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Sindrei [870]
4 years ago
9

A 15 kg block is on a ramp which is inclined at 20o above the horizontal. It is connected by a string to a 19 kg mass which hang

s over the top edge of the ramp. Assuming that frictional forces may be neglected, what is the magnitude of the acceleration of the 19 kg block
Physics
1 answer:
12345 [234]4 years ago
7 0

Answer:

The magnitude of the acceleration of the 19 kg block is 1.414 m/s²

Explanation:

From Newton's second law of motion;

F_{Net} = ma

where;

m is the mass of the objects involved, kg

a is the acceleration of the object, m/s²

different forces on the block and string

⇒force due to 15 kg block

=mgcosθ = 15×9.8×cos20 = 15×9.8×0.9396

= 138.12 N

⇒Tensional Force on 19 kg mass:

T = mg = 19×9.8 = 186.2 N

F_{Net} = T-mg = a(m_1+m_2)

186.2 - 138.12 = a(15+19)

48.08 =  a(34)

a = 48.08/34

a = 1.414 m/s²

Therefore, the magnitude of the acceleration of the 19 kg block is 1.414 m/s²

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3 years ago
After being struck by a bowling ball, a 1.7 kg bowling pin sliding to the right at 3.8 m/s collides head-on with another 1.7 kg
Alchen [17]

Answer:

3 m/s

Explanation:

Parameters given:

Mass of first bowling pin, m = 1.7 kg

Initial velocity of first bowling pin, u = 3.8 m/s

Final velocity of first bowling pin, v = 0.8 m/s

Mass of second bowling pin, M = 1.7 kg

Initial velocity of second bowling pin, U = 0 m/s

Let the final velocity of the second bowling pin be V

Using the principle of conservation of momentum:

Total initial momentum = Total final momentum

m*u + M*U = m*v + M*V

(1.7 * 3.8) + 0 = (1.7 * 0.8) + (1.7 * V)

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7 0
4 years ago
A 1800 kg car moving south at 17.5 m/s collide with a 2800 kg car moving north. The cars stick together and move as a unit after
weqwewe [10]

Answer:

   v₀₂ = -2.67 m / s

Explanation:

Let's use the conservation of the moment, for this we define a system formed by the two cars.

Consider the north direction as positive and the subscript 1 will be used for car 1 and the subscript 2 for the second car

Initial instant. Before the crash

          p₀ = -m₁ v₀₁ + m₂ v₀₂

Final moment. Right after the crash

          p_f = (m₁ + m₂) v

           p₀ = p_f

           -m₁ v_{o1} + m₂ v_{o2} = (m₁ + m2) v_{f}

         v₀₂ = \frac{(m_1 +m_2) v }{m_2}+  \frac{m_1 v_{o1}  }{m_{1} }

 

let's calculate

          v₀₂ = \frac{(1800 +2800) 5.22 + 1800 (-17.5)}{2800}

          v₀₂ = -2.67 m / s

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