A spring has 100 J of EPE when it is stretched 2 m. What is the spring constant?

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what d

Aleks04 [339]

I’m pretty sure it’s b

8 0

3 years ago

A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

7 0

3 years ago

4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx

frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

$Ix=.\frac{Req}{Rx} *Itotal$ (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

$Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2} }$ (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

$Req.=\frac{ 100*R1}{R1+100}$ (3)

Now substituting (3) in (1).

$25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}$

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0

3 years ago

A spacecraft is flying away from the moon toward earth.

Alekssandra [29.7K]

Answer:

it will decrease

Explanation:

According to the law of universal gravitation, the gravitational force exerted by the moon on the spacecraft is equal to the product of their masses and inversely proportional to the square of the distance that separates them. Therefore, as the spacecraft moves away, its distance increases and the force of attraction exerted by the moon decreases.

4 0

3 years ago

Dr. james van allen and his group of united states scientists discovered a set of ___________.

Lubov Fominskaja [6]

The answer is Radiation Belts.

6 0

4 years ago

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