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3 years ago

What is another way to describe the vector below?

Physics

1 answer:

yuradex [85]3 years ago

4 0

The correct choice is A .

(negative up) is the same direction as (positive down) .

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A rock of mass m is twirled on a string in a horizontal plane. The work done by the tension in the string on the rock is:

Varvara68 [4.7K]

Answer: c) 0

Explanation:

Work tension is zero because in any infinitesimal time interval, the tension force is directly perpendicular to the displacement.

7 0

3 years ago

A helicopter flies over the arctic ice pack at a constant altitude, towing an airborne 130-kg laser sensor that measures the thi

umka21 [38]

Answer:T=1316.21 N

Explanation:

The tension has two components: Vertical and Horizontal. The

horizontal component is ma, the vertical component is mg. Using

Pythagoras theorem, we can find the tension as:

T=((ma)^2 (mg)^2)^(1/2)

T=((129*2.84)^2 (129*9.8)^2)^(1/2)

T=1316.21 N

8 0

2 years ago

Pls pls help me out AHH, what is not true about MEIOSIS?

agasfer [191]

Answer:

Explanation:

This is not true as the number of chromosomes in the daughter cells are half the number in the parent's cells

4 0

3 years ago

Que cuerpos celestes observó laika durante su viaje

sergejj [24]

Do you want me to translate it?

4 0

2 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

3 years ago