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drek231 [11]
3 years ago
13

A car travels at a constant velocity of 70 mph for one hour. At the second hour, the car’s velocity was 60 mph. At the third hou

r, the car’s velocity was 80 mph. At what hour was their positive acceleration? At what hour was their negative acceleration?
Physics
2 answers:
katovenus [111]3 years ago
8 0

When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.

During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.

Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.


kati45 [8]3 years ago
5 0

<u>Answer:</u>

  Positive acceleration is in third hour and negative acceleration is in second hour.

<u>Explanation:</u>

  Velocity of car in first hour =  70 mph

  Velocity of car in second hour = 60 mph

  Velocity of car in third hour = 80 mph

   Acceleration = Change in velocity / Time

   Acceleration in second hour = (60 - 70)/1 = -10 mph²

   Acceleration in third hour = (80 - 60)/1 = 20 mph²

   So positive acceleration is in third hour and negative acceleration is in second hour.

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4 0
2 years ago
What is a point of view of an object used to determine another object's motion
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7 0
3 years ago
A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 10 degrees to the right of her orig
atroni [7]

Answer:

The plane is 2353.7 mi from the starting position.

Explanation:

Please, see the attached figure for a graphic representation of the problem.

We have 2 displacement vectors "a" and "b" and a vector "c" that is the sum of vectors "a" plus "b" (c = a + b). The module of "c" will be the distance of the plane from the starting point.

vector a = (xa, ya)

vector b = (xb, yb)

where “xa” and “xb” are the horizontal components of the vectors “a” and “b” respectively and “ya” and “yb” are the vertical components of each vector.

Then, the vector c = a + b will be:

c = (xa + xb, ya + yb)

The module of a vector is calculated using the following expression for a vector “v”:

module of v = \sqrt{x^{2} + y^{2} }

Then, the module of c will be:

module of c = \sqrt{(xa + xb)^{2} + (ya + yb)^{2}} = distance from starting point

Then, we have to find the components of vectors “a” and “b”

The distance traveled during the first 1.5 hours of the trip is the module of the vector “a”. Then:

module of a = \sqrt{xa^{2} + ya^{2} } = distance traveled during the first 1.5 hours.

The distance can be calculated using the equation of the position of an object moving in a straight line at constant speed:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = speed

t = time

Considering x0 as the starting point (x0 = 0)

x = 675 mi/h * 1.5 h = 1012.5 mi

Then:

module of a = \sqrt{xa^{2} + ya^{2} } = 1012. 5 mi

Since the plane moves only on the horizontal (see figure), the "y" component of the vector, "ya", will be 0.

Then:

(1012.5 mi)² = xa²

xa = 1012. 5mi

a = (1012.5 mi, 0)

In the same way, we have fo find the components of the vector “b”. The module of “b” will be the distance traveled during this part of the flight:

module of b = \sqrt{xb^{2} + yb^{2} } = x = x0 + v * t

Considering x0 as the point at which the plane turns (x0 = 0)

x = 675 mi / h * 2 h = 1350 mi

Using trigonometry, we can calculate xb and yb (see figure):

sin angle = opposite / hypotenuse

cos angle = adjacent / hypotenuse

In this case:

opposite = yb

adjacent = xb

hypotenuse = module of “b”

Then:

sin 10° = yb / module of “b”

sin 10° * module of “b” = yb

In the same way:

cos 10° * module of “b” = xb

Since module of “b” = 1350 mi

xb = 1329.5 mi

yb = 234.4 mi

b = (1329.5 mi, 234.4 mi)

The vector c = a+b can now be calculated:

c = (xa + xb, ya + yb)

c =(1012.5 mi + 1329.5 mi, 0 mi + 234.4 mi) = (2342 mi, 234.4 mi)

The module of c will be:

module of c = \sqrt{(2342 mi)^{2} + (234.4 mi)^{2} } = 2353.7 mi

The plane is 2353.7 mi from the starting position.

4 0
3 years ago
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yulyashka [42]

Answer:

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Explanation:

We have given that Buffalo has a latitude of 42.9°N

And Raleigh has a latitude of 35.8°N

Radius of the earth = 3960 miles

We have to calculate the distance between given two cities

Difference in their latitudes = 42.9-35.8=7.1^{\circ}

Now changing the angle in radian = 7.1\times\frac{\pi }{180}=0.12385radian

So distance will be =3960\times 0.12385=490.468miles

5 0
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Answer:  The younger elliptical and lenticular galaxies had results similar to spiral galaxies like the Milky Way. The researchers found that the older galaxies have a larger fraction of low-mass stars than their younger counterparts.

Explanation:

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