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timofeeve [1]
3 years ago
8

Danielle is applying 25.2 N to the right on her sled. The sled has a mass of 15.7 kg, and

Physics
1 answer:
Sever21 [200]3 years ago
4 0

The applied force is parallel to the ground, so the net vertical force on the sled is 0. By Newton's second law, this means the normal force and weight of the sled cancel out:

<em>n</em> - <em>w</em> = 0

<em>n</em> = <em>w</em>

<em>n</em> = <em>m g</em>

<em>n</em> = (15.7 kg) (9.80 m/s²)

<em>n</em> = 153.86 N

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3 years ago
A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 22.
Bogdan [553]

Answer:

A. 1.172 metres

B. 6.82 Ns

C. 4.796 m/s

Explanation:

The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.

The 1.40 kg block is at rest so velocity is zero and has no momentum.

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6.82 = 1.422v

V = 6.82/1.422

V = 4.796 m/s

How high the block will rise after the bullet is embedded is given by

H = (U²Sin²tita)/2g

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6 0
3 years ago
A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 de
jeyben [28]

Answer:

  c_{e1} = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded          Qh = m1 ce1 (T_{h} -T_{f})

Heat absorbed     Qc = m2 ce2 (T_{f} - T₀)

Body 1 is metal and body 2 is water .  Where m are the masses of the two bodies, ce their specific heat and T the temperatures

      Qh = Qc

      m₁ c_{e1} (T_{h}- T_{f}) = m₂  c_{e2} (T_{f} - T₀)

we clear the specific heat of the metal

      c_{e1} = m₂  c_{e2} (T_{f} - T₀) / (m₁ (T_{h}-T_{f}))

     c_{e1}= 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

      c_{e1} = 209.2 (9.36) / (75 78.85)

      c_{e1} = 1958.11 / 5913.75

     c_{e1} = 0.331 J / g ° C

5 0
3 years ago
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