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n200080 [17]
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?​

Physics
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
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