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3 years ago

7

Driving your car with a constant speed of 11 m/s, you encounter a hump in the road with a circular cross section. If the radius

of curvature of the hump is 54 m, what is the apparent weight of a 80 kg person in your car as you pass over the top of the hump?

1 answer:

Simora [160]3 years ago

6 0

Answer:604.74 N

Explanation:

Given

Speed of car=11 m/s

radius of curvature is 54 m

mass of person=80 kg

At Top

$mg-N=\frac{mv^2}{r}$

where N=normal reaction

$N=mg-\frac{mv^2}{r}$

$N=80\time 9.8-\frac{80\times 11^2}{54}$

N=604.74 N

Apparent weight will be equal to Normal reaction

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A square is 1.0 m on a side. Point charges of +4.0 μC are placed in two diagonally opposite corners. In the other two corners ar

finlep [7]

Answer:

<em>B) 1.0 × 10^5 V</em>

Explanation:

<u>Electric Potential Due To Point Charges </u>

The electric potential produced from a point charge Q at a distance r from the charge is

$\displaystyle V=k\frac{Q}{r}$

The total electric potential for a system of point charges is equal to the sum of their individual potentials. This is a scalar sum, so direction is not relevant.

We must compute the total electric potential in the center of the square. We need to know the distance from all the corners to the center. The diagonal of the square is

$d=\sqrt2 a$

where a is the length of the side.

The distance from any corner to the center is half the diagonal, thus

$\displaystyle r=\frac{d}{2}=\frac{a}{\sqrt{2}}$

$\displaystyle r=\frac{1}{\sqrt{2}}=0.707\ m$

The total potential is

$V_t=V_1+V_2+V_3+V_4$

Where V1 and V2 are produced by the +4\mu C charges and V3 and V4 are produced by the two opposite charges of $\pm 3\mu\ C$ . Since all the distances are equal, and the charges producing V3 and V4 are opposite, V3 and V4 cancel each other. We only need to compute V1 or V2, since they are equal, but they won't cancel.

$\displaystyle V_1=V_2=k\frac{Q}{r}=9\times 10^9 \frac{4\times 10^{-6}}{0.707}$

$V_1=V_2=50912\ V$

The total potential is

$V_t=50912\ V+50912\ V=1\times 10^5\ V$

$\boxed{V_t=1\times 10^5\ V}$

6 0

3 years ago

Volcanic Island Arcs, such as Japan, form when a(n) __________ plate is subducted below a(n) _________________ and takes seawate

MrMuchimi

Volcanic Island arcs form when an <u>oceanic tectonic</u> plate is subducted below a <u>tectonic plate </u> taking seawater with it.

Volcanic island arcs occur offshore during volcanoes. They occur due to the subduction of an oceanic plate under another tectonic plate. Volcanic Island arcs occur mostly in ocean basins.

Due to the subduction of oceanic plates under tectonic plates, Island Arcs are found, mostly along the margins of continent.

Hence we can conclude that Volcanic Island arcs Form when an <u>oceanic tectonic</u> plate is subducted below a <u>tectonic plate </u> taking seawater with it.

Learn more : brainly.com/question/18662349

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2 years ago

if a car has a constant acceleration of 4 m/s^2, starting from rest, how fast is it traveling after 5 seconds?

UkoKoshka [18]

Answer:

20 m/s

Explanation:

Recall that one of the equations of motions can be written:

v = u + at, (also see attached for reference)

Where,

v = final speed (we are asked to find this)

u = initial speed = 0 (because it starts from rest)

t = time taken = 5s

We simply substitute the given values into the equation:

v = u + at

v = 0 + (4)(5)

v = 20 m/s

3 0

3 years ago

Read 2 more answers

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug

Vaselesa [24]

Answer:

$1.57 * 10^{3} Q$

Explanation:

The volume charge density is defined by ρ = $\frac{Q}{V}$ (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are $\frac{Coulombs}{m^{3} }$ , so we have to express the radius in meters:

inner radius = $4 cm * \frac{1 m}{100 cm} = 0.04m$

outer radius = $6 cm * \frac{1m}{100cm} = 0.06m$

Now, we know that the volume of the sphere is calculated by the formula:

$V = \frac{4}{3}\pi r^{3}$ , and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = $\frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3})$ , where $r_{o}$ is the outer radius and $r_{i}$ is the inner radius.

Replacing the volume formula in the Equation A:

ρ = $\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}$

ρ = $\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }$

Replacing the values of the outer and inner radius whe have:

ρ = $\frac{3Q}{4\pi (1.52 * 10^{-4})}$

ρ = $1.57 * 10^{3} Q$

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3 years ago

The normal force always points ____ to the two surfaces in contact.

shepuryov [24]

Answer:

A..................

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3 years ago