Please help me with this! (:

Danny's mother believes her own fear of big dogs may have led to his phobia. This explanation is pointing to ______________.

matrenka [14]

<h2>Answer:</h2>

Cynophobia

<h3>Explanation:</h3>

Cynophobia originates from the Greek words that signify "dog" (cyno) and "fear" (phobis). An individual who has cynophobia encounters a dread of mutts that is both unreasonable and tenacious. It's something beyond feeling of scaredness whether a dog is barking or an individual is around dogs.

An individual who has cynophobia encounters a dread of dogs that is both silly and constant. Explicit fears, similar to cynophobia, influence somewhere in the range of 7 to 9 percent of the populace. They're regular enough that they're formally perceived in the Diagnostic and Statistical Manual of Mental Disorders,

6 0

2 years ago

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Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

kkurt [141]

Answer:

B. Steam burns the skin worse than hot water because the latent heat of vaporization is released as well.

Explanation:

It is given that both steam and the boiling water when in contact with the skin cools down from 100 to 34 degrees Celsius.

For any substance of mass m, the heat required to change the temperature by $\Delta T$ is $mC\Delta T$ (S.I. unit = Joules).

where C, the specific heat capacity is the same and a constant for both the condensed steam and the boiling water.

But, there is a "hidden" energy (heat) released by the steam called latent heat

(given by mL, L = specific latent heat) which allows the phase transition (gas to liquid). While both of them are at the same temperature, their energy (heat) is different, which is why steam causes burns worse than boiling water

3 0

2 years ago

A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force

Vlada [557]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec
vf = 0 m/sec 
Fμ = 7100 N (Force due to friction) 
Fg = 14700 N 
With the force due to gravity, you can find the mass of the car: 
F = ma 
14700 N = m (9.8 m/sec²) 
m = 1500 kg 
Now, we can use the equation again to find the deacceleration due to friction: 
F = ma 
7100 N = (1500 kg) a 
a = 4.73333333333 m/sec² 
And now, we can use a velocity formula to find the distance traveled: 
vf² = vo² + 2a∆d 
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d 
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d 
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d 
∆d = 66.0211267605634 m 
∆d = 66.02 m

7 0

3 years ago

At noon, ship A is 140 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast (in

Luba_88 [7]

Answer:

The rate of change of distance between the two ships is 18.63 km/h

Explanation:

Given;

distance between the two ships, d = 140 km

speed of ship A = 30 km/h

speed of ship B = 25 km/h

between noon (12 pm) to 4 pm = 4 hours

The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =

140 km - 120 km = 20 km

(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)

The displacement of ship B at 4pm = 25 km/h x 4h = 100 km

Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;

r² = a² + b²

r² = 20² + 100²

r = √10,400

r = 101.98 km

The rate of change of this distance is calculated as;

r² = a² + b²

r = 101.98 km, a = 20 km, b = 100 km

$2r(\frac{dr}{dt} ) = 2a(\frac{da}{dt} ) + 2b(\frac{db}{dt} )\\\\r(\frac{dr}{dt} ) = a(\frac{da}{dt} ) + b(\frac{db}{dt} )\\\\101.98(\frac{dr}{dt} ) = 20(-30 ) + 100(25 )\\\\101.98(\frac{dr}{dt} ) = -600 + 2,500\\\\101.98(\frac{dr}{dt} ) = 1900\\\\\frac{dr}{dt} = \frac{1900}{101.98} = 18.63 \ km/h$

5 0

2 years ago

Captain Hook is ghting Peter Pan, and they are about to step onto a tightrope strung horizontally between two maststhat are 16 m

andreyandreev [35.5K]

Complete Question

Captain Hook is fighting Peter Pan, and they are about to step onto a tightrope strung horizontally between two masts that are 16 m apart. When Pan and Hook are standing exactly halfway between the masts, the rope makes a 3 anglewith the horizontal. The rope has a diameter of 0.02 m and a Youngs modulus of 35 GPa.

What is the combined mass,in kilograms, of Peter Pan and Captain Hook?

Answer:

Their combined mass is $m= 161.2kg$