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2 years ago

9

3. During a race, a sprinter increases from 5.0 m/s to 7.5 m/s over a period of 1.25s. What is the sprinter’s average accelerati

on during this period?
Also please try and explain
Please and Thank you

1 answer:

Rasek [7]2 years ago

5 0

(7.5)-(5)
----------- = 2 m/s^2
(1.25-0)

Average acceleration is calculated using the equation V(f)-V(i) / t(f)-t(i). (Final velocity minus initial velocity divided by final time minus initial time)

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A skateboarder, starting from rest, rolls down a 12.8-m ramp. When she arrives at the bottom of the ramp her speed is 8.89 m/s.

melamori03 [73]

Answer:

a) a = 3.09 m/s²

b) aₓ = 2.60 m/s²

Explanation:

a) The magnitude of her acceleration can be calculated using the following equation:

$V_{f}^{2} = V_{0}^{2} + 2ad$

<u>Where</u>:

$V_{f}$ : is the final speed = 8.89 m/s

$V_{0}$ : is the initial speed = 0 (since she starts from rest)

a: is the acceleration

d: is the distance = 12.8 m

$a = \frac{V_{f}^{2}}{2d} = \frac{(8.89 m/s)^{2}}{2*12.8 m} = 3.09 m/s^{2}$

Therefore, the magnitude of her acceleration is 3.09 m/s².

b) The component of her acceleration that is parallel to the ground is given by:

$a_{x} = a*cos(\theta)$

<u>Where</u>:

θ: is the angle respect to the ground = 32.6 °

$a_{x} = 3.09 m/s^{2}*cos(32.6) = 2.60 m/s^{2}$

Hence, the component of her acceleration that is parallel to the ground is 2.60 m/s².

I hope it helps you!

7 0

3 years ago

If state law mandates that elevators cannot accelerate more than 4.80 m/s2 or travel faster than 19.8 m/s , what is the minimum

Rudik [331]

Answer:

23.0 s

Explanation:

Given:

v₀ = 0 m/s

v = 19.8 m/s

a = 4.80 m/s²

Find: Δx and t

v² = v₀² + 2aΔx

(19.8 m/s)² = (0 m/s)² + 2 (4.80 m/s²) Δx

Δx = 40.84 m

v = at + v₀

19.8 m/s = (4.80 m/s²) t + 0 m/s

t = 4.125 s

The elevator takes 40.84 m and 4.125 s to accelerate, and therefore also 40.84 m and 4.125 s to decelerate.

That leaves 291.3 m to travel at top speed. The time it takes is:

291.3 m / (19.8 m/s) = 14.71 s

The total time is 4.125 s + 14.71 s + 4.125 s = 23.0 s.

8 0

2 years ago

In a house 5 bulbs of 25 watts each are used for 6 hours a day.Calculate the unit of electricity consumed in a month of 30 days.

Svetradugi [14.3K]

(5 bulbs) x (25 watt/bulb) x (6 hour/day) x (30 day/month) =

   (5 x 25 x 6 x 30) watt-hour/month =

   22,500 watt-hour/month .

The most common unit of electrical energy used for billing purposes
is the 'kilowatt-hour' = 1,000 watt-hours .

   22,500 watt-hour/month = <em>22.5 kWh/month</em>.

   (22.5 kWh/month) x (1.50 Rs/kWh) = <em>33.75 Rs / month

</em>

4 0

2 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just bef

stich3 [128]

Answer:

v = 7.67 m/s for L= 1m

Explanation:

Let's use the conservation of mechanical energy, at the highest point and the lowest point

Initial. Vertical ruler

Em₀ = mg h

Final. Just before touching the floor

$Em_{f}$ = K = ½ I w²

Em₀ = $Em_{f}$

m g h = ½ I w²

The moment of inertia of a ruler that turns on one end is

I = 1/3 m L²

Let's replace

m g h = ½ (1/3 m L²) w²2

g h = 1/6 L² w²

They ask for the speed of the end so the height h is equal to the length of the ruler

g L = 1/6 L² w²

The linear and angular variables are related

v = w r

w = v / r

In this case the point of interest a in strangers r = L

g L = 1/6 L² v² / L²

v = √ 6 g L

Let's calculate

Assume that the length of the meter is L = 1 m

v = √ (6 9.8 1)

v = 7.67 m/s

7 0

2 years ago

A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels in the negative x direction with a speed of 30.0 m

lesya [120]

Answer:

a) A =0.021525m

b) $\phi=0.37869rad$

c) $v_{max}=5.4098\frac{m}{s}$

d) $y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

Explanation:

1) Notation

A= Amplitude

v= velocity

$\lambda$ = wavelength

k= wave number

$\omega$ = angular frequency

f= frequency

2) Part a and b

The equation of movement for a transverse sinusoidal wave is gyben by (1)

$y(t)=Acos(kx+ \omega t +\phi)$ (1)

At x=0 ,t=0 we have that:

$0.02=Acos(\phi)$

The velocity would be the derivate of the position, so taking the derivate of (1) respect to t we got (2)

$v(t)=-\omega Asin(kx+ \omega t+\phi)$ (2)

And replacing the conditions at x=0, t=0 we got

$-2\frac{m}{s}=-\omega Asin(\phi)$

Now we can find the angular frequency with equation (3)

$\omega =\frac{2\pi}{T}$ (3)

Replacing the values obtained we got:

$\omega =\frac{2\pi}{0.025s}=80\pi \frac{rad}{s}$

From equation (1) we have:

$Acos(\phi)=0.02$ (a)

$-2=-80\pi Asin(\phi)$ (b)

So from condition (b) we have:

$Asin(\phi)=\frac{1}{40\pi}$ (c)

If we divide condition (c) by condition (a) we got:

$\frac{Asin(\phi)}{Acos(\phi)}=tan(\phi)=\frac{1}{0.02x40\pi}=\frac{1}{0.8\pi}=0.39789$

If we solve for $\phi$ we got:

$\phi =tan^{-1}(0.39789)=0.37869$

And now since we have $\phi$ we can find A from equation (a)

$Acos(0.37869)=0.02$

So then Solving for A we got $A=\frac{0.02}{cos(0.37869)}=0.021525$

3) Part c

From equation (2) we can see that the maximum speed occurs when $sin(\omega t+\phi)=1$ , so on this case we have:

$v_{max}=\omega A=80\pi \frac{rad}{s}x0.021525m=5.4098\frac{m}{s}$

4) Part d

On this case we need an equation like (1), and we have everything except the wave number, and we can obtain this from the following expression:

$v=\lambda f=\frac{2\pi}{k}\frac{\omega}{2\pi}=\frac{\omega}{k}$ (4)

And solving for k from equation (4) we got

$k=\frac{\omega}{v}=\frac{80\pi \frac{rad}{s}}{30\frac{m}{s}}=\frac{8\pi}{3}m^{-1}}$

And with the k number we have everythin in order to create the wave function, given by:

$y(x,t)=(0.021525m)cos(\frac{8\pi}{3}x+80\pi t+0.37869)$

7 0

3 years ago