Which of the following occurs with both a cold front and a mountain breeze?

Tema [17]

Answer:

B.

1-3hours

Explanation:

This is because a diabetic patients have increased blood sugar or glucose concentration. After eating a meal, the blood glucose concentration will be increased as it is been accumulated . Therefore it is best diabetic patient exercise at that hour to reduce it's blood glucose concentration.

6 0

3 years ago

A ball is thrown downward with an initial speed of 7 m / s. The ball's velocity after 3 seconds is m / s. (g = -9.8m / s²)

yan [13]

Answer:

-36.4 m/s

Explanation:

final velocity= initial velocity + acceleration x time

7 + (-9.8)(3)= -36.4 m/s

5 0

3 years ago

What do each of the variables mean ? F=__________ m=_________ a=___________

wariber [46]

Answer:

F = Force (Measured in Newtons, N), m = Mass (Measured in kilograms, kg), and a = acceleration (Measured in metres per second squared, $m/s^{2}$

Explanation:

This is Newton's Second Law!

Hope this helps!

PLS mark as brainliest, hope this helps!

8 0

3 years ago

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

Answer:

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

Explanation:

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago

A car travels north along a straight highway at an average speed of 85 km/h. After driving 2.0 km, the car passes a gas station

jenyasd209 [6]

Constant speed is the answer

5 0

3 years ago