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3 years ago

What force is needed to give a 4.5-kg bowling ball an acceleration of 9 m/s2?

Physics

2 answers:

Sloan [31]3 years ago

8 0

F=ma
f=4.5*9

40.5 N
hope this help

topjm [15]3 years ago

7 0

What force is needed to give a 4.5-kg bowling ball an acceleration of 9 m/s2?

$F = ma

F = 4.5 x 9$

40.5 N

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a city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in

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We have that for the Question "A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?" it can be said that

OC=2

From the question we are told

A city government received a $1 million grant to build swimming pools and skating rinks for youth. based on the data provided in the graph, what is the opportunity cost of building one swimming pool?

Generally the equation for the Opportunity cost is mathematically given as

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3 years ago

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What would happen if you touched an object that is the same temperature as your hand? Would any heat transfer happen? Explain yo

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2 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

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$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

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