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3 years ago

A woman is standing on a steep hillside in the rain and is not moving. A

Physics

1 answer:

NemiM [27]3 years ago

3 0

Answer:

The correct option is;

A. The component of the woman's weight along the hillside is larger than the kinetic friction between the woman and the ground

Explanation:

Given that the coefficient of static friction is always larger than the coefficient of kinetic friction, we have that before the wind blew, the component of the woman's weight along the hillside was lesser than the static friction between the woman and the ground, when the wind blew the total force of the wind and the component of the woman's weight put her into motion such that the acting frictional force was then the kinetic frictional force which was lesser than the kinetic frictional force so the woman continues to slide down the hillside without the wind.

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10. The shape of the earth is<br> A taco<br> b. A carrot<br> C. A sphere<br> d. A potato

expeople1 [14]

Answer:

C. A sphere.

Explanation:

I'm 100% sure.

6 0

3 years ago

1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational

enyata [817]

Answer:

d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

x = vox t

Y Axis

y = $v_{oy}$ t - ½ g t²

It's displacement is

d = x i ^ + y j ^ + z k ^

Substituting

d = (23 i ^ + 33 j ^ + 0) m

Using your notation

d = <23, 33, 0> m

The force of gravity is the weight of the body

W = m g

W = 1 9.8 = 9.8 N

In vector notation, in general the upward direction is positive

W = (0 i ^ - 9.8 j ^ + 0K ^) N

W = <0, -9.8, 0>

Work is defined

W = F. dy

W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

Cos 180 = -1

W = -F y

W = - 9.8 (33-0)

W = -323.4 J

6 0

3 years ago

Cientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.

scZoUnD [109]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back.

Explanation:

3 0

3 years ago

In gym class you run 22 m horizontally, then climb a rope vertically for 6.2 m. What is the direction angle of your total displa

miskamm [114]

Im 99% sure
tan^-1 (6.2/22)= 15.7º
with 2 significant figures 16º

7 0

3 years ago

Read 2 more answers

An adiabatic nozzle has an inlet area of 1 m^2 and an outlet area of 0.25 m^2. Water enters the nozzle at a rate of 5 m^3/s and

svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

$V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s$

velocity at the exit=20m/s

for entry

$V=\frac{5}{1} =5m/s$

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

$\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}$

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

$(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81 =p2\\P2=-37.5kPa$

the pressure at exit is -37.5kPa

7 0

3 years ago