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Reil [10]
3 years ago
14

Jenise is buying a car for $13,085. She wants $9,400 in upgrades to the car.

Physics
2 answers:
VARVARA [1.3K]3 years ago
5 0
I think it will be 405
inysia [295]3 years ago
3 0

Answer:

$55.6

Explanation:

sorry sorry sorry sorry sorry

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A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t
marishachu [46]

Answer:

The impact occured at a distance of 2478.585 meters from the person.

Explanation:

(After some research on web, we conclude that problem is not incomplete) The element "Part A" may lead to the false idea that question is incomplete. Correct form is presented below:

<em>A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 6.4 seconds apart. How far away did the impact occur? (Sound speed in the air: 343 meters per second, sound speed in concrete: 3000 meters per second)</em>

Sound is a manifestation of mechanical waves, which needs a medium to propagate themselves. Depending on the material, sound will take more or less time to travel a given distance. From statement, we know this time difference between air and concrete (\Delta t), in seconds:

\Delta t = t_{A}-t_{C} (1)

Where:

t_{C} - Time spent by the sound in concrete, in seconds.

t_{A} - Time spent by the sound in the air, in seconds.

By suposing that sound travels the same distance and at constant speed in both materials, we have the following expression:

\Delta t = \frac{x}{v_{A}}-\frac{x}{v_{C}}

\Delta t = x\cdot \left(\frac{1}{v_{A}}-\frac{1}{v_{C}}  \right)

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  } (2)

Where:

v_{C} - Speed of the sound in concrete, in meters per second.

v_{A} - Speed of the sound in the air, in meters per second.

x - Distance traveled by the sound, in meters.

If we know that \Delta t = 6.4\,s, v_{C} = 3000\,\frac{m}{s} and v_{A} = 343\,\frac{m}{s}, then the distance travelled by the sound is:

x = \frac{\Delta t}{\frac{1}{v_{A}}-\frac{1}{v_{C}}  }

x = 2478.585\,m

The impact occured at a distance of 2478.585 meters from the person.

7 0
3 years ago
What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?
Lady bird [3.3K]

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

4 0
3 years ago
A force of 6N and another of 8N can be applied together to produce the effect of how much newtons?
const2013 [10]
12N because you are just adding those two up on the same side
5 0
3 years ago
Read 2 more answers
How many doctors are ther in Edmonton​
taurus [48]

Answer:

There are nine PCNs in the Edmonton area. They work alongside more than 1,100 family doctors in over 330 clinics to provide care for 1.2 million patients. PCN teams include more than 370 nurses, mental health clinicians and other health professionals.

Explanation:

3 0
3 years ago
A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N
crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

Unbalanced Force = Horizontal Component Applied Force - Frictional Force

Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

f = frictional force  = ?

F = Applied force  = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Therefore,

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

<u>f = 485.62 N</u>

6 0
3 years ago
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