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krek1111 [17]
3 years ago
6

Vector A is 3 units length and points along the positive x-axis vector B is 4 units in length and points along negative y-axis .

find magnitude and direction of the vector (a) A+B (b)A-B
Physics
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

a) < 3 , -4 >

b) < 3 , -4 >

Explanation:

a) If you can imagine this, adding vectors is like putting them "tip to tail", where you put the beginning point of vector B to the end point of vector A (or vice versa).  Your new vector (A+B) would be from the "tip" of vector A to the "tail" of vector B.

Mathematically, this is the same as adding the x-components of each vector together as well as the y-components.

Vector A: 3 units along the positive x-axis: < 3 , 0 >

Vector B: 4 units along the negative y-axis: < 0 , -4 >

A+B = < 3 , 0 > + < 0 , -4 > = < (3+0) , (0+(-4)) > = < 3 , -4 >

b) Subtracting is like adding a negative, so you could use the same "tip to tail" visual by adding the negative of vector B instead (which is B in the opposite direction).

Vector A: < 3 , 0 >

Vector B: < 0 , -4 >

Vector -B: < 0 , 4 >

A-B = A+(-B) = < 3 , 0 > + < 0 , 4 > = < (3+0) , (0+4) > = < 3 , 4 >

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A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0
3 years ago
At the moment t = 0, a 24.0-v battery is connected to a 5.00-mh coil and a 6.00-ω resistor. (a) immediately thereafter, how does
agasfer [191]
At the moment the answer is Yeet
5 0
3 years ago
a spring gun initially compressed 2cm fires a 0.01kg dart straight up into the air. if the dart reaches a height it 5.5m determi
Vikki [24]

Answer:

2697.75N/m

Explanation:

Step one

This problem bothers on energy stored in a spring.

Step two

Given data

Compression x= 2cm

To meter = 2/100= 0.02m

Mass m= 0.01kg

Height h= 5.5m

K=?

Let us assume g= 9.81m/s²

Step three

According to the principle of conservation of energy

We know that the the energy stored in a spring is

E= 1/2kx²

1/2kx²= mgh

Making k subject of formula we have

kx²= 2mgh

k= 2mgh/x²

k= (2*0.01*9.81*5.5)/0.02²

k= 1.0791/0.0004

k= 2697.75N/m

Hence the spring constant k is 2697.75N/m

7 0
3 years ago
A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
bogdanovich [222]

Answer:

A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

5 0
3 years ago
Question 18 (2 points)
fomenos

Answer:

A.Moving electric charges (electrons in a circuit) creates a magnetic field and

a magnetic field can cause an electric charge to move (electricity).

Explanation:

4 0
3 years ago
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