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3 years ago

15

Lucia kicks a ball on a level playing field with an initial velocity of 11.3 m/s at an angle of 35° above the horizontal. Find:

Time of flight? Horizontal distance traveled? Maximum height?

1 answer:

grandymaker [24]3 years ago

6 0

Explanation:

Given that,

Initial velocity, u = 11.3 m/s

Angle above the horizontal, $\theta=35^{\circ}$

Time of flight :

$t=\dfrac{2u\sin\theta}{g}\\\\t=\dfrac{2\times 11.3\times \sin(35)}{9.8}\\\\t=1.32\ s$

Horizontal distance traveled is given by :

x = ut

x = 11.3 m/s × 1.32 s

x = 14.916 m

Maximum height is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(11.3)^2\times \sin^2(35)}{2\times 9.8}\\\\H=2.14\ m$

Hence, time of flight is 1.32 s, horizontal distance is 14.916 m and maximum height is 2.14 m.

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3 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.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

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

(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:

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

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

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

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3 years ago

The chart shows data for a moving object. A 2-column table with 3 rows. Column 1 is labeled time in seconds with entries 2, 4, 6

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3 years ago

A racecar driver steps on the gas, and his racecar travels 20 meters in 2 seconds starting from rest. The acceleration of the ra

givi [52]

Answer:

2.5m/s^2

Explanation:

Step one:

given

distance = 20meters

time = 2 seconds

initial velocity u= 0m/s

let us solve for the final velocity

velocity = distance/time

velocity= 20/2

velocity= 10m/s

$v^2=u^2+2as\\\\10^2=0^2+2*a*20\\\\100=40a$

divide both sides by 40

$a= 100/40\\\\a=10/4\\\\a= 5/2\\\\a= 2.5m/s^2$

5 0

3 years ago

Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

3 0

3 years ago