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Masja [62]
2 years ago
9

Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half

an hour. Then, she drives for 100 more kilometers, which takes her an hour and a half. What is Sandy's average velocity?
Physics
1 answer:
lutik1710 [3]2 years ago
0 0
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
v= \frac{S}{t}

The total distance covered is
S=200 km+0+100 km=300 km
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h
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An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary
Assoli18 [71]

Answer:

The tension in the cable when the craft was being lowered to the seafloor is 4700 N.

Explanation:

Given that,

When the craft was stationary, the tension in the cable was 6500 N.

When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N.

The drag force of 1800 N will act in the upward direction. As it was lowered or raised at a steady rate, so its acceleration is 0. As a result, net force is 0. So,

T + F = W

Here, T is tension

F = 1800 N

W = 6500 N

Tension becomes :

T=W-F\\\\T=6500-1800\\\\T=4700\ N

So, the tension in the cable when the craft was being lowered to the seafloor is 4700 N.

7 0
2 years ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0
2 years ago
Some vehicles, like _____, have different rules for registration and renewal. A. trailers B. coupes C. sedans D. pickup trucks
ycow [4]

Answer:

trailers

Explanation:

7 0
2 years ago
On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the
EastWind [94]

Answer:

ω = √(2T / (mL))

Explanation:

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The x-components of the tension forces cancel each other out, so the net force is in the y direction:

∑F = -2T sin θ, where θ is the angle from the horizontal.

For small angles, sin θ ≈ tan θ.

∑F = -2T tan θ

∑F = -2T (Δy / L)

(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m).  (This is derived by solving a second order differential equation.)

In this case, k = 2T/L, so the frequency is:

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6 0
2 years ago
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Rainbow [258]

Answer:

v = √(Lsinθ tanθ)

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From the diagram attached,

v = the tangential speed.

r = the radius of the horizontal circle.

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From geometry,

r = Lsin θ

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a = v²/r = v²/Lsin θ

Force = mass x acceleration

Thus,

Centripetal force = mv²/Lsin θ

Let's balance the vertical forces to obtain,

T cosθ - mg = 0

Thus, T cosθ = mg - - - - (eq1)

Similarly, let's balance the horizontal forces to obtain;

T sinθ = F = mv²/Lsin θ

So, T sinθ = mv²/Lsin θ - - - - (eq2)

Let's divide eq 2 by eq 1 to get;

Tsinθ/Tcosθ = (mv²/Lsin θ)/mg

tanθ = gv²/Lsinθ

Thus, v² =  Lsinθ tanθ

v = √(Lsinθ tanθ)

4 0
1 year ago
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