Is Mercury a heavier element than tin

The current world-record motorcycle jump is 77.0 m, set by Jason Renie. Assume that he left the take-off ramp at 13.0° to the ho

kotegsom [21]

Answer:

The take-off speed is 41.48 $\frac{m}{s}$

Explanation:

Given :

Range $R = 77$ m

Projectile angle $\alpha =$ 13°

From the formula of range,

$R = \frac{v_{o} ^{2} }{g} \sin 2\alpha$

Find the velocity from above equation,

$v_{o} = \sqrt{\frac{gR}{\sin 2 \alpha } }$

$v_{o} = \sqrt{\frac{9.8 \times 77}{\sin 26} }$ ( ∵ $g = 9.8$ $\frac{m}{s^{2} }$ )

$v_{o} = 41.48$ $\frac{m}{s}$

Therefore, the take-off speed is 41.48 $\frac{m}{s}$

7 0

3 years ago

Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha

Inessa05 [86]

Answer:

$+1.11\mu C$

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

$Q_1$ is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to $Q_2$ and $Q_3$ at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of $Q_1$ .

Let the electric field intensity due to $Q_2$ be + $E_2$ and that due to $Q_3$ be - $E_3$ since the charge is negative. Hence at the origin;

$+E_2-E_3=0..................(1)$

From equation (1) above, we obtain the following;

$E_2=E_3.................(2)$

From Coulomb's law the following relationship holds;

$+E_2=\frac{kQ_2}{r_2^2}\\$

$-E_3=\frac{kQ_3}{r_3^2}$

where $r_2$ is the distance of $Q_2$ from the origin, $r_3$ is the distance of $Q_3$ from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

$\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)$

k can cancel out from both side of equation (3), so that we finally obtain the following;

$\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)$

Given;

$Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m$

Substituting these values into equation (4); we obtain the following;

$\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\$

$Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C$

6 0

3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

Umnica [9.8K]

Answer:

Explanation:

<u>Simple Harmonic Motion</u>

A mass-spring system is a common example of a simple harmonic motion device since it keeps oscillating when the spring is stretched back and forth.

If a mass m is attached to a spring of constant k and they are set to oscillate, the angular frequency of the motion is

$\displaystyle w=\sqrt{\frac{k}{m}}$

The equation for the motion of the object is written as a sinusoid:

$\displaystyle X=A\ cos\ w\ t$

Where A is the amplitude.

The instantaneous speed is computed as the derivative of the distance

$\displaystyle X'=V=-A\ w\ sin\ w\ t$

And the maximum speed is

$\displaystyle V_{max}= A\ w$

Solving for the amplitude

$\displaystyle A= \frac{V_{max}}{w}$

Computing w

$\displaystyle w =\sqrt{\frac{120}{0.2}}=24.5\ rad/ s$

Calculating A

$\displaystyle A=\frac{1.7}{24.5}=0.069\ m$

$\displaystyle \boxed{A=6.9\ cm}$

7 0

3 years ago

Kirchhoff's loop rule for circuit analysis is an expression of which of the following? Conservation of charge Conservation of en

Irina18 [472]

Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in the lumped element model of electrical circuits. They were first described in 1845 by German physicist Gustav Kirchhoff. This generalized the work of Georg Ohm and preceded the work of Maxwell.

6 0

3 years ago

How do I solve for time if I have initial velocity and final speed?

xeze [42]

Final speed = initial speed + (acceleration x time)

(final speed - initial speed) = acceleration x time

Time = (final speed - initial speed) / acceleration

4 0

3 years ago