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2 years ago

What is the maximum negative displacement a dog could have if it started its motion at +3 m?

Physics

1 answer:

raketka [301]2 years ago

7 0

Answer:

- 3 meter

Explanation:

A dog has started motion from +3 meter. ...(Given)

∴ maximum positive distance = + 3 meter

Magnitude of distance = 3

Maximum negative distance = (-) (magnitude of distance)

Maximum negative distance = (-) (3)

Maximum negative distance = -3 meter

Hence, maximum negative distance is -3 meter.

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Which group of people was most directly affected by the twenty sixth amendment

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An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.

zhenek [66]

Answer:

2 m/s and -2 m/s

Explanation:

The object travels with an angle of

60.0°

with the positive direction of the y-axis: this means that it lies either in the 1st quadrant (positive x) or in the 2nd quadrant (negative x).

If it lies in the 1st quadrant, the value of vx (component of v along x direction) is:

$v_x = v cos \theta = (4.0 m/s) cos 60.0^{\circ}=2 m/s$

If it lies in the 2nd quadrant, the value of vx (component of v along x direction) is:

$v_x = -v cos \theta = -(4.0 m/s) cos 60.0^{\circ}=-2 m/s$

5 0

3 years ago

A stone takes 5.4 seconds to fall from the top of a cliff. The cliff is

Allushta [10]

Answer:

143

Explanation:

Using one of the 3 fundamental equations in physics, y=vo*t+1/2gt^2, we can use this equation to find the total distance that was traveled.

Acceleration due to gravity is always 9.8m/s^2 and time is 5.4s, we also have no initial velocity.

Given this, we can plug in the known variables.

y=0t+1/2*9.8*5^2

simplify,

y=4.9*5.4^2

y=4.9*29.16

y=142.884m which we can round up to 143 meters

Final Answer: 143 meters

4 0

2 years ago

A wheel with a tire mounted on it rotates at the constant rate of 2.73 revolutions per second. Find the radial acceleration of a

Lostsunrise [7]

Answer:

110.9 m/s²

Explanation:

Given:

Distance of the tack from the rotational axis (r) = 37.7 cm

Constant rate of rotation (N) = 2.73 revolutions per second

Now, we know that,

1 revolution = $2\pi$ radians

So, 2.73 revolutions = $2.73\times 2\pi=17.153\ radians$

Therefore, the angular velocity of the tack is, $\omega=17.153\ rad/s$

Now, radial acceleration of the tack is given as:

$a_r=\omega^2 r$

Plug in the given values and solve for $a_r$ . This gives,

$a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]$

Therefore, the radial acceleration of the tack is 110.9 m/s².

4 0

3 years ago