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3 years ago

11

Tara's cell phone plan costs $39.00 a month, which includes 100 text messages. After she uses all of her text messages, it will

cost her $.15 per text message

1 answer:

vekshin13 years ago

5 0

I think it is $585.hope it is right

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What type of friction occurs when you are trying to move an object, but the object isnt moving?

Tasya [4]

the answer is static friction

7 0

4 years ago

On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

3 years ago

Read 2 more answers

What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical posi

maksim [4K]

Answer:

ΔU = 2 mg h

Explanation:

In a spring mass system the potential energy is U = m g h

where h is measured from the equilibrium point of the spring

the potential energy at the highest point is

U₁ = m g h

the potential energy at the lowest point is

U₂ = m g (-h)

instead in this energy it is

ΔU = 2 mg h

In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero

4 0

3 years ago

How do I solve this

lesantik [10]

multiply grav pull by mass of astro maybe with a calculator

7 0

4 years ago

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert

cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, $k=4.2\times 10^5\ N/m$

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

$mgh=\dfrac{1}{2}kx^2$

x is the compression in spring

$x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m$

So, the compression in the spring is 5.88 meters.

6 0

3 years ago