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elena-14-01-66 [18.8K]

3 years ago

12

What happened to the amplitude from wave A to wave B?

1 answer:

Ad libitum [116K]3 years ago

6 0

The amplitude was doubled

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A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu

Pepsi [2]

Answer:

$v = 10.89\ m/s$

Explanation:

given,

radius of loop = 12.1 m

to find the minimum speed transverse by the rider to not to fall out upside down

centripetal force = $\dfrac{mv^2}{r}$

gravitational force = m g

computing both the equation]

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{rg}$

$v = \sqrt{12.1 \times 9.8}$

$v = \sqrt{118.58}$

$v = 10.89\ m/s$

5 0

3 years ago

You are at the controls of a particle accelerator, sending a beam of 2.10×107 m/s protons (mass m) at a gas target of an unknown

Kipish [7]

Answer:

a

The mass is $m_2 =21.75*10^{-27} \ kg$

b

The velocity is $v_2 = 3.0*10^{6} m/s$

Explanation:

From the question we are told that

The speed of the protons is $u_1 = 2.10*10^{7} m/s$

The mass of the protons is $m$

The speed of the rebounding protons are $v_1 = -1.80 * 10^{7} \ m/s$

The negative sign shows that it is moving in the opposite direction

Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as

$m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1$

Where $m_1$ is the mass of a single proton

So substituting values

$m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1$

$m_2 =13 m_1$

The mass of on proton is $m_1 = 1.673 * 10^{-27} \ kg$

So $m_2 =13 ( 1.673 * 10^{-27} )$

$m_2 =21.75*10^{-27} \ kg$

Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as

$m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2$

Now $u_2$ because before collision the the nucleus was at rest

So

$m_1 u_1 = m_1 v_1 + m_2v_2$

=> $v_2 = \frac{m_1(u_1 -v_1)}{m_2}$

Recall that $m_2 =13 m_1$

So

$v_2 = \frac{m_1(u_1 -v_1)}{13m_1}$

=> $v_2 = \frac{(u_1 -v_1)}{13}$

substituting values

$v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}$

$v_2 = 3.0*10^{6} m/s$

7 0

3 years ago

Suppose you are sitting on a rotating stool holding a 2 kgmass

Aloiza [94]

Answer:Increase

Explanation:

Given

You are holding 2 kg mass in each outstreched hand

If the masses are dropped then Moment of inertia will decease by $2mr^2$

Where m=2 kg

r=length of stretched arm

Since angular momentum is conserved therefore decrease in Moment of inertia will result in increase of angular velocity

as I $\omega$ =constant

I=Moment of inertia

$\omega$ =angular velocity

3 0

3 years ago

What are the general characteristics of a solid?

Arisa [49]

Fixed volume and shape
-Particles are locked into place
Not easily compressible
-Little free space between particles
Does not flow easy
-Particles cannot move past each other

4 0

3 years ago

A spring balance used to weigh candy is built with a spring the spring stretches 3.00 cm when a 15 n weight is placed in the pan

ira [324]

Answer: 5.9cm

Explanation:

15N/3cm=5

Convert 3kg to N (29.42N)

Use the 5 from earlier to divide 29.42/5=5.884

Round to 5.9

5.9cm is the answer

(I don’t know if I solved this the correct way but I got it right. 5.9 is the correct answer)

6 0

3 years ago