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Setler [38]
3 years ago
8

two technicians are discussing fuses technician a says that current flows should be 20% less than fuse rating technician B says

that an orange blade fuse will have an amperage rating of nine which technician is correct
Physics
1 answer:
MrMuchimi3 years ago
3 0
The technician A is correct. The current should be 20%  or any given percentage lower than the current for which the fuse is created.

The technician B argues that orange blade fuses have a 9 Amp rating. But they usually have a 40A rating
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3 A rocket of mass 1200 kg is travelling at 2000 m/s. It fires its engine for 1 minute. The forward thrust provided by the rocke
natita [175]

The impulse shared by the object equals the difference in momentum of the object. In equation form,

F • t = m • Δ v. In a collision, objects experience an impulse; the impulse causes and is equal to the difference in momentum.

<h3>How to calculate  thrust provided by the rocket engines is 10 kN (10 000 N).?</h3>

a)There is this impulse-momentum change equation.

where m$ is the mass of a body, $F$ is a force acting to the body, $t$ is time and $D E L A T A N\}=V_{2}-V_{1}$ is the change of velocity.We consider everything is happen along a straight line, and gravitation does not participate.So, the increase of momentum is $\mathrm{F}^{*} \mathrm{t}=10000 \mathrm{~N} * 60$ seconds $=600000 \mathrm{~N}^{*} \mathrm{~s}=600000\left(\mathrm{~kg}^{*} \mathrm{~m}\right)^{*} \mathrm{~s} / \mathrm{s}^{\wedge} 2=600000 \mathrm{~kg}{ }^{*} \mathrm{~m} / \mathrm{s}$.

We consider everything exits happen along a straight line, and gravitation does not participate.

So, the increase of momentum is F×t = 10000 N × 60 seconds = 600000 N*s = 600000 (kg*m)*s/s^2 = 600000 kg*m/s.

$$\Delta(\mathrm{V})=\frac{\mathrm{F.t}}{\mathrm{m}}=\frac{600000}{1200}=500 \mathrm{~m} / \mathrm{s} .$$

New velocity after  engine was firing during 60 seconds is 2000 + 500 = 2500 m/s.

To learn more about Impulse-momentum, refer

brainly.com/question/20586658

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2 years ago
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Anastaziya [24]
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Answer:

0.8c and -0.14c

Explanation:

The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c

Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.

The Lorentz transform for velocity is:

u = (u' + v) / (1 + (u' * v) / c^2)

u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c

The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.

u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c

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