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densk [106]
3 years ago
9

If A vector = i^-j^+k^ then unit vector in the direction of A vector

Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0

If <em>A</em> = <em>i</em> - <em>j</em> + <em>k</em>, then the magnitude of <em>A</em> is

||<em>A</em>|| = √(1² + (-1)² + 1²) = √3

Then the unit vector in the direction of <em>A</em> is 1/||A|| multiplied by <em>A</em> :

<em>u</em> = <em>A</em>/||<em>A</em>|| = (<em>i</em> - <em>j</em> + <em>k</em>)/√3

(choice D)

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A 56 kg woman climbs the stairs to a height of 5 m. Determine: a) what mechanical work
Aleonysh [2.5K]

Answer:

a) W = 2746.8[J]

b) W = 2992.05 [J]

Explanation:

Work is defined as the product of force by distance. We must bear in mind that the force that performs the work is the one that acts in the same direction of displacement.

For this case, we must calculate the weight of the woman, the weight is defined as the product of mass by gravity.

w=m*g\\

where:

w = weight [N] (units of Newtons]

m = mass = 56 [kg]

g = gravity acceleration = 9.81 [m/s²]

w=56*9.81\\w=549.36[N]

a)

W=F*d

where:

F = weight = 549.36[N]

d = distance = 5 [m]

W = 549.36*5\\W = 2746.8[J]

b)

The new mass will be the combination of the mass of the woman plus that of the load.

m_{new} = 56+5\\m_{new}=61[kg]

w_{new}=61*9.81\\w_{new}=598.41[N]

The new work done.

W =598.41*5\\W=2992.05[J]

4 0
3 years ago
Where is the epicenter (it's about earthquakes)
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Answer:

The epicenter is the point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins.

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Explain why wedges and screws are actually types of inclined planes.
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<span>The screw is really just an inclined plane covered around with a tiny pole. The wedge is definitely an inclined plane, since it starts with a point, then rises getting thicker, as an inclined plane. </span>
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4 years ago
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Drosera magnifica is an organism that was first discovered in 2015.
mestny [16]

Answer:

The correct answer is - Plantae.

Explanation:

Drosera m<em>agnifica</em> is discovered in 2015 for the first time and the characteristics this organism's cell shows are -

- permanent vacuoles

- surrounded by cellulose layer

Vacuoles are present in both Plantae and Animalia kingdom of the eukaryotic organism but in animal cells, there are small and numerous vacuoles present and they are not permanent whereas in plant cells vacuoles are present permanently.

The cell of an animal cell has no surrounding layer other than cell membrane while in the plant cell there is a supporting and protecting layer of cellulose cell wall present.

On the basis of the given characteristics, it is confirmed that the Drosera magnifica belongs to Plantae kingdom.

3 0
3 years ago
A 12.0 kg block rests on an inclined plane. The plane makes an angle of 31.0° with the horizontal, and the coefficient of fricti
allochka39001 [22]

Answer:

The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

Explanation:

Given that,

Mass = 12.0 kg

Angle = 31.0°

Friction coefficient = 0.158

Mass of second block = 38.0 kg

Using formula of frictional force

f_{\mu} = \mu N....(I)

Where, N = normal force

N = mg\cos\theta

Put the value of N into the formula

N =12\times9.8\times\cos 31^{\circ}

N=100.80\ N

Put the value of N in equation (I)

f_{mu}=0.158\times100.80

f_{mu}=15.9264\ N

Now, Weight of second block

W = mg

W=38.0\times9.8

W=372.4\ N

The horizontal force is

F = mg\sintheta

F=12\times9.8\times\sin 31^{\circ}

F=60.5684\ N....(II)

(I). We need to calculate the acceleration

a=m_{2}g-\dfrac{f_{\mu}+mg\sin\theta}{m_{1}+m_{2}}

a=\dfrac{372.4-(15.9264+60.5684)}{12+38}

a=5.92\ m/s^2

(II). We need to calculate the tension in the rope

m_{2}g-T=m_{2}a

-T=38\times5.92-38\times9.8

T=147.44\ N

Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

5 0
4 years ago
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