Ask question

Ask question

All categories

3 years ago

7

A parachute on a racing dragster opens and changes the speed of the car from 85 m/sec to 45 m/sec in a period of 4.5 seconds. Wh

at is the acceleration of the dragster?

1 answer:

Serggg [28]3 years ago

6 0

Acceleration is change is velocity over time. So the change in velocity is 40 and time is 4.5 sec. So the acceleration is 40/4.5=8.89 m/s^2

You might be interested in

Which of the following is the most appropriate description of an electric current?

Olenka [21]

Electric current is the flow of charge due to the potential difference between two terminals per unit time. It is denoted by I and its unit is amp. It can be mathematically expressed as I=Q/t.

If this isn’t what your looking for I need the options they gave you

6 0

2 years ago

The trajectory of a projectile always ________________.

sattari [20]

Answer:

c) curves downward, below the initial velocity vector

Explanation:

A projectile is usually launched from a height, where it is launched with an initial velocity. From that point the gravitational force begins to act on the projectile causing it to decay. As time passes, the projectile advances but its height decreases. So its trajectory is curved downward, below the initial velocity vector.

8 0

2 years ago

A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p

Vesnalui [34]

The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that: </span>
<span>F * 3.0 = 150 x 9.81 x 1.20
Then solve for F, the result should be in newtones = 588N

Do hope it makes sense.</span>

3 0

2 years ago

What is the Creation Mandate?

Ilia_Sergeevich [38]

Is the divine injunction found in Genesis 1:28, in which God, after having created the world and all in it, ascribes to humankind the tasks of filling, subduing, and ruling over the earth.

3 0

3 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.

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

5 0

3 years ago