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andreyandreev [35.5K]
3 years ago
7

A cylindrical object can roll down an incline, as shown in Figure 1. The incline is slightly less than one meter in length. A gr

oup of students wants to determine the acceleration of the object while it is rolling
down the incline. The students have access to the following equipment.
• A stopwatch which can measure time intervals up to 999 s with a precision of 0.01s
• A clock, which can measure time intervals up to 12 hours with a precision of 1 minute (60 s)
A meterstick, which can measure lengths up to 1 m with a precision of 1 mm
• A pair of calipers, which can measure lengths up to 10 cm with a precision of 0.05 mm
(1) Assume the object moves with a constant acceleration as it rolls down the incline Write an equation that includes acceleration and quantities that can be measured or obtained from measurements by
using the available equipment in the list.
Physics
1 answer:
worty [1.4K]3 years ago
8 0

Answer:

a = 2d / t²

or

a = 2gh / (3d)

Explanation:

One method is to use the equation:

Δx = v₀ t + ½ at²

d = (0) t + ½ at²

d = ½ at²

a = 2d / t²

By measuring the length of the incline d, and the time it takes to reach the bottom t, the students can calculate the acceleration, using only the meter stick and the stopwatch.

Another method is to use conservation of energy to find the final velocity.

Initial potential energy = final rotational energy + kinetic energy

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For a solid cylinder, I = ½ mr².  For rolling without slipping, ω = v/r.

mgh = ½ (½ mr²) (v/r)² + ½ mv²

mgh = ¼ mv² + ½ mv²

mgh = ¾ mv²

4gh/3 = v²

Using constant acceleration equation:

v² = v₀² + 2aΔx

4gh/3 = 0² + 2ad

a = 2gh / (3d)

Using this equation, the students can measure the height of the incline h, and the length of the incline d, to calculate the acceleration.  The only equipment needed is the meter stick.

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a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

8 0
3 years ago
A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f
anzhelika [568]

Answer:

\phi_i = BA

Explanation:

magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

\phi = \vec B. \vec A

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

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6 0
3 years ago
A close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject
kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

Radius_{mean} = 60\ mm

Diameter_{mean} = 120\ mm

Load = 200 N

We need to calculate the deflection

Using formula of deflection

\delta=\dfrac{8pD^3n}{Cd^4}

Put the value into the formula

\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}

\delta =34.56\ mm

Hence, The deflection of the spring is 34.56 mm.

4 0
3 years ago
M = 30.3kg<br>M = 40.17kg 9<br>R = 0.5m<br>G = 6. 67x10^11<br>F ?​
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Answer:

m¹=30.3kg

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R=0.5m

G=6.67*10¹¹

F=Gm¹m²/R²

=160.68

4 0
3 years ago
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