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Sergeu [11.5K]
2 years ago
10

I need help with those 2.

Physics
1 answer:
Nimfa-mama [501]2 years ago
5 0

Hi there!

Question 1:

We know that Impulse = Δp = mΔv, so:

I = 0.045(70) = 3.15 Ns

I = F · t, so:

3.15/0.1 = F

F = 31.5 N

Question 2.

We can find the impulse using:

I = Ft

I = 300 · 0.04 = 12 Ns

Find the change in velocity using:

I = mΔv

12/0.13 = Δv = 92.31 m/s

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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
A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure
Alex787 [66]
V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

       t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

     v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
     speed = |-2| m/s = 2 m/s
 
Moving right
     V > 0 => t^2 - 9t + 18 > 0
     (t - 6)(t - 3) > 0

     => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)

    

Going faster and slowing dowm

acceleration, a = v' = 2t - 9
     a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
     Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
     


4 0
3 years ago
A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?
Rama09 [41]
The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J
4 0
3 years ago
Read 2 more answers
Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo
Vlad1618 [11]

Answer:

Tension T1 is less than tension T2.

T1 < T2

Explanation:

According to given data,

mass of box A ( mA) is grater than mass of box B (mB)

we can write,

m(A) > m(B)

Newton's second law states that:

Tension of object is directly proportional to the mass of the system.

T ∝ m

here Boxes A and B are being pulled to the right on a frictionless surface,

so Tension T1 generates due to the mass of box A m(A)

and Tension T2 arises due to mass of the system m(A) + m(B)

Thus tension T1 will be less than tension T2

T1 < T2

learn more about Tension force here:

<u>brainly.com/question/13175014</u>

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#SPJ4

8 0
2 years ago
4. A 1,000-kilogram satellite completes a uniform circular orbit of radius 8.0 x 10 meters as
lesantik [10]

Answer:

Zero

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement of the object

In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so \theta=90^{\circ} and cos \theta =0. Therefore, the work done by gravity is also zero.

5 0
3 years ago
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