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3 years ago

Displacement vectors of 3 m and 5 m in the same direction combine to ma

Physics

1 answer:

andrey2020 [161]3 years ago

6 0

Answer:

The magnitude of the displacement vector will be 8 meters.

Explanation:

Given the vectors:

r= 3 m

s= 5 m

The resulting vector is the sum of the vectors r and s. When we add vectors, the vectors that point in the same direction are added and the vectors that point in opposite directions are subtracted. Therefore, the result will be:

R= r + s = 3 m + 5 m = 8 m

The resulting vector will be 8 m in magnitude and the same direction as the previous vectors.

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Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?A. Fluorescent lamps are more

mr_godi [17]

Answer:

B. Fluorescent lamps operate at a higher temperature than incandescent

Explanation:

Fluorescent lamps have a number of advantages over incandescent lamps which are given in the options given in A, C and D. The option available in B is a drawback, not an advantage. This is because it can give out and radiate more heat as a result of working at a higher temperature. Hence B option is correct.

8 0

3 years ago

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

3 years ago

Can someone tell me how to do half life or something because I am so confused and it is due in a couple of hours

Genrish500 [490]

Do you have the equation for half life

6 0

2 years ago

Read 2 more answers

Find an expression for the kinetic energy of the car at the top of the loop.Express the kinetic energy in terms of m, g, h, and

lyudmila [28]

Answer:

K.E₂ = mg(h - 2R)

Explanation:

The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:

K.E₁ + P.E₁ = K.E₂ + P.E₂

where,

K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)

P.E₁ = Initial Potential Energy = mgh

K.E₂ = Final Kinetic Energy at the top of the loop = ?

P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)

Therefore,

0 + mgh = K.E₂ + mg(2R)

3 0

3 years ago

What is 6.02 x 10^4 in standard notation

Vesnalui [34]

I believe you mean 6.02*10^7 but you want to shift the decimal 7 times to the right which would be 60200000 (:

4 0

2 years ago