Ask question

Ask question

All categories

bearhunter [10]

3 years ago

10

List three reasons why knowing how to graph movement can help you on a practical level.

1 answer:

valina [46]3 years ago

6 0

<span>-1.0 m/s2
1.0 m/s2
-5.0 m/s2
5.0 m/s2
0.24 m/s<span>2</span></span>

You might be interested in

A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

aleksklad [387]

Answer:

$V=9.2565m/s$

Explanation:

From the question we are told that:

Force $F = 34 N$

Time $t = 0.6 s$

Length of pedal $l_p=16.5cm \approx0.165m$

Radius of wheel $r = 33 cm = 0.33 m$

Moment of inertia, $I = 1200 kgcm2 = 0.12 kg.m2$

Generally the equation for Torque on pedal $\mu$ is mathematically given by

$\mu=F*L\\\mu=34*0.165$

$\mu=5.61N.m$

Generally the equation for angular acceleration $\alpha$ is mathematically given by

$\alpha=\frac{\mu}{l}$

$\alpha=\frac{5.61}{0.12}$

$\alpha=46.75$

Therefore Angular speed is \omega

$\omega=\alpha*t$

$\omega=(46.75)*(0.6)$

$\omega=28.05rad/s$

Generally the equation for Tangential velocity V is mathematically given by

$V=r\omega$

$V=(0.33)(28.05)$

$V=9.2565m/s$

5 0

3 years ago

You are driving at 90 km/h. How many meters are you covering per second?

andreev551 [17]

This question wants a conversion
90km = 90000m
1hr = 3600 seconds
90000m/3600s= 900/36
= 25m/s

8 0

3 years ago

50 POINTS HELP ME PLEASE!!!!!!!!!!!!

Mariana [72]

Answer:

W and X

Explanation:

When escaping a rip current, one should always walk to the side until you escape from the rip current. If you walk towards the shore, you have the ability to keep getting dragged toward the current, such as with X and Y.

4 0

4 years ago

Read 2 more answers

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

4 years ago

What voltage is needed to create a current of 43 mA in a circuit containing only a 1.3-μF capacitor, when the frequency is 3.9 k

Leviafan [203]

Answer:

1.35 V

Explanation:

Given:

Capacitance of the capacitor (C) = 1.3 μF = $1.3\times 10^{-6}\ F\ \ [1\mu F=10^{-6}\ F]$

Frequency (f) = 3.9 kHz= 3.9 × 1000 Hz = 3900 Hz [1 kHz = 1000 Hz]

Current flowing in the circuit (I) = 43 mA = 0.043 A [1 mA = 0.001 A]

Now, as only capacitor is there in the circuit, the impedance of the circuit is equal to the reactance of the capacitor.

So, $Z=X_C$

The reactance of the capacitor is given as:

$X_C=\frac{1}{2\pi fC}$

Plug in the values given and solve for $X_C$ . This gives,

$X_C=\frac{1}{2\pi\times 3900\times 1.3\times 10^{-6}}\\\\X_C=31.39\ \Omega$

Therefore, the impedance is, $Z=31.39\ \Omega$

Now, the voltage needed to create the given current is obtained using the formula from Ohm's law and is given as:

$V=IZ\\\\V=(0.043\ A)(31.39\ \Omega)\\\\V=1.35\ V$

Therefore, the voltage needed to create a current of 43 mA is 1.35 V.

3 0

3 years ago