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3 years ago

What is the role of equations in this course?

1 answer:

4 0

That will depend on which course you're talking about. It will be a minor role in, say, Maritime Law or Comparitive Religion, but a major one in, say, Particle Physics or Linear Algebra.

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

3 years ago

A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials’ boat pulls

babunello [35]

Answer:

They are 7.4m apart.

Explanation:

Here we have a parabolic motion problem. we need the time taken to land so:

$Y=Yo+Vo*t+\frac{1}{2}*a*t^2$

considerating only the movement on Y axis:

$0=4.6-(9.81)*t^2\\t=0.68s$

Because we have a contant velocity motion on X axis:

$xs=vs*t\\xs=17m/s*(0.68s)\\xs=11.6m$

and

$xg=vg*t\\xg=28m/s*(0.68s)\\xg=19m$

the distance between them is given by:

$d=|xg-xs|\\d=7.4m$

7 0

3 years ago

The unit for measuring the rate at which light energy is radiated from a source is the

zzz [600]

The correct answer is letter D. candela. The unit for measuring the rate at which light energy is radiated from a source is the candela. L<span>umen is the unit for measuring the total amount of visible light emitted by a source. Lux is lumen per square meter. </span>

5 0

3 years ago

A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont

Viktor [21]

Answer:

1.826m/s

Explanation:

E=1/2*k*(∆L)^2

E=1/2*mV^2

6 0

3 years ago

A bar magnet is held above the center of a conducting ring in the horizontal plane. The magnet is dropped so it falls lengthwise

Alenkinab [10]

Explanation:

Since, it is given that the magnet drops and falls lengthwise towards the canter of the ring. As a result, change in magnetic flux will occur which tends to induce an electric current in the ring.

Therefore, a magnetic field is also produced by the ring itself which will actually oppose or repel the magnet.

Thus, we can conclude that the falling magnet be repelled by the ring due to the magnetic interaction of the magnet and the ring.

7 0

3 years ago