71.415) J

= -209.415 J ≅ -209.42J

5 0
2 years ago
Based on the emf value measured at frame 700, what is the average magnitude of the magnetic field inside the magnet assembly? No
Nadusha1986 [10]

Answer:

The average magnitude of magnetic field B= 0.0433/ d Tesla

(You have not provided length of side of loop, so if you divide this value by length you will get value of magnetic field.)

Explanation:

Induced emf

where B= magnetic field  

d= breadth of rectangular piece

V= velocity with which the rectangular piece = o.o6m/s

n= no of turns  = 10

EMF = 26mV

since d (breadth of the frame) is not given, I will use it as a variable

EMF= n×B×d×V ------------------(1) (EMF induced due to multiple turns)

From eq 1, we get

B= (EMF)/(n d V)

B= (26 X 0.001) / (10 d 0.06)

B= 0.0433/ d Tesla

4 0
2 years ago
f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance
Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

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so

stopping distance increase = (1.6)²

\frac{D40}{D25} = (1.6)²

\frac{D40}{D25} = 2.56

so here

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

5 0
2 years ago
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