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KengaRu [80]
2 years ago
7

during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which f

orce stops you from sliding all the way home? a friction b gravity c pull d push
Physics
1 answer:
vovangra [49]2 years ago
3 0

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

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F(x)= 10x-5<br>What is the value of f-1(-4) ?​
Afina-wow [57]

Answer:

f^{-1}(-4) = \frac{1}{10}

Explanation:

Firstly finding f^{-1}(x)

So,

f(x) = 10x-5

Substitute y = f(x)

y = 10x-5

Exchange the values of x and y

x = 10y-5

Solving for y

x = 10y-5

Adding 5 to both sides

10y = x+5

Dividing both sides by 10

y = \frac{x+5}{10}

Replace y = f^{-1}(x)

f^{-1}(x) = \frac{x+5}{10}

For x = -4

f^{-1}(-4) = \frac{-4+5}{10}

f^{-1}(-4) = \frac{1}{10}

6 0
2 years ago
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zubka84 [21]
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7 0
3 years ago
Sketch the electric field around these two objects if they have the same sign of charge. Make a separate drawing showing equipot
diamong [38]

Answer:

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

Explanation:

The electric field around two objects with charge of the same sign, for simplicity suppose that the objects have positive point spherical charges,

          E = k q / r2

bold letters indicate vectors, therefore the total electric field is

           E_total = E1 + E2

the module of this field is

           E_total = E1- E2

therefore we can outline this field

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

An outline of these shows in Attachment A

The equipotential surfaces are defined as being perpendicular to the electric field lines since the electric field and the power difference are related

              E = \frac{dV}{dx} i^ + \frac{dV}{dy} j^ + \frac{dV}{dz} k^ = \Delta V

We can schematize some characteristics of these surfaces

* very close to each load are spherical surfaces

* very far from the load is an elliptical surface, which envelops the loads

* between them there is a point of zero potential point C

See attached part B

5 0
2 years ago
Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m
aev [14]

Answer:

Explanation:

Given

charge of first body q_1=2.5\ mu C

charge of second body q_2=-3.5\ mu C

Particle 1 is at origin and particle 2 is at x=0.6\ m

third Particle which charge +q must be placed left of 2.5\mu C because it will repel the q charge while -3.5\mu C will attract it

suppose it is placed at a distance of x m

F_{1q}=\frac{kq(2.5)}{x^2}

F_{2q}=\frac{kq(-3.5)}{(0.6+x)^2}

F_{1q}+F_{2q}=0

\frac{kq(2.5)}{x^2}+\frac{kq(-3.5)}{(0.6+x)^2}=0

\frac{kq(2.5)}{x^2}=\frac{kq(3.5)}{(0.6+x)^2}

\frac{0.6+x}{x}=(\frac{3.5}{2.5})^{0.5}

0.6+x=1.1832x

x=3.27\ m

5 0
3 years ago
What waves have high amplitudes
Olegator [25]

-- loud sounds

-- bright lights

-- strong radio signals

-- Slinkies that can pinch you painfully

-- a tsunami in the ocean

-- earthquakes above Richter 5 or 6

5 0
3 years ago
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