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3 years ago

14

How do you do this? I know that the Density equation is D=m/v but these are in kilograms and Liters...so do I convert it to gram

s and ml? Or do I just solve the equation like I normally would?

1 answer:

BARSIC [14]3 years ago

8 0

Explanation:

The problem doesn't specify that the units have to be g/mL, so you can calculate the density in kg/L without converting the mass or volume.

Just make sure that either way, you write the units.

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To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago

the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a

inysia [295]

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

The initial velocity of the car, u = 0 m/s

The displacement of the ca, S = 1100 m

The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

S = ut + ½ at²

Substituting the given values,

1100 = 0 + ½ x a x 14²

a = 11.22 m/s²

Using the III equation of motion

v² = u² + 2 as

v = √(2as) (∵ u = 0)

Substituting,

v = √(2 x 11.22 x 1100)

= 157.11 m/s

The average speed of the car,

$V=\frac{0+65+157.11}{3}$

V = 74.04 m/s

Hence, the average velocity of the car is, V = 74.04 m/s

4 0

3 years ago

When a transformer is step down then

Vera_Pavlovna [14]

Answer:

Hey!

Your answer should be D!

Explanation:

In a transformer Np / Ns is called the voltage ratio. If Ns is less than Np then Vs is less than Vp. This is called a step-down transformer as the voltage is reduced.

(source from google.com!)

3 0

3 years ago

Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

3 years ago

What is the most appropriate SI unit to express the speed of a cyclist in a 10-km race? km/s cm/h km/h mm/s

Murljashka [212]

You sure wouldn't want something like cm/s or (yikes cm/hr). You want a reasonable number for sports usually between 0 and 100

Km / hour would be a good choice.

The next town to where I live is 25 km away. On a good day, I can make it there in about 3/4 of an hour.

Speed = 25 km / 0.75 hour = 33.3 km/hour. That's actually a little fast most of the time. But you should understand what I mean.

5 0

3 years ago