Can you guys please help me on this one?

Why the unit of power is called derived unit?

Lilit [14]

A :-) it was given the name Newton (N). from this, the derived unit of energy (or work) is defined ,as the work produced when the unit of force causes a displacement equal to the unit of length of its point of application along its direction . It was given the name Joule (J).

7 0

3 years ago

Find the radius of the circle formed by the intersection of a sphere of diameter 26 units and a plane that is 5 units away from

horrorfan [7]

Answer:

12 units

Explanation:

This problem can be solved if we take into account the equation for a sphere

$x^{2}+y^{2}+z^{2}=r^{2}\\x^{2}+y^{2}+z^{2}=(\frac{26}{2})^{2}=13^{2}$

where we took that the radius is 13 units. If we take z=5 and we replace this value in the equation of the sphere we have

$x^{2}+y^{2}+(5)^{5}=13^{2}\\x^{2}+y^{2}=144=(12)^{2}$

where we have taken x2 +y2 because if the equation of a circunference.

In this case the intersection is made when we take z=5, for this value the sphere and the plane coincides in values.

Hence, the radius is 12 units

I hope this is useful for you

regards

8 0

3 years ago

NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU

daser333 [38]

Acceleration = (change in speed)/(time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (10 m/s) - (20 m/s) = -10 m/s

Time for the change = 5.00 seconds

Acceleration = (-10 m/s) / (5 sec)

Acceleration = -2 m/s²

That's choice-A .

8 0

3 years ago

Read 2 more answers

(b) Fig. 1.1 shows two airports A and C. north SE с sea land क WE not to scale Fig. 1.1 An aircraft flies due north from A for a

GalinKa [24]

The addition of vectors and the uniform motion allows to find the answers for the questions about distance and time are:

The distance to go between airports A and C is 373.6 10³ m

The time to go from airport A to B is 2117 s

Vectors are quantities that have modulus and direction, so their addition must be done using vector algebra.

In this case the plane flies towards the North a distance of y = 360 10³ m at an average speed of v = 170 m / s, when arriving at airport B it turns towards the East and travels from x = 100 10³ m, until' it the distance reaches the airport C

Let's use the Pythagoras theorem to find the distance traveled

R = Ra x² + y²

R = 10³

R = 373.6 10³ m

They indicate the average speed for which we can use the uniform motion ratio

v = $\frac{\Delta y }{t}$

t = $\frac{\Delta y}{v}$

They ask for the time in in from airport A to B, we calculate

t = 360 10 ^ 3/170

t = 2.117 10³ s

In conclusion we use the addition of vectors the uniform motion we can find the answer for the question of distance and time are:

The distance to go between airports A and C B is 373.6 10³ m

The time to go from airport A to B is 2117 s

Learn more here: brainly.com/question/15074838

5 0

3 years ago

A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a

Elena L [17]

Answer:

Impulse = 88 kg m/s

Mass = 8.8 kg

Explanation:

We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec.

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

$F=ma$ Eqn. (1)

where

$F$ : is the Net Force in Newtons ( $N$ )

$m$ : is the mass ( $kg$ )

$a$ : is the acceleration ( $m/s^2$ )

We also know that the acceleration is denoted by the velocity ( $v$ ) of an object as a function of time ( $t$ ) with

$a=\frac{v}{t}$ Eqn. (2)

Now substituting Eqn. (2) into Eqn. (1) we have

$F=m\frac{v}{t}\\ \\Ft=mv$ Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. $Ft$ is in fact the Impulse $J$ of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ $22N$ and $t=4 sec$ (thus just before the cut-off time of the force acting).