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Veseljchak [2.6K]
3 years ago
13

1. Calculati greutatea unui sac cu 5 kg de cartofi într-o zonă in care acceleratia gravitatională

Physics
1 answer:
harkovskaia [24]3 years ago
6 0

Answer:

the weight is 49.1 N

Explanation:

The computation of the weight is shown below:

As we know that

= 5kg of potatoes × gravitational acceleration

= 5kg of potatoes × 9.82 m/s

= 49.1 N

Hence, the weight is 49.1 N

We simply applied the above formula in order to determine the weight

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The 1994 Winter Olympics included the aerials competition in skiing. In this event skiers speed down a ramp that slopes sharply
kykrilka [37]

Answer: Got It!

<em>Explanation:  </em>let s = speed at launch

v = 0 at top = s sin 63 - g t

so at top

t = s sin 63/g = .0909 s

h = 13.6 = s sin 63 t - 4.9 t^2

13.6 = .081s^2 - .0405 s^2

s^2 = 336

s = 18.3 m/s

0  0

4 0
3 years ago
The density of atmosphere (measured in kilograms/meter3) on a certain planet is found to decrease as altitude increases (as meas
alexgriva [62]

Answer:

B.  inverse plot, 0.51 kilograms/meter3

Explanation:

First of all, we note that the relationship between the altitude and the atmospheric density is an inverse relationship. In fact, an inverse relationship is a relationship between the x-variable and the y-variable of the form

y \propto \frac{1}{x}

Therefore, as the x increases, the y decreases, and as the x decreases, they increases. This is exactly what occurs with the altitude and the atmospheric density in this plot: as the altitude increases, the density decreases, and vice-versa.

Moreover, we can infer the value of the atmospheric density at an altitude of 1,291 km. This point is located between point A (2550 km) and point B(1000 km), so the density must have a value between 0.30 kg/m^3 and 0.54 kg/m^3, so the correct choice is

B.  inverse plot, 0.51 kilograms/meter3


5 0
3 years ago
to move a resting box of 100 Newton on the ground with kinetic friction coeficient of 0,250 is applied a force of 60 N horizonta
krok68 [10]
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
                          100N x 0.250 = 25.0 N
the work done is,
                        W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J. 
6 0
4 years ago
How will the solubility of a gas solute change if the pressure above the solution is reduced?
Hoochie [10]
If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.
6 0
4 years ago
Read 2 more answers
A 0.105- kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on
Veseljchak [2.6K]

Answer:

<u><em>0.03 m/s</em></u>

Explanation:

<em>Applying law of conservation of momentum, </em>

  • <em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
  • <em>0.105(24) + 75(0) = (0.105 + 75)v</em>
  • <em>75.105v = 2.52</em>
  • <em>v = 2.52/75.105</em>
  • <em>v = </em><u><em>0.03 m/s</em></u>
3 0
3 years ago
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