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2 years ago

10

jason hits a baseball off a tee toward right field. the ball has a horizontal velocity of 10 m/s and lands 5 meters from the tee

. what is the height of the tee? show your work, including formula(s) and units.

1 answer:

Leni [432]2 years ago

8 0

Answer:

The height is 1,225 meters

Explanation:

DistanceX= speedX × time ⇒ time= (5 meters) ÷ (10 meters/second) = 0,5 seconds

DistanceY= high= (1/2) × g × (time^2) = (1/2) × 9,8 (meters/(second^2)) × 0,25 (second^2) = 1,225 meters

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Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions

iren [92.7K]

Answer:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

Explanation:

For this case we have the following info given:

Number of Na+ ions $5.7 x10^{11} ions$

Each ion have a charge of +e and the crage of the electron is $1.6 x10^{-19}C$

The time is given $t = 7 ms$ if we convert this into seconds we got:

$t = 7ms * \frac{1s}{1000 ms}= 0.007s$

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

$Q = Ne$

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

$Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C$

And from the general definition of current we know that:

$I =\frac{Q}{t}$

And since we know the total charge Q and the time we can replace:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

The current during the inflow charge in the meter axon for this case is $13.02 \mu A$

3 0

2 years ago

Based on the law of conversation of energy how can we reasonably improve a machines ability to do work?

tensa zangetsu [6.8K]

MARK ME BRAINLIEST!!

your answer should be “C”.

4 0

2 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

2 years ago

A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i

galben [10]

Answer:

350 F to 100 F it take approx 87.33 min

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling

$\frac{dT}{dt}$ = -k(T-Ta)

$\frac{dy}{dt}$ = $\frac{d}{dt}$ (T(t) -Ta)

= $\frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt}$ = -k(T-Ta)

-ky $\frac{dy}{dt}$ = -ky

T(t) -Ta = (To -Ta) $e^{-kt}$ T(t) = Ta+ (To -Ta) $e^{-kt}$

put her value for time 30 min and T(t) = 200◦F and To =350◦F and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) $e^{-k30}$

k = 0.025575

so here for T(t) = 100F

100 = 70 + ( 350 - 70 ) $e^{-0.025575*t}$

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min

5 0

2 years ago

Which statement is true of a glass lens that diverges light in air?

Arisa [49]

C it is uniformly thick

3 0

1 year ago

Read 2 more answers