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3 years ago

Find the dimensional formulas=ut+1/2at^2

Physics

1 answer:

Airida [17]3 years ago

4 0

Explanation:

Dimensions of u is [LT-1]

Dimensions of a is [LT-2]

Dimensions of ut+1/2(at²) is

[ constant are dimensionless]

[LT-1 . T ] + [LT-² . T²]

[ L + L ]

[ L ]

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A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have

DiKsa [7]

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

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3 years ago

Jack and jill are married. this year jack earned $72,000 and jill earned $80,000 and they received $4,000 of interest income fro

MaRussiya [10]

<span>If Jack is filing married-filing-separate he would report $76,000 gross income as head of household.</span>

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3 years ago

A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?

jok3333 [9.3K]

Answer:

$S_a_v_e_r_a_g_e=48km/h$

Explanation:

Ok, the average speed can be calculate with the next equation:

$S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}$ (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

$Total\hspace{3}distance=2*d$

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

$Total\hspace{3}time=t1+t2$

From basic physics we know:

$t=\frac{d}{S1}$

so:

$t1=\frac{d}{S1}$

$t2=\frac{d}{S2}$

Using the previous information in equation (1)

$S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }$

Factoring:

$S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}$ (2)

Finally, replacing the data in (2)

$S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h$

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3 years ago

Doc Brown has calculated his Delorean can accelerate at a rate of 2.52 m/s/s. How

GREYUIT [131]

Answer:

304.89m

Explanation:

Given

acceleration a = 2.52m/s²

final speed v = 39.2m/s

initial speed = 0m/s (car accelerates from rest)

Using the equation of motion below to get the distance of Doc brown from Marty;

v² = u²+2as

substitute the given parameters

39.2² = 0²+2(2.52)s

1536.64 = 0+5.04s

divide both sides by 5.04

1536.64/5.04 = 5.04s/5.04

rearrange the equation

5.04s/5.04 = 1536.64/5.04

s = 304.89m

Hence He and Marty must stand at 304.89m to allow the car to accelerate from rest to a speed of 39.2 m/s?

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4 years ago

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$

$Y=(24cm/tan(32.12))$

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0

3 years ago

Find the dimensional formulas=ut+1/2at^2​

Find the dimensional formulas=ut+1/2at^2