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NNADVOKAT [17]
2 years ago
9

Gerard is measuring the speed of a remote-controlled airplane. He measures that the airplane travels 110 m in 5.4 seconds, givin

g a speed of 20 m/s. What will Gerard need to change, if he wants to get a more precise value for speed
Physics
1 answer:
netineya [11]2 years ago
3 0

Gerard need to change the Acceleration.

<h3>What is speed?</h3>

Speed is defined as the distance travelled by an object in  a unit time.

Mathematically ,

Speed =\frac{Distance}{Time}

Here the given values are:

Distance = 110m and  time = 5.4 s

Speed = \frac{110}{5.4} = 20 m/s

For increasing the speed he need to change the acceleration.

Which is define as the  rate at which velocity changes with time, in terms of both speed and direction.

Any change in the velocity of an object results in an acceleration , increasing speed (what people usually mean when they say acceleration).

learn more about speed: brainly.com/question/7359669

#SPJ1

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iogann1982 [59]

Answer: a) 8.2 * 10^-8 N or 82 nN and b) is repulsive

Explanation: To solve this problem we have to use the Coulomb force for two point charged, it is given by:

F=\frac{k*q1*q2}{d^{2}}

Replacing the dat we obtain F=82 nN.

The force is repulsive because the points charged have the same sign.

5 0
3 years ago
In regions where few species existed before or where species were wiped out what occurs?
marysya [2.9K]
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4 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
4 years ago
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6 0
3 years ago
What portion (division) of a meter stick is a centimeter?
Anastasy [175]
A meter is 100 meters. So a hundredth of a meter stick is a centimeter.<span />
8 0
4 years ago
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