In can expermient why the can is chrushed

It usually takes more force to start an object sliding than it does to keep an object sliding because static friction is usually

labwork [276]

The answer is already in the blank for, its was greater

5 0

4 years ago

A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the secon

Gemiola [76]

The average speed of a moving object is the rate of change of a certain distance with respect with time. It is equal to the total distance that was traveled by the object over the total time it takes to travel that distance. For this problem we need to assume that the total distance that was traveled would be equal to 120 miles. So, for the first half of the distance or 60 miles at a speed of 30 miles per hour, the time taken would be two hours. For the remaining 60 miles at a speed of 60 miles per hour, 1 hour is total time traveled. So, we calculate the average speed as follows:

Average speed = total distance / total time
Average speed = 120 miles / 2 hr + 1 hr
Average speed = 40 mi / hr

5 0

4 years ago

When pushed with the same amount of force a ross a desk which toy car would most likely travel the fastest A car weighing 25 gra

Arada [10]

Answer:

A car weighing 15 grams

Explanation:

According to the equation that shows the relationship between mass (m), force (F), and acceleration (a) i.e. F = ma, it can be seen that with a constant force, the mass is inversely proportional to the acceleration.

This means that an increase in mass of an object denotes a decrease in acceleration of that object and vice versa. Hence, according to this question, when pushed with the same amount of force, a car with the least mass will move faster. This car is the car weighing 15grams.

3 0

3 years ago

Action and reaction forces are described by which of Newton’s laws of motion?

nikdorinn [45]

Answer:

third

Explanation:

for every action there is an equal and opposite reaction

example

bouncing ball

ball receives a force from the ground due to collision

that same force causes the ball to bounce off the ground

first law:

inertia

example

stuff in space doesn't move unless its pushed

then it keeps going unless something stops it

second law

force = mass times acceleration

example

harder you push a cart the faster it goes

the less a cart weighs the faster a cart goes

studysmarterus

spacecenterorg

energywavetheory

5 0

3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: