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eduard
3 years ago
11

Substance a is a solid ionic compound substance b solid covalent compound what would happen if you put them both in the same con

tainer
Physics
1 answer:
sukhopar [10]3 years ago
7 0
I would one big ionic compound

You might be interested in
A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13
Irina18 [472]

The work done by the gravitational force = 0

Given the mass of the box = 40 kg

The box is initially at rest.

Distance moved by the applied force = 5m

Force applied = 130 N

Co-efficient of friction between the box and floor = 0.3

The box is moved only in the horizontal direction by the applied force.

Gravitational force is applied in a direction perpendicular to the applied force. hence it doesn't do any work on the box.

Therefore, the work done by the gravitational force is zero.

Learn more about the gravitational force at brainly.com/question/862529

#SPJ4

3 0
1 year ago
a ball of mass 100g moving at a velocity of 100m/s collides with another ball of mass 400g moving at 50m/s in same direction, if
klio [65]

Answer:

Velocity of the two balls after collision: 60\; \rm m \cdot s^{-1}.

100\; \rm J of kinetic energy would be lost.

Explanation:

<h3>Velocity</h3>

Because the question asked about energy, convert all units to standard units to keep the calculation simple:

  • Mass of the first ball: 100\; \rm g = 0.1\; \rm kg.
  • Mass of the second ball: 400\; \rm g = 0.4 \; \rm kg.

The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.

Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.

The momentum of an object of mass m and velocity v is: p = m \cdot v.

Momentum of the two balls before collision:

  • First ball: p = m \cdot v = 0.1\; \rm kg \times 100\; \rm m \cdot s^{-1} = 10\; \rm kg \cdot m \cdot s^{-1}.
  • Second ball: p = m \cdot v = 0.4\; \rm kg \times 50\; \rm m \cdot s^{-1} = 20\; \rm kg \cdot m \cdot s^{-1}.
  • Sum: 10\; \rm kg \cdot m \cdot s^{-1} + 20 \; \rm kg \cdot m \cdot s^{-1} = 30 \; \rm kg \cdot m \cdot s^{-1} given that the two balls are moving in the same direction.

Based on the assumptions, the sum of the momentum of the two balls after collision should also be 30\; \rm kg \cdot m \cdot s^{-1}. The mass of the two balls, combined, is 0.1\; \rm kg + 0.4\; \rm kg = 0.5\; \rm kg. Let the velocity of the two balls after the collision v\; \rm m \cdot s^{-1}. (There's only one velocity because the collision had sticked the two balls to each other.)

  • Momentum after the collision from p = m \cdot v: (0.5\, v)\; \rm kg \cdot m \cdot s^{-1.
  • Momentum after the collision from the conservation of momentum: 30\; \rm kg \cdot m \cdot s^{-1}.

These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about v:

0.5\, v = 30.

v = 60.

In other words, the velocity of the two balls right after the collision should be 60\; \rm m \cdot s^{-1}.

<h3>Kinetic Energy</h3>

The kinetic energy of an object of mass m and velocity v is \displaystyle \frac{1}{2}\, m \cdot v^{2}.

Kinetic energy before the collision:

  • First ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.1\; \rm kg \times \left(100\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Second ball: \displaystyle \frac{1}{2} \, m \cdot v^2 = \frac{1}{2}\times 0.4\; \rm kg \times \left(50\; \rm m \cdot s^{-1}\right)^{2} = 500\; \rm J.
  • Sum: 500\; \rm J + 500\; \rm J = 1000\; \rm J.

The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.

  • Mass of the two balls, combined: 0.5\; \rm kg.
  • Velocity of the two balls right after the collision: 60\; \rm m\cdot s^{-1}.

Sum of the kinetic energies of the two balls right after the collision:

\displaystyle \frac{1}{2} \, m \cdot v^{2} = \frac{1}{2}\times 0.5\; \rm kg \times \left(60\; \rm m \cdot s^{-1}\right)^2 = 900\; \rm J.

Therefore, 1000\; \rm J - 900\; \rm J = 100\; \rm J of kinetic energy would be lost during this collision.

7 0
3 years ago
How is the speed of an object related to the amount of<br> kinetic energy the object has?
xenn [34]

Answer:

directly proportional to the square of its speed.

Explanation:

;)

3 0
1 year ago
E) VAVC
gogolik [260]

Answer:

b) tan a = 3

Explanation:

Draw a free body diagram (see attached).

There are three forces acting on the insect.  Weight downwards, normal force towards the center of the hemisphere, and friction tangent to the surface.

Sum of the forces in the radial direction:

∑F = ma

N − mg cos α = 0

N = mg cos α

Sum of the forces in the tangential direction:

∑F = ma

μN − mg sin α = 0

Substituting:

μ(mg cos α) − mg sin α = 0

μ cos α − sin α = 0

μ cos α = sin α

μ = tan α

The maximum possible value of the angle is such that its tangent is equal to the coefficient of friction.

7 0
3 years ago
What are the units for the spring constant, k?
Mnenie [13.5K]

C. newtons/meter

Explanation:

The unit of spring constant K is newtons/meter.

The spring constant is also known as the stiffness.

    F = k e

F is the force

k is the spring constant

e is the extension.

 The unit of F is newtons

                    e is meters

Therefore, k becomes newtons/meter

  • Stiffness is the resistance to stretch of a body.

learn more:

Force brainly.com/question/3820012

#learnwithBrainly

4 0
3 years ago
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