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3 years ago

5

Çok soyut mu somutmu

1 answer:

Licemer1 [7]3 years ago

8 0

Answer:

Soyut

Explanation:

ÇÜNKÜ elle dokunamazsın hissedemezsin

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What type of acceleration does an object moving with constant speed in a circular path experience?Select one:A) constant acceler

olga_2 [115]

ANSWER:

D) centripetal acceleration.

STEP-BY-STEP EXPLANATION:

When a body performs a uniform circular motion, the direction of the velocity vector changes at every instant. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circumference that gives rise to the centripetal acceleration.

Therefore, the answer is centripetal acceleration.

3 0

1 year ago

Technology can best be defined as what?

jasenka [17]

B

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3 years ago

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Which theory was first proposed by Albert Einstein? A. the theory of special relativity B. the theory of universal gravitation C

balu736 [363]

A. the theory of special relativity

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3 years ago

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A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in

motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

$F_{initial} = \frac{kq_1q_2}{r^2}$

Here,

k = Coulomb's constant

$q_{1,2}$ = Charge at each object

r = Distance between them

As the distance is doubled so,

$F_{final} = \frac{kq_1q_2}{( 2r )^2}$

$F_{final} = \frac{ kq_1q_2}{ 4r^2}$

$F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}$

$F_{final} = \frac{1}{4} F_{initial}$

$\frac{F_{final}}{ F_{initial}} = \frac{1}{4}$

Therefore the factor is 1/4

6 0

3 years ago