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Sedbober [7]
3 years ago
14

Two vectors A and B are at right angles to each other. The magnitude of A is 3.00. What should be the length of B, so that the m

agnitude of their vector sum is 5.00?
Physics
1 answer:
AfilCa [17]3 years ago
7 0

Answer: length of B =4.00

Explanation:

for  the vectors A and B and the angle between them as  x.

Magnitude of the sum of A and B is  given as = √(A²+B²+2ABcosx

where

Magnitude of A  = 3.00

Magnitude of the sum of A and B is  5.00

5.00=√(A²+B²+2ABcos90°

5.00= √3² +b² +0

5²= 3² +b²

25=9+b²

b²= 25-9

b² = 16

b=  √16

b= 4

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Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as \mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}.

Where, {V}_{i} is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

{V}_{i} =3 m/s, V = 0 m/s, and  ∆s = 2 m

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0 = 9 - 4a

4a = 9

a=2.25 \mathrm{m} / \mathrm{s}^{2}

a=2.25 \mathrm{m} / \mathrm{s}^{2} is the acceleration.

Calculating Coefficient of friction:

\mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}

Compare the above equation

\mu \times m \times g=m \times a

Cancel "m" common term in both L.H.S and R.H.S

\text { Equation becomes, } \mu \times g=a

\text { Coefficient of friction } \mu=\frac{a}{g}

\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}

\mu=\frac{2.25}{9.8}

\mu=0.229

Hence coefficient of friction is 0.229.

calculating force:

\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}

\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

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iii)26.5 m/s or 95.4 km/hr

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Firstly convert 90km/hr to m/s

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(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

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Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

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In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

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(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

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Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

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