How are progression and variation similar?

A football is kicked at ground level with a speed of 15.0 m/s at an angle of 31.0° to the horizontal. How much later does it hit

svetoff [14.1K]

I would say two seconds but I'm not to sure if that is correct.

7 0

4 years ago

2

Arte-miy333 [17]

Answer:

the answer is C ----- it is the rate of change of velocity per unit time

Explanation:

acceleration

$= \frac{velocity(v)}{time(t)}$

3 0

3 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

7 0

3 years ago

The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant

KIM [24]

Answer: 50 km, 0, 27.78 m/s

Explanation:

Given

Circumference of the track is $12.5\ km$

Speed of car is $100\ km/h$

Car drives for $30\ \text{minute}\ or\ 0.5\ hr$

(a)Distance traveled is

$\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km$

(b)displacement of the car

It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.

Therefore, the displacement of the car is zero.

(c)To convert kmph to m/s, multiply the entity by $\frac{5}{18}$

$\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s$

5 0

3 years ago

How to classify an inner planet vs. outer planet

zvonat [6]

Outer planets are farther away and made up of gases. Inner planets closer. It's pretty much self explanatory. Hope this helps.

7 0

4 years ago