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patriot [66]
3 years ago
15

A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplie

s power to an emergency beacon. The power draw from the load causes the capacitor to lose 9.9% of its charge every minute. When the charge drops below 8 hundred millivolts the beacon will cease functioning. How long will t
Physics
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes

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Help pls, see picture. Will mark Brainliest
Leno4ka [110]

Answer:

B

Explanation:

the graph shows the line going up (accelerating) and it isn't curving like d so it doesn't stop accelerating

Hope this helps :)

4 0
3 years ago
Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10
gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

\Sigma \tau=0

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

8 0
3 years ago
A ball of mass 2 kg is kept on the hill of height 3 km. Calculate the potential energy possessed by it ?
zysi [14]
We know that -

P.E=m*g*h

Where,

m = mass

g = acceleration due to gravity

h=height

First we convert height into meters.

1 km = 1000 meters

3 km = 1000 * 3 meters = 3000 meters

So, putting the values in the above formula, and by taking 'g' = 9.8 m/s², we get-

P.E.= 2*3000*9.8

P.E.= 58800 Joules

P.E.= 58.8 kJ

7 0
4 years ago
Read 2 more answers
A conducting loop has an area of 0.065 m2 and is positioned such that a uniform magnetic field is perpendicular to the plane of
aev [14]

Answer:

initial magnetic field  1.306 T

Explanation:

We have given area of the conducting loop A=0.065m^2

Emf induced = 1.2 volt

Initial magnetic field B = 0.3 T

Time dt = 0.087 sec

We know that induced emf is given by e=\frac{d\Phi }{dt}=-A\frac{db}{dt}

1.2=0.065\times \frac{db}{0.087}

db=1.606T

So initial magnetic field = 1.606-0.3= 1.306 T

5 0
3 years ago
if a car travels 30 miles north for one half hour, 50 miles east for one hour, and 30 miles south for 30 minutes, what is the to
egoroff_w [7]
50 miles east. because the 30 miles north then south cancel each other out. 
5 0
3 years ago
Read 2 more answers
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