Ask question

Ask question

All categories

3 years ago

14

What is the final velocity of a runner who is accelerators at 2 feet per second squared for 3 seconds with an initial velocity o

f 4 feet per secound

1 answer:

Kryger [21]3 years ago

6 0

The runner's final velocity is 10 m/s

You might be interested in

What is the degree Celsius temperature of a person with 95 degree Fahrenheit fever

My name is Ann [436]

Answer:

35° Celsius

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperatu

PSYCHO15rus [73]

Answer:

a) $I(95,50) = 73.19$ degrees

b) $I_{T}(95,50) = -7.73$

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

$I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}$

a) Calculate I at (T ,H) = (95, 50).

$I(95,50) = 45.33 + 0.6845*(95) + 5.758*(50) - 0.00365*(95)^{2} - 0.1565*95*50 + 0.001*50*95^{2} = 73.19$ degrees

(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is $I_{T}(T,H)$ . So

$I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}$

$I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H$

$I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73$

8 0

3 years ago

The base of a triangle is seven less than twice its height. If the area of the triangle is 102 cm2, find the length of the base.

Shalnov [3]

Answer:

195

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

!!please answer quickly and correctly!!

Valentin [98]

Large=40 pounds

small=30 pounds

and this is the solution

3 0

2 years ago

What is the distance between (2,-1) and (2,5) rounded to the nearest 10th

xxMikexx [17]

Answer:

<h3> 5.0</h3>

Step-by-step explanation:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\d=\sqrt{(2-2)^2+(5+1)^2}=\sqrt{0^2+5^2}=\sqrt{25}=5$

5 0

3 years ago