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NARA [144]
3 years ago
14

Please help me with this assignment!

Physics
2 answers:
makkiz [27]3 years ago
5 0

Answer:

lol you got the answer just need points sorry

Explanation:

Licemer1 [7]3 years ago
3 0

Answer:

Ice cube, molecules, melting, zero degrees, liquid, faster, temperature, 100 degrees, steam

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A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.
finlep [7]

Answer:

The acceleration expressed in the new units is 114.048 Km/h^{2}

Explanation:

To convert from m/s^{2} to Km/h^{2} it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{3600s}{1h})^{2}

8.8\frac{m}{s^{2}}.(\frac{1Km}{1000m}).(\frac{12960000s^{2}}{1h^{2}})

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

114.048 Km/h^2

So the acceleration expressed in the new units is 114.048 Km/h^2.

6 0
4 years ago
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
⚠️ PLEASE HELP ⚠️
Svetllana [295]

Answer:

-79.6

-80

-81.2

Explanation:

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