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3 years ago

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An Olympic diver springs off of a high dive that is 3 m above the surface of the water. When she lands in the water she is trave

ling at a speed of 8.90 m/s at an angle of 75.0° with respect to the horizontal. What was her take off speed and direction, up from the horizontal?

1 answer:

klemol [59]3 years ago

7 0

Answer: $u=4.51 m/s\ at\ angle\ of\ \theta =59.34^{\circ}$

Explanation:

Given

height of building h=3 m

Landing velocity of diver $v=8.90 m/s$ at an angle of $75^{\circ}$

Let u be the initial velocity of diver at an angle of \theta with horizontal

Since there is no acceleration in horizontal direction therefore horizontal component of velocity will remain same

$u\cos \theta =8.9\cos (75)$ ---- -----1

Considering Vertical motion

$v^2-u^2=2as$

here $v=8.9\sin (75)$

$u=u\sin \theta$

$s=3 m$

$a=9.8 m/s^2$

$(8.9\sin (75))^2-(u\sin \theta )^2=2\times 9.8\times 3$

$u\sin \theta =\sqrt{(8.9\sin 75)^2-(2\cdot 9.8\cdot 3)}$

$u\sin \theta =3.886$ ----------------2

Divide 2 and 1 we get

$\tan \theta =\frac{3.886}{8.9\cos (75)}$

$\tan \theta =1.687$

$\theta =59.34^{\circ}$

Thus $u\cos (59.34)=8.9\cos (75)$

$u=4.51 m/s$

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3 years ago

A helicopter is lifting two crates simultaneously. One crate with a mass of 160 kg is attached to the helicopter by cable A. The

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Answer

given,

mass of crate attached by cable A = 160 Kg

mass of crate attached by cable B = 73 Kg

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tension in the cable when the move up

F = m (a + g)

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3 years ago

At each corner of a square of side there are point charges of magnitude Q, 2Q, 3Q, and 4Q

Bad White [126]

Answer:

$\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}$

$\displaystyle \theta =68^o$

Explanation:

Electrostatic Force

It's the force that appears between two electrical charges q1 q2 when they are placed at a certain distance d. The force can be computed by using the Coulomb's law:

$\displaystyle F=\frac{KQ_1Q_2}{d^2}$

We have an arrangement of 4 charges as shown in the image below. We need to calculate the total force exerted on the charge 2Q by the other 3 charges. The free body diagram is also shown in the second image provided. The total force on 2Q is the vectorial sum of F1, F2, and F3. All the forces are repulsive, since all the charges have the same sign. Let's compute each force as follows:

$\displaystyle |F_1|=\frac{KQ(2Q)}{l^2}=\frac{2KQ^2}{l^2}$

$\displaystyle |F_2|=\frac{K(2Q)(4Q)}{l^2}=\frac{8KQ^2}{l^2}$

The distance between 3Q and 2Q is the diagonal of the rectagle of length l:

$\displaystyle |d_3|=\sqrt{l^2+l^2}=\sqrt{2}\ l$

The force F3 is

$\displaystyle |F_3|=\frac{K(3Q)(2Q)}{(\sqrt{2l)}^2}=\frac{3KQ^2}{l^2}$

Each force must be expressed as vectors. F1 is pointed to the right direction, thus its vertical components is zero

$\displaystyle \vec{F_1}=\left \langle |F_1|,0 \right \rangle=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle$

F2 is pointed upwards and its horizontal component is zero

$\displaystyle \vec{F_2}=\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle$

F3 has two components because it forms an angle of 45° respect to the horizontal, thus

$\displaystyle \vec{F_3}=\left \langle \frac{3KQ^2}{l^2}\ cos45^o,\frac{3KQ2}{l^2} sin45^o\right \rangle$

$\displaystyle \vec{F_3}=\left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

Now we compute the total force

$\displaystyle \vec{F_t}=\vec{F_1}+\vec{F_2}+\vec{F_3}$

$\displaystyle \vec{F_t}=\left \langle \frac{2KQ^2}{l^2},0 \right \rangle +\left \langle 0,\frac{8KQ^2}{l^2} \right \rangle + \left \langle \frac{3\sqrt{2}KQ^2}{2l^2},\frac{3\sqrt{2}KQ^2}{2l^2}\right \rangle$

$\displaystyle \vec{F_t}=\left \langle \left(2+\frac{3\sqrt{2}}{2}\right)\frac{KQ^2}{l^2},\left(8+\frac{3\sqrt{2}}{2}\right) \frac{KQ^2}{l^2}\right \rangle$

$\displaystyle F_t=\left \langle 4.121,10.121 \right \rangle \frac{KQ^2}{l^2}$

Now we compute the magnitude

$\boxed{\displaystyle |F_t|=10.9\ \frac{KQ^2}{l^2}}$

The direction of the total force is given by

$\displaystyle tan\theta =\frac{10.121}{4.121}=2.4558$

$\boxed{\displaystyle \theta =68^o}$

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3 years ago

A parallel circuit contains four identical lamps. The current through the energy source is 4 A. The total resistance of the circ

TEA [102]

Answer:

b. 40V , 40V

Explanation:

Connections are as per the figure.

As total current through source is 4A , current through each lamp is 1A.

As total resistance of the circuit is 10Ω ,resistance of each bulb is 40Ω because in case of a parallel circuit in which identical objects are connected , $R_{eff} = \frac{R}{n}$ where R is the resistance of each bulb and n is the number of bulbs.

As per Ohm's law , voltage of the source =IR = 4×10 =40V.

We can see from the figure that if the voltage across the source is 40V , the voltage across each bulb is also 40V.

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