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Anastasy [175]
3 years ago
5

After initially picking the stone up off the ground, a man exerts a 148 N force upwards on a 10 kg stone. What is the accelerati

on of the stone?
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

5m/s²

Explanation:

The force exerted on an object is calculated using the formula ;

F = ma

Note that the upward force (F) is 148N

Where;

F = force (N)

m = mass (kg)

a = acceleration (m/s²)

The downward force towards gravity is F' = mg

F' = 10 × 9.8m/s²

F' = 98N

Net force (F - F') = m × a

148 - 98 = 10 × a

50 = 10a

a = 50/10

a = 5m/s²

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7 0
3 years ago
Will mark as BRAINLIEST.....
Jlenok [28]

Answer: initially the packet was ascending up with the balloon.

Taking upward as positive direction;

initial velocity, u = 4.9 m/s

final velocity = v m/s

initial height, h₁ = 245 m  

final height, h₂ = 0

a = -9.8 m/s²

time taken = t seconds

s = ut + 0.5at²

⇒ (h₂-h₁) = ut + 0.5at²

⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²

⇒ -245 = 4.9t - 4.9t²

⇒ 4.9t² -4.9t -245 =0

Solving it, we get  t = 7.59s

v = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/s

So velocity is 69.48 m/s downward

Explanation:

5 0
4 years ago
A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for
Elena-2011 [213]

Answer:

1) 10.1 s  2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

  • In the vertical direction, as the bomb is dropped, its initial velocity is 0.
  • So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

       \Delta y = \frac{1}{2}*a*t^{2} (1)

  • where Δy = -500 m (taking the upward direction as positive).
  • a=-g=-9.8 m/s²
  • Replacing these values in (1), and solving for t, we have:

       t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s

  • The time required for the bomb to reach the earth is 10.1 s.

2)

  • In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
  • As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

       x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.

3)

  • The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

  • In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

        v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s

5)

  • If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
  • So, when the bomb hits the ground, the helicopter will be exactly over it.

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Explanation:

8 0
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