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3 years ago

8

A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landin

g is angled at 60 degree with the roof.
a) find the horizontal distance d it travels.
b) what is the magnitude of the balls initial velocity?
c) what is the angle (relative to the horizontal) of the balls initial velocity?

1 answer:

grandymaker [24]3 years ago

3 0

Answer: 20

Explanation:

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Batteries are rated in terms of ampere-hours (A·h). For example, a battery that can produce a current of 2.00 A for 3.00 h is ra

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(a) 423 J

The power of the battery is the ratio between the total energy stored (E) and the time elapsed (t):

$P=\frac{E}{t}$

However, the power is also the product of the voltage (V) and the current (I):

$P=VI$

Linking the two equations together,

$\frac{E}{t}=VI\\E=VIt$

Since we know:

V = 9.0 V

$I \cdot t = 47.0 A\cdot h$

We can calculate the total energy:

$E=(9.0 V)(47 A \cdot h)=423 J$

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The battery has a total energy of E = 423 J. (2)

1 Watt (W) is equal to 1 Joule (J) per second (s):

$1 W = \frac{1 J}{1 s}$

so 1 kW corresponds to 1000 J/s:

$1 kW = \frac{1000 J}{1 s}$

Multiplying both side by 1 hour (1 h):

$1 kW \cdot h = \frac{1000 J}{1 s} 1 h$

and $1 h = 3600 s$ , so

$1 kWh = \frac{1000 J}{1 s}\cdot 3600 s =3.6\cdot 10^6 J$

So we find the conversion between kWh and Joules. So now we can convert the energy from Joules (2) into kWh:

$1 kWh = 3.6\cdot 10^6 J = x : 423 J\\x=\frac{1 kWh \cdot 423 J}{3.6\cdot 10^6 J}=1.18\cdot 10^{-4}kWh$

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Seema knows the mass of basketball. What other information is needed to find the balls potential energy

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Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field. In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy $U$ will be:

$U=mgh$

Where:

$m$ is the mass of the ball

$g=9.8 m/s^{2}$ is the acceleration due gravity (assuming the ball is on the Earth surface)

$h$ is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

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A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

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Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

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$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

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