How many number of electron shells does calcium have ?

When a moving car is brought to a stop with the brakes, its acceleration vector is 29. (a) in the same direction as its velocity

professor190 [17]

To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is

$\vec{a} = \frac{\vec{v_2}-\vec{v_1}}{\Delta t}$

If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.

Therefore the correct answer is B.

7 0

3 years ago

Virginia beach is 15 kilometers wide and 50 kilometers long. if 2 cm of rain falls on virginia beach, how many cubic meters of r

icang [17]

Assume that the shape of Virginia beach is rectangular.

Note that
1 km = 10³ m
1 cm = 10⁻² m

The area is
A = (15 km)*(50 km)
= (15 x 10³ m)*(50 x 10³ m)
= 7.5 x 10⁸ m²

Because 2 cm of rain fell, the volume is
V = (7.5 x 10⁸ m)*(2 x 10⁻² m) = 1.5 x 10⁶ m³

Answer: 1.5 x 10⁶ m³

4 0

3 years ago

Prove that the weight of an object on moon is 1/6th of that on earth

Elena L [17]

Answer:

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th its weight on the earth.

4 0

2 years ago

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

$E_{A}=E_{B}$

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

$E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}$

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

$E_{B}=m*g*h+\frac{1}{2}*m*v^{2}$

Therefore we will have the following equation:

$(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]$

The kinetic energy can be easily calculated by means of the kinetic energy equation.

$KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]$

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

$E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]$