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3 years ago

6

Two steamrollers begin 105 m apart and head toward each other, each at a constant speed of 1.30 m/s . At the same instant, a fly

that travels at a constant speed of 2.40 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.What distance does the fly travel?

1 answer:

Alisiya [41]3 years ago

4 0

Answer:Kobe

Explanation:

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4 years ago

Read 2 more answers

A loaded truck is going at a speed Of 36km/hr and a car starts at a point 3 km behind the truck with an acceleration of 2 m/s^2

Andrew [12]

Answer:

60 sec

3.6 Km

Explanation:

Given that the truck is going at a speed of 36 km/hr = $36\times \frac{5}{18}$ = 10 m/sec

Given that the car starts at a point 3km=3000 m behind the truck

Initial velocity (u) of the car = 0 m/sec

Acceleration (a) of the car = 2 m/ $sec^2$

We know when they meet let the distance travelled by the truck be d Then the distance travelled by the car is d + 3000

For truck

$speed=\frac{distance}{time}$

$10 =\frac{d}{t}$

d = 10t

For car

$distance = ut +\frac{1}{2}at^2$

d+3000 = $\frac{1}{2}2t^2$

10t + 3000 = $t^2$

On solving we get t = 60 sec

The car will overtake the truck at 60 sec

The distance travelled by car is $\frac{1}{2}2t^2$ = 3600 m= 3.6 Km

7 0

3 years ago

A ball thrown vertically upward is caught by the thrower after 2.00 s. Find (a) the initial velocity of the ball and (b) the max

ankoles [38]

Answer:

a) 9.8 m/s

b) 4.9 m

Explanation:

This problem is a good example of Vertical motion, where the main equations for this situation are:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

$V^{2}={V_{o}}^{2}-2gy$ (2)

Where:

$y$ is the height of the ball at a given time

$y_{o}=0m$ is the initial height of the ball (assuming the hand of the thrower the origin of the system)

$V_{o}$ is the initial velocity of the ball

$V$ is the final velocity of the ball

$t=2s$ is the time it takes for the ball to make the complete movement (from the moment it is thrown until it falls back into the pitcher's hands)

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

Knowing this, let's begin with the answers:

<h3>a) Initial velocity </h3>

In order to find the initial velocity $V_{o}$ of the ball, we will use equation (1) and $t=2s$ , taking into account that $y=0 m$ and $y_{o}=0m$ at this given time:

$0=0+V_{o}t-\frac{1}{2}gt^{2}$ (3)

Isolating $V_{o}$ :

$V_{o}=\frac{1}{2}gt$ (4)

$V_{o}=\frac{1}{2}(9.8 m/s^{2})(2 s)$ (5)

Then:

$V_{o}=9.8 m/s$ (6)

<h3>b) Maximum height </h3>

In this part, we will use equation (2), knowing the value of the height is maximum when $V=0$ . So, we will name this height as $y_{max}$ :

$0={V_{o}}^{2}-2gy_{max}$ (7)

Isolating $y_{max}$ :

$y_{max}=\frac{{V_{o}}^{2}}{2g}$ (8)

$y_{max}=\frac{{(9.8 m/s)}^{2}}{2(9.8 m/s^{2})}$ (9)

Finally:

$y_{max}=4.9 m$

4 0

4 years ago