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2 years ago

If a snail moves 6.5in/min, how many miles per hour does the snail travel?

Physics

1 answer:

Arturiano [62]2 years ago

7 0

Answer:

Unit Conversion

6.5in/min× 1ft/12in× 1 mile/5280ft= 0.001231 m/min
0.001231 m/min× 60 min/1hr= 0.0738 miles/hr

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Mass of object is 50g moves in a circular path of radius 10cm find work done

topjm [15]

Answer:

Work done = 0.3142 Nm

Explanation:

Mass of Object is 50 g

Circular path of radius is 10 cm ⇒ 0.1 m

Work done = Force × Distance = ?

*Distance moved (circular path) ⇒ Circumference of the circular path

2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m

*Force that is enough to move a 50 g must be equal or more than its weight.

therefore convert 50 grams to newton = 0.5 N

Recall that; work done is force times distance

∴ 0.5 N × 0.6284 m

Work done = 0.3142 Nm

3 0

3 years ago

Once an object enters orbit, what keeps the object moving sideways?

yaroslaw [1]

It's the natural tendency of things to keep going unless there's something trying to stop them.

It's usually called "inertia".

Don't get the idea from all of this that things stop unless there's something to keep them going. The truth is exactly the opposite: Things keep going unless there's something to make them stop.

7 0

3 years ago

Read 2 more answers

A student is running to catch the campus shuttle bus, which is stopped at the bus stop. The student is runnign at a constant spe

Serggg [28]

Answer:

Part a)

$t = 16.8 s$

$d = 100.8 m$

Part b)

$v_f = 2.86 m/s$

Explanation:

Part a)

Constant speed by which the student will run is given as

$v = 5 m/s$

now after some time if student is going to overtake the position of bus

so here the final positions will be same

so we have

$x_{bus} = x_{student}$

$0 + \frac{1}{2}at^2 + d = v_{student} t$

$\frac{1}{2}(0.170)t^2 + 60 = 5 t$

$0.085 t^2 - 5t + 60 = 0$

so it is

$t = 16.8 s$

So student will run the total distance

$d = vt$

$d = (6)(16.8)$

$d = 100.8 m$

Part b)

Speed of bus when student reach the bus is given as

$v_f = v_i + at$

$v_f = 0 + (0.170)(16.8)$

$v_f = 2.86 m/s$

4 0

3 years ago

A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

Learn more about vertical velocity here:

brainly.com/question/24949996

#SPJ1

8 0

1 year ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$