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3 years ago

8

2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe

s it take for the football to rise to the highest point of its trajectory?

1 answer:

ValentinkaMS [17]3 years ago

5 0

Answer: 2.04 s

Explanation:

Let the initial velocity be v, Angle of projectile be

Then the horizontal component = v cos θ = 16 m/s

Vertical component of velocity = v sin θ = 20 m/s

Time taken to reach the highest point is half the time taken for total flight.

Time for total flight,

$t = \frac{2vsin \theta}{g}$

$t'=\frac{vsin \theta}{g} = \frac {20 m/s}{9.8 m/s^2} = 2.04 s$

Thus, the football takes 2.04 s to rise to the highest point of its trajectory.

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Answer:

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Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr

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Answer:

If I double the current in the inductor, the new total energy will become 4E (option f).

Explanation:

The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

$E=\frac{1}{2} *L*I^{2}$

where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

$E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}$

Then:

$E'=4*E$

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

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3 years ago

Find electric field at point p which is a distance l away from the both +q and -q

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Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

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3 years ago

A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform

madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is $\epsilon_{max}= 26.8 V$

The emf induced at t = 1.00 s is $\epsilon = 24.1V$

The maximum rate of change of magnetic flux is $\frac{d \o}{dt}|_{max} =26.8V$

Explanation:

From the question we are told that

The number of turns is N = 44 turns

The length of the coil is $l = 15.0 cm = \frac{15}{100} = 0.15m$

The width of the coil is $w = 8.50 cm =\frac{8.50}{100} =0.085 m$

The magnetic field is $B = 745 \ mT$

The angular speed is $w = 64.0 rad/s$

Generally the induced emf is mathematically represented as

$\epsilon = \epsilon_{max} sin (wt)$

Where $\epsilon_{max}$ is the maximum induced emf and this is mathematically represented as

$\epsilon_{max} = N\ B\ A\ w$

Where $\o$ is the magnetic flux

N is the number of turns

A is the area of the coil which is mathematically evaluated as

$A = l *w$

Substituting values

$A = 0.15 * 0.085$

$= 0.01275m^2$

substituting values into the equation for maximum induced emf

$\epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0$

$\epsilon_{max}= 26.8 V$

given that the time t = 1.0sec

substituting values into the equation for induced emf $\epsilon = \epsilon_{max} sin (wt)$

$\epsilon = 26.8 sin (64 * 1)$

$\epsilon = 24.1V$

The maximum induced emf can also be represented mathematically as

$\epsilon_{max} = \frac{d \o}{dt}|_{max}$

Where $\o$ is the magnetic flux and $\frac{d \o}{dt}|_{max}$ is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

$\frac{d \o}{dt}|_{max} =26.8V$

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3 years ago