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UNO [17]
3 years ago
7

what is the mass of a cannon ball if a force 2500 N gives the cannon ball an acceleration of 200 m/s squared?

Physics
1 answer:
nignag [31]3 years ago
8 0

Using Newton's second law of motion:

F=ma ;  [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)


Given:                      Find:                   Formula:                  Solve for m:

F: 2500N                 mass:?                F=ma Eq.1              m=F/a  Eq. 2

a= 200m/s^2  


Solution:

Using Eq.2

m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg

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strojnjashka [21]

Explanation:

We know

  • C/5 = F -32/9
  • C/5 = 75 -32/9
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  • C ≈ 25°C
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2 years ago
Who is the father of electricity?
Tpy6a [65]
William Gilbert is known as the father of electricity.
7 0
3 years ago
Read 2 more answers
A +26.3 uC charge qy is repelled by a force
Musya8 [376]

Answer:

+1.46×10¯⁶ C

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C

Force (F) = 0.615 N

Distance apart (r) = 0.750 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Charge 2 (q₂) =?

The value of the second charge can be obtained as follow:

F = Kq₁q₂ / r²

0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²

0.615 = 236700 × q₂ / 0.5625

Cross multiply

236700 × q₂ = 0.615 × 0.5625

Divide both side by 236700

q₂ = (0.615 × 0.5625) / 236700

q₂ = +1.46×10¯⁶ C

NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C

5 0
2 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
The CERN particle accelerator is circular with a circumference of 7.0 km.
Contact [7]

Answer:

a_c=2.0196\times 10^{13}\ m/s^2

F=3.37273\times 10^{-14}\ N

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

v = Speed of proton = 0.5c = 0.5\times 3\times 10^8=1.5\times 10^8\ m/s

Circumference of the colider is 7 km

P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m

a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2

Centripetal acceleration is 2.0196\times 10^{13}\ m/s^2

F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N

Force on protons is 3.37273\times 10^{-14}\ N

8 0
3 years ago
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