What nuclear weapon is the biggest and who currently owns this bomb

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

3 years ago

What do we call fixed point around which a lever pivots?

xxTIMURxx [149]

Answer is A) Fulcrum

The fixed point that a lever rotates around is called the fulcrum.

4 0

1 year ago

Read 2 more answers

Q3. You throw a ball into the air, it reaches a certain height and then comes back to you.

Grace [21]

The If a car is going round a curve , there is an acceleration because the direction of the velocity changes.

<h3>What is the direction of the velocity?</h3>

Now we know that if you throw the ball upwards, the motion is in opposite direction to gravity thus the ball is experiencing deceleration and the speed decreases. The velocity decreases and the acceleration is negative.

If the ball is coming down, then the ball is accelerated thus it speeds up and the direction of the acceleration is positive.

If a car is going round a curve, the vehicle is accelerating because the direction of the velocity changes even if its amount remains constant.

When a board is moving down a hill at 2 ms-1, it is experiencing an acceleration because the motion is in the same direction as gravity.

If a car is coming to a stop at a point, it experiences a deceleration and not an acceleration since the change of velocity with time is negative as the car comes to rest.

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

5 0

1 year ago

You have a pick-up truck that weighed 4,000 pounds when it was new. you are modifying it to increase its ground clearance. when

jok3333 [9.3K]

U have to *modify it to increase its ground clearance*

4 0

3 years ago

Read 2 more answers

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.

Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0

3 years ago