A car traveling at 35 m/s starts to decelerate steadily. It comes to a complete stop in 5 seconds. What is its acceleration.

Shawn and his bike have a total mass of

Trava [24]

Answer:

Shawn's kinetic energy is 61.45 J.

Explanation:

Given;

Mass of the system is, $m=52.3\ kg$

Displacement of the bike is, $d=1.6\ km=1.6\times 1000=1600\ m$

Time taken is, $t=17.4\ min=17.4\times 60=1044\ s$

Let the constant velocity be $v$ .

For constant velocity, magnitude of velocity is given as distance by time.

Therefore, $v=\frac{d}{t}=\frac{1600}{1044}=1.533\ m/s$

Now, kinetic energy of a body is given as:

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

Here, $m=52.3\ kg,\ v=1.533\ m/s$

$\therefore KE=\frac{1}{2}\times 52.3\times (1.533)^2\\KE=0.5\times 52.3\times 2.35\\KE=61.45\ J$

Therefore, Shawn's kinetic energy is 61.45 J.

3 0

3 years ago

In a bi-prism experiment the eye-piece was placed at a distance 1.5m from the source. The distance between the virtual sources w

Vladimir [108]

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

6 0

3 years ago

On a velocity vs. time graph, what does it mean when the line crosses over the x-axis?

Phantasy [73]

On a velocity - time graph, if the line crosses the x - axis it depicts that the object has started moving in the opposite direction.

8 0

2 years ago

Orange Juice mass = 18 G Volume = ?? Density 6 g/ ML <br><br>What is the Volume?

ohaa [14]

Answer is 3. Volume= mass divided by density

4 0

3 years ago

Read 2 more answers

You toss a ball straight up with an initial speed of 30m/s. How high does it go, and how long is it in the air (neglecting air r

Brut [27]

Explanation:

Given that,

A ball is tossed straight up with an initial speed of 30 m/s

We need to find the height it will go and the time it takes in the air.

At its maximum height, its final speed, v = 0 and it will move under the action of gravity. Using equation of motion :

v = u +at

Here, a = -g

v = u -gt

i.e. u = gt

$t=\dfrac{u}{g}\\\\t=\dfrac{30\ m/s}{9.8\ m/s^2}\\\\t=3.06\ s$

So, the time for upward motion is 3.06 seconds. It means that it will in air for 3.06×2 = 6.12 seconds

Let d is the maximum distance covered by it.

$d=ut-\dfrac{1}{2}gt^2$

Putting all values

$d=30(3.06)-\dfrac{1}{2}\times 9.8\times (3.06)^2\\\\d=45.91\ m$

Hence, it will go to a height of 45.91 m and it will in the air for 6.12 seconds.

8 0

3 years ago