A team of scientists is studying a volcanic hot spring. Which task would the physical scientist most likely do?

Part C: Quantitative Problems when vf is not 0

Alina [70]

Answer:

(a)

$\triangle v=-8\ m/s\\\triangle mv=-56\ kg.m/s$

(b)

1120 N

Explanation:

Change in velocity, $\triangle v$ is given by subtracting the initial velocity from the final velocity and expressed as $\triangle v= v_f -v_i$

Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

$\triangle v=0-8=-8\ m/s$

To find $m\triangle v$ then we substitute 7 kg for m and -8 m/s for $\triangle v$ therefore $\triangle\ v=7 Kg\times -8 m/s=-56\ Kg.m/s$

(b)

The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

$a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}$

Substituting t with 0.05 s then $a=\frac {\triangle v}{t}=\frac { v_f -v_i}{t}=\frac {-8}{0.05}=-160 m/s^{2}$

Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N

Therefore, the impact force is equivalent to 1120 N

3 0

4 years ago

if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were discovered?

Anna11 [10]

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

The surprising thing was that quasars and active galaxies have a lot of mass in the center of the very small volume of the space.

Therefore, the surprising thing about quasars was that due to this mass and energy they are 100 times more luminous than Milky Way which means they have high recession velocity and a very large amount of red-shifting.

To learn more about quasars, refer: brainly.com/question/9965257

#SPJ4

8 0

1 year ago

Hello plz help meeeee plz I need helppp if u know it plz help

tamaranim1 [39]

The person does not sink into the snow because the force acts on a larger area so that the pressure is less
The edge of the sharp knife has a smaller area so the force acting on the knife produces a larger pressure

6 0

3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have: