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3 years ago

A helicopter ascends at a constant speed. If it rises 173.5 m in 17.9 s, what is it’s speed?

1 answer:

5 0

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V2:(I. other words..) In what terms do you want the answer expressed?

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m/s
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A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o

Nitella [24]

Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

7 0

3 years ago

A carpet is to be installed in a room of length 9.72 meters and width 17.30 of the room retaining the proper number of significa

Usimov [2.4K]

Answer:

168 m^2, 380 m^2

Explanation:

length of the room, l = 9.72 m

width of the room, b = 17.30 m

Area of teh rectangle is given by

A = length x width

So, A = 9.72 x 17.30 = 168.156 m^2

the significant digits should be 3 in the final answer

So, A = 168 m^2

Now length = 72 m

width = 17.39 feet = 5.3 m

Area, A = 72 x 5.3 = 381.6 m^2

There should be two significant digits in the answer so, by rounding off

A = 380 m^2

3 0

3 years ago

At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she

insens350 [35]

Answer:

$1.34 * 10^{3}$ m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, $v_{x}$ = 990m/s

Surface charge density, σ = 2.34 x $10^{-9}$ C/ $m^{2}$

We need to calculate the speed at time, t = 7.0 * $10^{-8}$ s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

$v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E = 132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/ $s^{2}$

Therefore, at time, t = 7.0 * $10^{-8}$ , we can use one of the equations of linear motion to find the velocity in the y axis:

$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$ )

$v_{y}$ = 910 m/s

∴ $v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

v = 1344.69 m/s = $1.34 * 10^{3}$ m/s

8 0

4 years ago

Lee knew that certain events occur to create new material in the process of sea-floor spreading. Rearrange the steps into the co

Marina86 [1]

Answer:

Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other. ... As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense.

Explanation:

did this help???

8 0

3 years ago

. Potential energy is a vector quantity True or false?

DedPeter [7]

False. Energy in any form is a scalar quantity.

6 0

3 years ago