A cool down is considered the activity performed

Mark and Paul are in a race. Mark is 20 meters from the finish line and running at a constant 3.5 m/s. Paul is 5 meters behind h

USPshnik [31]

Answer:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

Explanation:

For this case we have an illustration for the problem on the figure attached.

And we can solve this problem analyzing each one of the runner. Let's begin with Mark

Mark

For this case we know that $V_M = 3.5 m/s$ and th velocity is constant. The distance from Mark and the finish line is $D_M = 20 m$

Since the velocity is constant we can create the following relation:

$D_M = V_M t_M$

And solving for $t_M$ we got:

$t_m = \frac{D_M}{V_M}= \frac{20m}{3.5 m/s}= 5.71 s$

So then Mark will nd the race after 5.71 seconds

Paul

We know that the initial velocity for Paul is given $V_{iP}= 2.7 m/s$ we also know that the total distance between Paul and the finish line is 25 m and we want to find the acceleration that Paul needs to apply in order to tie the race, and Paul have 5.71 sconds in order to reach the finish line.

We can use this formula in order to find the acceleration (because we assume that the acceleration is constant) that he needs to apply:

$x_f = x_i + v_i t + \frac{1}{2} a t^2$

And since $\Delta x = x_f - x_i$ we have this:

$\Delta x= v_i t + \frac{1}{2} a t^2$

And if we replace we have this:

$25 m= 2.7 m/s * (5.71 s)+ \frac{1}{2} a (5.71s)^2$

And solving for a we got:

$9.583 m = \frac{1}{2} a (5.71s)^2$

$a = 0.588 \frac{m}{s^2}$

And the final velocity for Paul using this acceleration would be:

$V_{fP}= V_{iP}+ a_P t = 2.7m/s + 0.588 m/s^2 (5.71s)= 6.057 m/s$

3 0

4 years ago

A 65-kg ice skater stands facing a wall with his arms bent and then pushes away from the wall by straightening his arms. At the

Marrrta [24]

Our values can be defined like this,

$m = 65kg$

$v = 3.5m / s$

$d = 0.55m$

The problem can be solved for part A, through the Work Theorem that says the following,

$W = \Delta KE$

Where

KE = Kinetic energy,

Given things like that and replacing we have that the work is given by

W = Fd

and kinetic energy by

$\frac {1} {2} mv ^ 2$

So,

$Fd = \frac {1} {2} m ^ 2$

Clearing F,

$F = \frac {mv ^ 2} {2d}$

Replacing the values

$F = \frac {(65) (3.5)} {2 * 0.55}$

$F = 723.9N$

B) The work done by the wall is zero since there was no displacement of the wall, that is d = 0.

6 0

3 years ago

As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.1 s. If the train's acceleration remains constant

vodomira [7]

Acceleration is the rate of change of the velocity of a moving object. To determine the speed the object has at a certain time, we need to determine the acceleration first by dividing the speed with the time given. We do as follows:

acceleration = 4.9 m/s / 5.1 s = 0.96 m/s^2

To determine the velocity after 7.1 s, we multiply 7.1 s to the acceleration.

speed or velocity = 0.96 x 7.1 = 6.52 m/s

3 0

4 years ago

Read 2 more answers

If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f

Novay_Z [31]

Answer:

E = ρ ( R1²) / 2 ∈o R

Explanation:

Given data

two cylinders are parallel

distance = d

radial distance = R

d < (R2−R1)

to find out

Express answer in terms of the variables ρE, R1, R2, R3, d, R, and constants

solution

we have two parallel cylinders

so area is 2 $\pi$ R × l

and we apply here gauss law that is

EA = Q(enclosed) / ∈o ......1

so first we find Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( $\pi$ R1² × l )

so put all value in equation 1

we get

EA = Q(enclosed) / ∈o

E(2 $\pi$ R × l) = ρ ( $\pi$ R1² × l ) / ∈o

so

E = ρ ( R1²) / 2 ∈o R

6 0

4 years ago

In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to

Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

$\beta = \frac{\lambda D}{d}$

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

4 0

3 years ago