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3 years ago

A 100 g wire is held under a tension of 250 n with one

Physics

1 answer:

sashaice [31]3 years ago

6 0

So??
what is the question??

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A rocket is fired vertically upwards starting frkm rest. It accelerates at 30m/s for 4secs. At the end of 4secs it runs out of f

HACTEHA [7]

Answer:

t = 16.5 s

Explanation:

First we apply first equation of motion to the accelerated motion of the rocket:

$v_{f1} = v_{i1} + at_{1}$

where,

vf₁ = final speed of rocket during accelerated motion = ?

vi₁ = initial speed of rocket during accelerated motion = 0 m/s

a = acceleration of rocket during accelerated motion = 30 m/s²

t₁ = time taken during accelerated motion = 4 s

Therefore,

$v_{f} = 0\ m/s + (30\ m/s^2)(4\ s)\\\\v_{f} = 120\ m/s$

Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:

$v_{f2} = v_{i2} + g t_{2}$

where,

vf₂ = final speed of rocket after engine is off = 0 m/s

vi₂ = initial speed of rocket after engine is off = Vf₁ = 120 m/s

g = acceleration of rocket after engine is off = - 9.8 m/s² (negative sign for upward motion)

t₂ = time taken after engine is off = ?

Therefore,

$0\ m/s = 120\ m/s + (- 9.8\ m/s^2)(t_{2})\\\\t_{2} = \frac{120\ m/s}{9.8\ m/s^2}\\\\t_{2} = 12.25\ s$

So, the time taken from the firing position till the stopping position is:

$t = t_{1} + t_{2}\\\\t = 4 s + 12.5 s$

8 0

3 years ago

What kind of modulator is used in this scenario?

sladkih [1.3K]

The unmodulated carrier wave is going into the box, and when it comes out, its AMPLITUDE has been modulated.

3 0

3 years ago

Read 2 more answers

An inclined plane allows you to do ________ work with ________ force.

Alecsey [184]

Answer is C. Less, more

7 0

3 years ago

Read 2 more answers

Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!

Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

$\theta_8 = 1.12^{\circ}$ = angle to 8th max (note how m = 8 since we're comparing this to the form $\theta_m$ )

$x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}$ (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

$d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}$

Make sure your calculator is in degree mode.

-----------------

Method 2

$\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\$

-----------------

Method 3

$\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\$

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0

3 years ago

Object A weighs 750 N on earth. Object B weighs 750 N on Jupiter.

Katarina [22]

The object A has the greater mass compared to object B.

<u>Explanation: </u>

The weight of any object on any planet is the measurement of gravity’s influence acting on the mass of the object. So for Earth, the acceleration will be acting on the object A’s mass (m) in Earth leading to the weight of the object A as 750 N.

While the acceleration of Jupiter will be acting on the object B’s mass kept in Jupiter to attain the weight of 750 N. So, the mass of both the objects at their respective planet will vary depending on the acceleration of each planet. We can check this as below:

$\text { object A's weight }=\text { m of object A } \times \text { Acceleration of Earth}$