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disa [49]
2 years ago
5

Calculate the x and y components for the vector below. Show all work. d = 43.2 m [43.0° W of N]​

Physics
1 answer:
Archy [21]2 years ago
8 0

The x and y components of  the vector is 31.59 m and 29.46 m respectively.

<h3>X component of the vector</h3>

The x component is calculated as follows;

v(x) = v cos(43)

v(x) = 43.2 m x cos(43) = 31.59 m

<h3>Y component of the vector</h3>

The y component is calculated as follows;

v(y) = v sin(43)

v(y) = 43.2 m x sin(43) = 29.46 m

Thus, the x and y components of  the vector is 31.59 m and 29.46 m respectively.

Learn more about vectors here: brainly.com/question/25811261

#SPJ1

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What is the maximum distance that a 60.-watt motor may vertically lift a 90.-newton?weight in 7.5 seconds?a. 2.3mb. 5.0mc. 140md
Naddik [55]

Answer:

b. 5.0 m

Explanation:

Power: Power can be defined as the rate at which work is done.

The S.I unit of power is watt (W).

P = W/t ............................................... Equation 1

Where p = power, W = work, t = time.

But

W = F×d..................................... Equation 2

Where F = force, d = maximum distance.

Substituting equation 2 into equation 1

P = F×d/t

making d the subject of the equation above,

d = Pt/F......................................................... Equation 3

<em>Given: F = 90 Newton, P = 60 watt, t = 7.5 seconds.</em>

<em>Substituting these values into equation 3,</em>

<em>d = 60×7.5/90</em>

<em>d = 5 m</em>

Thus the maximum distance is = 5 m

The right option is b. 5.0 m

5 0
3 years ago
A ball of mass is released from rest at a height of 30 how fast is it going when it hits the ground
elena55 [62]

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If the height is in metres, the speed is 24.25m/s

7 0
3 years ago
A 2100 g block is pushed by an external force against a spring (with a 22 N/cm spring constant) until the spring is compressed b
Vilka [71]

Answer:

6.5e-4 m

Explanation:

We need to solve this question using law of conservation of energy

Energy at the bottom of the incline= energy at the point where the block will stop

Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=\frac{1}{2} kx^{2} +PE1

Energy at the point where the block will stop consists of only gravitational potential energy=PE2

Hence from Energy at the bottom of the incline= energy at the point where the block will stop

⇒\frac{1}{2} kx^{2} +PE1=PE2

⇒PE2-PE1=\frac{1}{2} kx^{2}

Also PE2-PE2=mgh

where m is the mass of block

g is acceleration due to gravity=9.8 m/s

h is the difference in height between two positions

⇒mgh=\frac{1}{2} kx^{2}

Given m=2100kg

k=22N/cm=2200N/m

x=11cm=0.11 m

∴2100*9.8*h=\frac{1}{2}*2200*0.11^{2}

⇒20580*h=13.31

⇒h=\frac{13.31}{20580}

⇒h=0.0006467m=6.5e-4

7 0
4 years ago
When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc
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Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

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          w² = mg d / I

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        I = ¼ m r2 + ⅓ m L2

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now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

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let's substitute

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        w² = g L / (½ r² + 2/3 L²)

        L >> r

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4 years ago
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Mamont248 [21]
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First find the notations:

2×10^-3
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Then divide:

0002÷25000=8E-9

Your answer:
=8 x 10-8
3 0
3 years ago
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