Newton is derived unit why<br>

How is the control group and experimental group different

zzz [600]

The control group is the independent variable and the experimental group is the dependent due to change during the experiment. The experimental group will usually rely on another variable in the experiment for change.

4 0

4 years ago

A snail travels 300 cm in 4 minutes.calculate speed of snail in m/s

LekaFEV [45]

Answer:

$\frac{0.0125m}{s}$

Explanation:

In order to solve this question we need to know that $speed = \frac{distance}{time}$ . Then we need to convert 4 minutes into seconds and cm into m. We can do that by multiplying the number of minutes by 60 (because there is 60 seconds in one minute) and dividing the number of cm by 100 (because there is 100 cm in one m). So.......

4min = 4 x 60s = 240s

300cm = 300/100 m = 3m

Now we know that distance = 300m, and that the time = 4min = 240s ⇒

⇒ $speed=\frac{distance}{time} = \frac{3m}{240s} = \frac{0.0125m}{s}$

5 0

3 years ago

Given the force diagram of the basketball, what can you tell about the basketball motion?

VARVARA [1.3K]

Newton's laws allow to find the result for the movement of the basketballl:

On the vertical axis the ball is on the ground.

On the horizontal axis the ball is accelerating in the direction of the pushing force.

Newton's laws establish the relationship between the forces on objects:

The 1st law states that if the net force is zero the object is stationary or with constant speed.

The 2nd law gives a relation of the force with the mass and the acceleration of the body.

The 3rd. Law states that the force appears in pairs, one on each body with the same magnitude, but in the opposite direction.

Let's apply these principles to the ball's motion diagram.

The two vertical forces are in the opposite direction, one is due to the weight of the body and the other is the attraction of the earth to the support of the ball, they are of equal magnitude, not their action-reaction force and reluctant because it is applied to the same body

In conclusion we can say that the ball is on the ground.

The two horizontal forces are in the opposite direction, the thrust force is greater than the friction therefore using Newton's second law the ball must be accelerating in the direction of the thrust force.

In conclusion we can say that the ball is accelerating in the direction of the pushing force.

In conclusion using Newton's laws we can find the result for the motion of the basketball:

On the vertical axis the ball is on the ground.

On the horizontal axis the ball is accelerating in the direction of the pushing force.

Learn more about Newton's laws here: brainly.com/question/3715235

8 0

3 years ago

Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.

faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = $\sqrt{1037 ft^2}$ = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

$\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt} + \frac{dy^{2} }{dt}$

→ $2s\frac{ds }{dt} = 2x\frac{dx}{dt} + 2y\frac{dy }{dt}$ which gives

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

$\frac{dy }{dt}$ = 12 ft/sec and

$\frac{dx}{dt}$ = 8 ft/sec

$s\frac{ds }{dt} = x\frac{dx}{dt} + y\frac{dy }{dt}$ becomes

$32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec} + 26ft\times 12\frac{ft}{sec}$ = 464 ft²/sec

$\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.}$ = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

3 0

3 years ago

The day VFR visibility and cloud clearance requirements to operate over the town of Cooperstown, after departing and climbing ou

pshichka [43]

Answer:

The correct option is;

C. 1 mile clear of clouds

Explanation:

Given that the indicated airspace location is at or below 700 feet AGL therefore, it is taken as being in the region of a class G airspace which covers the airspace regions from the base up to and equal to 1,200 feet beneath the class E airspace and the requirement for VFR flight for class G are 1 mile and clear of clouds.

4 0

3 years ago

Newton is derived unit why​

Newton is derived unit why