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Alex17521 [72]
2 years ago
12

A car travels straight for 20 miles on a road that is 30° north of east. What is the east component of the car’s displacement to

the nearest tenth of a mile?
A.) -17.3 miles
B.) -10.0 miles
C.) 10.0 miles
D.) 17.3 miles

Physics
2 answers:
Virty [35]2 years ago
4 0
17.3 would be the correct answer.

Oduvanchick [21]2 years ago
4 0

Answer: The correct answer is option(D).

Explanation:

Distance travel by car in north of east = 20 miles

In a triangle ABC

Cos\theta =\frac{BC}{AC}

Cos 30^o=\frac{\sqrt{3}}{2}=\frac{BC}{20}

BC=17.32

The east component of the car’s displacement to the nearest tenth of a mile 17.32 miles.

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A longitudinal wave is observed to be moving along a slinky. Adjacent crests are 2.4 m apart. Exactly 6 crests are observed to m
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5

Explanation:

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1 year ago
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
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Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

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So,

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Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0
1 year ago
A certain signal molecule S in heart tissue is degraded by two different biochemical pathways: when only Path 1 is active, the h
Misha Larkins [42]

Answer:

Half life of S = 3.76secs

Explanation:

The concept of half life in radioactivity is applied. Half life is the time taken for a radioactive material to decay to half of its initial size.

For part 1 - How much signal will be degraded in 1secs = 1/3.9 = 0.2564

for part 2 - How much signal will be degraded in 1secs = 1/104 = 0.009615

Simply say = 1/3.9 + 1/104 = 0.266015

So both part 1 and part 2 took 1/0.266015 = 3.76secs is the half life of S when both pathways are active

6 0
2 years ago
A student lifts a set of books off a table and places them in the upper shelf of a book case which is 2 meters above the table.
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The work done is the product between the intensity of the force applied F, the amount of the displacement d of the book and the cosine of the angle \theta between the direction of the force and the direction of the displacement:
W=Fd \cos \theta
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6 0
2 years ago
Read 2 more answers
An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th
pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

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Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

5 0
2 years ago
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