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ratelena [41]
3 years ago
11

A 100 kg bag of sand has a weight on 100 N. When dropped its acceleration is what?

Physics
2 answers:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

<h2>Gravity.</h2>

Explanation:

If you are holding the bag of sand, the only acceleration in the system is gravity, because is a vertical movement. So, a = 9.81 \frac{m}{s^{2} }

We are assuming that is vertical movement because is say ''dropped'', which is a term used in Free-fall models, where acceleration is constant.

Lyrx [107]3 years ago
7 0
100N describes the weight of the sandbag, while 100kg is the mass of the sandbag.

To calculate acceleration, divide your weight by the mass, thus the accleration is:

100N/100kg = 1(m/s^2)
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What is the overall charge of 1.5 x 10^10 electrons?
Troyanec [42]

Answer: Charge = -2.4x10^-9 Coulombs

Explanation:

The charge of one electron is e = -1.6x10^-19 C

Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:

Here i will use the relation (a^b)*(a^c) = a^(b + c)

Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C  

Charge = -2.4x10^-9 C

7 0
3 years ago
Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the
scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = \frac{kQ_{1}Q_{2} }{r^{2} }

F= \frac{K(Q/2)(3Q/4)}{r^{2} }

how \frac{KQ^{2} }{r^{2} } = 0.42

Then

F = \frac{0.42*3}{8}

F = 0.1575 N

3 0
3 years ago
Which is a factor that affects size of mineral crystals formed in magma
Natali5045456 [20]
The volcanic ashes from the volcano
6 0
3 years ago
A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2
Ahat [919]

Answer:

Approximately 0.608 (assuming that g = 9.81\; \rm N\cdot kg^{-1}.)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

  • Let m represent the mass of this car.
  • Let r represent the radius of the circular track.

This answer will approach this question in two steps:

  • Step one: determine the centripetal force when the car is about to skid.
  • Step two: calculate the coefficient of static friction.

For simplicity, let a_{T} represent the tangential acceleration (1.90\; \rm m \cdot s^{-2}) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to 90^\circ or \displaystyle \frac{\pi}{2} radians.

The angular acceleration of this car can be found as \displaystyle \alpha = \frac{a_{T}}{r}. (a_T is the tangential acceleration of the car, and r is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity (u and v) to (tangential) acceleration a_{T} and displacement x:

v^2 - u^2 = 2\, a_{T}\cdot x.

The idea is to solve for the final angular velocity using the angular analogy of that equation:

\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta.

In this equation, \theta represents angular displacement. For this motion in particular:

  • \omega(\text{initial}) = 0 since the car was initially not moving.
  • \theta = \displaystyle \frac{\pi}{2} since the car travelled one-quarter of the circle.

Solve this equation for \omega(\text{final}) in terms of a_T and r:

\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}.

Let m represent the mass of this car. The centripetal force at this moment would be:

\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}.

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, m\, g.

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let \mu_s denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g.

The size of this force should be equal to that of the centripetal force when the car is about to skid:

\mu_s\, m\, g = \pi\, m\, a_{T}.

Solve this equation for \mu_s:

\mu_s = \displaystyle \frac{\pi\, a_T}{g}.

Indeed, the expression for \mu_s does not include any unknown letter. Let g = 9.81\; \rm N\cdot kg^{-1}. Evaluate this expression for a_T = 1.90\;\rm m \cdot s^{-2}:

\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608.

(Three significant figures.)

7 0
3 years ago
When you drop a object on earth, earth gets pulled slightly toward that object as well. How do I calculate the force if the eart
kiruha [24]

The gravitation force with which the earth is being pulled can be determined by applying Newton's law of universal gravitation.

<h3>What is gravitation force?</h3>

According Newton's law of universal gravitation, the force exerted between two objects in the universe is directly proportional to the product of masses of the two objects and inversely proportional to the square of the distance between the two objects.

Mathematically, the formula for gravitation force is given as;

F = GmM/R²

where;

  • m is the mass of the object
  • M is mass of earth
  • R is the distance of the object from earth
  • G is universal gravitation constant

If the mass of the object is know and the distance between earth  and the object is also known, the force with which the earth is being pulled can be calculated by applying Newton's law of universal gravitation as shown in the above equation.

Thus, the force with which the earth is being pulled can calculated as well.

Learn more about gravitation force here: brainly.com/question/27943482

#SPJ1

7 0
1 year ago
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