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3 years ago

How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?

1 answer:

7 0

Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.

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Explain how to measure volume using a graduated cylinder

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Put object into cylinder and notice the volume increased. Difference of volume is the object's volume

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3 years ago

What is one disadvantage of sending information over long distances

shusha [124]

Answer:

A is the correct answer.

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3 years ago

Read 2 more answers

A wave has a wavelength of 4. 9 m and a velocity of 9. 8 m/s. The medium through which this wave is traveling is then heated so

tatiyna

The wavelength of a wave is obtained by taking the ratio of wave speed and frequency.

The wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

What is frequency of a wave?

The number of oscillations completed by a wave in one second is known as the frequency of a wave. It is expressed as the ratio of the velocity of the wave to its wavelength.

Given data-

The wavelength of the wave is, $\lambda = 4.9 \;\rm m$ .

The velocity of the wave is, v = 9.8 m/s.

The mathematical expression for the frequency of the wave is,

$f = \dfrac{v}{\lambda}$

Solving as,

$f = \dfrac{9.9}{4.9}\\\\f =2 \;\rm Hz$

Now, with constant frequency and the double magnitude of velocity (v' = 2 × 9.8 = 19.6 m/s). The wavelength of the heated wave is calculated as,

$f = \dfrac{v'}{\lambda'}$

here,

$\lambda'$ is the wavelength of the heated wave.

Solving as,

$2 = \dfrac{19.6}{\lambda'}\\\\\lambda' = \dfrac{19.6}{2}\\\\\lambda' = 9.8 \;\rm m$

Thus, we can conclude that the wavelength of the heated wave is 9.8 m. Hence, option (b) is correct.

Learn more about the frequency of wave here:

brainly.com/question/1324797

3 0

3 years ago

Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.

8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

$V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

The distance from the center of the square to one of the corners is $\sqrt2 L/2 = 0.035m$

$V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0$

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) $V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}$

$r_1 = 0.05\sqrt2m\\r_2 = 0.05m$

$V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between energies. This is the work-energy theorem. So,[tex]W = U_b - U_a$

$U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3$

$W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}$

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3 years ago

Consider a constant density gas flowing steadily over an airfoil. Far upstream the velocity is V0. Halfway along the top surface

Mashutka [201]

Answer:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

Explanation:

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3 years ago