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kvasek [131]
3 years ago
14

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45° relative to edge of the sidewalk. If it has

traveled 1 m perpendicular to the edge of the sidewalk, how long has it been walking?
A.) 0.07 s


B.) 5 s


C.) 14 s


D.) 45 s

Physics
1 answer:
Marat540 [252]3 years ago
3 0

Answer:14 s

Explanation:

Given

Velocity of ant is 0.1 m/s in a direction 45^{\circ}

if it has traveled 1 m perpendicular to the edge of the sidewalk

i.e. from diagram

\sin 45=\frac{1}{L}

L=\sqrt{2}

time=\frac{Distance}{speed}

t=\frac{\sqrt{2}}{0.1}

t=14.14 s

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Answer:

D

Explanation:

Water in itself is a bad conductor of electricity. Any compound that dissociates in water into ions and charged molecules, will increase conductivity of water. In this case ionic compounds weak basis and acids will put in free ions into the water. The ions can pass electricity because they are attracted to respective poles of electricity depending on their charges. In water, these ions are free moving unlike when they are immobilized in their lattice in solid form.

Organic compounds are mainly made of covalent bonds (around carbon) hence do not dissociate in water.

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3 years ago
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A 21.7kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.2m/s. How much thermal energy due to frictio
nadya68 [22]

Answer:

Subtract the kinetic energy at the bottom from the potential energy loss. The remainder becomes frictional heat.

Potential energy loss:

M g H = 21.7*9.81*3.5 = 745.1 J

Kinetic energy at bottom of slide:

= (1/2) M v^2 = 52.5 J

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3 years ago
What is the resultant of a pair of forces, 100N, upward and 75N, downward?
soldi70 [24.7K]

Answer:

25N

Explanation:

100 - 75 = 25

That should be right if im not dumb...

3 0
3 years ago
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h
mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(8.89² - 6.11²)

                                                    = 20.85m J

 

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

                                                    = 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

6 0
2 years ago
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