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inn [45]
3 years ago
14

A bug completes one lap along the edge of a circular planter of radius 18.3 cm in 13.81 s. answer to the nearest thousandth in u

nits of meters per second
Physics
1 answer:
algol133 years ago
7 0

Answer:

Speed of the bug is 0.0832 m/s.

Explanation:

We have,

Radius of the circular planet is 18.3 cm or 0.183 m and time taken by the bug to complete one lap is 13.81 s.

It is required to find the speed of the bug in the circular path. It is given by the displacement in one lap divided by time taken. So,

v=\dfrac{d}{t}\\\\v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 0.183 }{13.81}\\\\v=0.0832\ m/s

So, the speed of the bug is 0.0832 m/s.

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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan
Nesterboy [21]

Answer:

Compression distance: d \approx 0.102\,m

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (K), in joules, at the expense of elastic potential energy (U), in joules, and overcoming work losses due to friction (W_{l}), in joules:

K + W_{l} = U (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2} (2)

Where:

m - Mass of the block, in kilograms.

v - Final velocity of the block, in meters per second.

\mu - KInetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s - Width of the rough patch, in meters.

k - Spring constant, in newtons per meter.

d - Compression distance, in meters.

If we know that m = 1.2\,kg, v = 2.3\,\frac{m}{s}, \mu = 0.44, g = 9.807\,\frac{m}{s^{2}}, s = 0.05\,m and k = 730\,\frac{N}{m}, then the compression distance of the spring is:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2}

m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}

d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }

d \approx 0.102\,m

4 0
3 years ago
Now, let's see what happens when the cannon is high above the ground. Click on the cannon, and drag it upward as far as it goes
kipiarov [429]

Answer:

<h2>B. 20°</h2>

Explanation:

Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g

U is the initial velocity of the body (in m/s)

Ф is the angle of projection

g is the acceleration due to gravity.

Given U = 14m/s, g = 9.8m/s and range R = 15 m

we will substitute this value into the formula to get the projection angle Ф as shown;

15 = 15²sin2Ф/9.8

15*9.8 = 15²sin2Ф

147 = 225sin2Ф

sin2Ф = 147/225

sin2Ф = 0.6533

2Ф = sin⁻¹0.6533

2Ф = 40.79°

Ф = 40.79°/2

Ф = 20.39° ≈ 20°

Hence, the range is greatest at angle 20°

5 0
3 years ago
Plot StartRoot 1.5 EndRoot and StartRoot 1.9 EndRoot on the number line to find which inequalities are true. Check all that appl
Vera_Pavlovna [14]

Answer:

A, B, C, E

Explanation:

now gimme thanks please

4 0
3 years ago
Read 2 more answers
Calculate the object's velocity as shown on the position-time graph,
Cloud [144]

Answer:

10 m/s

Explanation:

The following data were obtained from the question:

Initial Displacement (d1) = 10 m

Final Displacement (d2) = 60 m

Initial time (t1) = 0 s

Final time (t2) = 5 s

Velocity (v) =.?

Next, we shall determine the change in displacement of the object and likewise the change in time.

This can be obtained as follow:

Initial Displacement (d1) = 10 m

Final Displacement (d2) = 60 m

Change in displacement (Δd) = d2 – d1

Change in displacement (Δd) = 60 – 10

Change in displacement (Δd) = 50 m

Initial time (t1) = 0 s

Final time (t2) = 5 s

Change in time (Δt) = t2 – t1

Change in time (Δt) = 5 – 0

Change in time (Δt) = 5 s

Finally, we shall shall calculate the velocity of the object as illustrated below:

Change in displacement (Δd) = 50 m

Change in time (Δt) = 5 s

Velocity (v) =.?

v = Δd/Δt

v = 50/5

v = 10 m/s

Therefore, the velocity of the object is 10 m/s.

4 0
3 years ago
Esta Bien Hecho?<br> desde alli gracias
SVETLANKA909090 [29]

Answer:yeah

Explanation:its good

4 0
3 years ago
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