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3 years ago

13

Do the two cars ever have the same velocity at one instant of time? If so, between which two frames? Check all that apply. Check

all that apply. Cars have the same velocity at one instant of time between dots 1 and 2. Cars have the same velocity at one instant of time between dots 2 and 3. Cars have the same velocity at one instant of time between dots 3 and 4. Cars have the same velocity at one instant of time between dots 4 and 5. Cars have the same velocity at one instant of time between dots 5 and 6. Cars never have the same velocity at one instant of time.

1 answer:

alexgriva [62]3 years ago

8 0

Answer:

Cars have the same velocity at one instant of time between dots 4 and 5.

Explanation:

after looking at the image it is visible that same distance is covered by both cars in the frame between 4 and 5

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A fire hose has an inside diameter of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1.

inessss [21]

Answer:

The Reynolds numbers for flow in the fire hose.

Explanation:

Given that,

Diameter = 6.40 cm

Rate of flow = 40.0 L/s

Pressure $P=1.62\times10^{6}\ N/m^2$

We need to calculate the Reynolds numbers for flow in the fire hose

Using formula of rate of flow

$Q=Av$

$v=\dfrac{Q}{A}$

Where, Q = rate of flow

A = area of cross section

Put the value into the formula

$v=\dfrac{40.0\times10^{-3}}{3.14\times(3.2\times10^{-2})^2}$

$v=12.44\ m/s$

We need to calculate the Reynolds number

Using formula of the Reynolds number

$n_{R}=\dfrac{2\rho\times v\times r}{\eta}$

Where, $\eta$ =viscosity of fluid

$\rho$ =density of fluid

Put the value into the formula

$n_{R}=\dfrac{2\times100\times12.44\times3.2\times10^{-2}}{1.002\times10^{-3}}$

$n_{R}=7.945\times10^{5}$

Hence, The Reynolds numbers for flow in the fire hose.

3 0

3 years ago

A solid cylinder with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with anangular speed of 3.5 rad/s. Wha

Lynna [10]

Answer:

kE=0.0735 J

Explanation:

Given that

Radius ,R=10 cm = 0.1 m

Mass ,m= 3 kg

Angular speed ,ω = 3.5 rad/s

We know that moment of inertia for solid sphere given as

$I=\dfrac{2}{5}mR^2$

Kinetic energy

$KE=\dfrac{1}{2}I\omega^2$

Now by putting the values

$KE=\dfrac{1}{2}\times \dfrac{2}{5}mR^2\times \omega^2$

$KE=\dfrac{1}{2}\times \dfrac{2}{5}\times 3\times 0.1^2\times 3.5^2\ J$

kE=0.0735 J

Therefore the kinetic energy will be 0.0735 J

3 0

3 years ago

My model is supposed to have

bearhunter [10]

Answer:

yes you can submit

:):):)

4 0

3 years ago

During a softball game, a shortstop catch a ground ball. The action forces is the ball pushing on the glove. What is the reactio

marusya05 [52]

<span>The reaction force would be the exact opposite of the action. In this case, choice (c) would be the most correct. If the action is the ball pushing the glove, the reaction would then be the glove pushing back on the ball.</span>

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3 years ago

if pete (mass =90.00kg) weighs himself and finds that he weighs 30.0 pounds, how far away from the earth is he?

Svet_ta [14]

Here we know that mass of the person is 90 kg

His weight is given as 30 lbf

so here we can convert it into Newton as we know that

1 lbf = 4.45 N

Now from above conversion

$30 lbf = 30 \times 4.45 = 133.45 N$

now we can use this to find the gravity at this height

$mg = 133.45$

$90\times g = 133.45$

$g = 1.49 m/s^2$

now we know that with height gravity varies as

$g' = g(\frac{R}{R+H})^2$

$1.49 = 9.8(\frac{6.37\times 10^6}{6.37\times 10^6 + h})^2$

$h = 1.0 \times 10^7 m$

so above is the height from the surface of earth

4 0

3 years ago