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leva [86]
3 years ago
9

A bug is moving along a meter stick in the negative direction at a constant speed of 0.85cm/s. After 42 seconds have passed the

bug is located at 27.0 cm. What was the bug’s starting position.
Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Given that,

bug speed, v= 0.85 m/s

time, t =42 s

Final position of bug on meter stick was 27 cm

Starting position of bug on meter stick = ?

Since we know that,

s = vt

s= 0.85*42 = 35.7 cm

this is the distance covered by bug in the given time and velocity.

since the bug is moving in negative direction, starting point will be:

27.0 cm+ 35.7 cm = 62.7 cm

The bugs starting position on meter stick was 62.7 cm.

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algol [13]

Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>.  That description is a little more direct.

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A crate remains stationary after it has been placed on a ramp inclined at an angle with the horizontal. Which of the following s
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Answer:

d. It is equal to the component of the gravitational force acting down the ramp.

Explanation:

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Water from a vertical pipe emerges as a 10-cm-diameter cylinder and falls straight down 7.5 m into a bucket. The water exits the
cupoosta [38]

The diameter of the column of the water as it hits the bucket is 4.04 cm

The equation of continuity occurs in the fluid system and it asserts that the inflow and the outflow of the volume rate at the inlet and at the outlet of the system are equal.

By using the kinematics equation to determine the speed of the water in the bucket and applying the equation of continuity to estimate the diameter of the column, we have the following;

Using the kinematics equation:

\mathbf{v_f ^2 = v_i^2 + 2gh}

\mathbf{v_f ^2 =(2.0)^2 + 2\times 9.8 \times 7.5}

\mathbf{v_f ^2 =151 m/s}

\mathbf{v_f  =\sqrt{151 m/s}}

\mathbf{v_f  =12.29 \ m/s}  

From the equation of continuity:

\mathbf{A_iV_i = A_fV_f}

\mathbf{\pi r^2_iV_i = \pi r^2_fV_f}

\mathbf{ r^2_iV_i =  r^2_fV_f}

\mathbf{ (\dfrac{10}{2})^2\times 2.0 =  r_f^2 \times 12.29}

\mathbf{ 50 = 12.29 \times r_f^2}

\mathbf{ r_f=  \sqrt{\dfrac{50}{12.29} }}

\mathbf{ V_f= 2.02 \ cm }

Since diameter = 2r;

∴

The diameter of the column of the water is:

= 2(2.02) cm

= 4.04 cm

Learn more about the equation of continuity here:

brainly.com/question/10822213

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