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3 years ago

7

A car travels up a hill at a constant speed of 44 km/h and returns down the hill at a constant speed of 74 km/h. Calculate the a

verage
Speed for the round trip.

1 answer:

Hunter-Best [27]3 years ago

5 0

Answer:

calculate it by yourself

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How do you calculate the volume and density of something that weighs 20 grams?

Nostrana [21]

Mass divided by volume is density, while mass times density is volume. you cannot calculate density without volume and you cannot calculate volume without density.
i believe that's the answer.
oh gosh i didn't realize my middle school education involved high school stuff.

6 0

3 years ago

Ill mark as brainliest.

rewona [7]

Answer: 75V

Explanation:

Given that,

total resistance (Rtotal) = 150Ω

Current (I) = 0.5A

Change in electric potential (V) = ?

Recall that potential difference is the product of amount of current and the amount of resistance in the circuit. And its unit is volts.

So, apply the formula V = I x Rtotal

V = 0.5A x 150Ω

V = 75V

Thus, the change in electric potential across the circuit is 75 Volts

8 0

3 years ago

Four aqueous solutions and their concentrations are shown in the above illustration. which of the solutions is most likely to be

Elza [17]

The solution that would most likely be a strongest conductor of electricity is the solution that is most saturated or concentrated. This is because the atoms that are found within the aqueous solutions have become positively charged resulting to the attraction of negatively charged ions that are found in electricity. On the other hand, the least conductive from the aqueous solutions would be the most unsaturated one because of less conductive ions present.

3 0

3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

4 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

3 years ago