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-Dominant- [34]

3 years ago

15

Mike Powell holds the record for the long jump of 8.95 m, established in 1991. If he left the ground at an angle of 18.8°, what

was his initial speed (in m/s)?

1 answer:

maksim [4K]3 years ago

7 0

Answer:

<em>12 m/s</em>

Explanation:

<u>Projectile Motion</u>

It's also known as 2D motion because the movement takes place in both axis x and y. The x-axis motion is at a constant speed since in absence of friction, no external force stops or accelerates the object. The y-axis motion is at variable speed, which is changed by the acceleration of gravity that makes the object to reach a maximum height and then go back to ground level.

The maximum horizontal distance reached (also called Range) is given by

$\displaystyle X_m=\frac{V_{o}^2sin2\theta}{g}$

Knowing that $\theta=18.8^o, X_m=8.95\ m$ , we solve for Vo

$\displaystyle V_o=\sqrt{\frac{gX_m}{sin2\theta}}$

$\displaystyle V_o=\sqrt{\frac{9.8\cdot 8.95}{sin(2\cdot 18.8^o)}}=12\ m/s$

Thus, the initial speed of Mike Powell was 12 m/s

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What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240

Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

$V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC$

For (b) charge density

As charge density σ is given as:

σ=q/(4πR²)

σ=(5.333×10⁻⁹) / (4π×(0.20)²)

σ=10.62 nC/m²

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3 years ago

A foot is 1/3 of a yard. What part of a meter is a millimeter?

N76 [4]

Answer:

Explanation:

1/1000

6 0

3 years ago

Does a wall or door have more inertia?

suter [353]

Answer:

wall

Explanation:

4 0

3 years ago

Read 2 more answers

The resistance of a very fine aluminum wire with a 20 μm × 20 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by w

notsponge [240]

Explanation:

The given data is as follows.

Resistance (R) = 1200 ohm, Area (A) = $20 \times 10^{-6}$ m (as $1 \mu m = 10^{-6} m$ )

Diameter (d) = 2.3 mm = $2.3 \times 10^{-3}$ m

First, we will calculate the length as follows.

R = $\rho \frac{L}{A}$

Here, $\rho$ = resistivity of aluminium = $2.65 \times 10^{-8}$

Putting the given values above and we will calculate the value of length as follows.

R = $\rho \frac{L}{A}$

1200 = $2.65 \times 10^{-8} \times \frac{L}{20 \times 10^{-6}}$

L = $9.056 \times 10^{5}$

As the circumference of circular wire = $2 \pi r$

or, = $2 \times \pi \times \frac{d}{2}$

= $\pi \times d$

And, number of turns will be calculated as follows.

No. of turns × Circumference = Length of wire

No. of turns × $3.14 \times 2.3 \times 10^{-3} = 9.056 \times 10^{5}$

= $1.25 \times 10^{8}$

Thus, we can conclude that $1.25 \times 10^{8}$ turns of wire are needed.

6 0

3 years ago

Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for

viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

$\omega_f$ = Final angular velocity = $\dfrac{1}{6}78=13\ rad/s$

$\omega_i$ = Initial angular velocity = 78 rad/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2$

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875$

Frictional force is given by

$F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N$

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

5 0

3 years ago