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2 years ago

On a hypothetical scale X The ice point is 40° and steam point is 120°.

Physics

1 answer:

arlik [135]2 years ago

7 0

Answer:

The reading of Y is -10°.

Explanation:

For scale X, the ice point is 40° and steam point is 120°.

Difference between the two extremes for scales X = 120 - 40 = 80

For scale X, the ice point and steam points are -30° and 130° respectively.

Difference between the two extremes for scales X = 130 - (-30) = 160

Comparing both scales:

One unit of scale X = x

One unit of scale Y = y

Scale X has 80 divisions while scale Y has 160

80x = 160y

x = 2y

50° in scale X = 10x + ice point in X scale

10 divisions in Y scale = 20y

Reading of Y scale = ice point of Y + 20y

= -30° + 20°

= -10°

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A slender uniform rod 100.00 cm long is used as a meter stick. Two parallel axes that are perpendicular to the rod are considere

Nataliya [291]

Answer:

The correct answer is D $I_{30}$ / $I_{50}$ = 1.5

Explanation:

In this exercise the moment of inertia equation should be used

I = ∫ r² dm

In addition to the parallel axis theorem

I = $I_{cm}$ + M D²

Where $I_{cm}$ is the moment of the center of mass, M is the total mass of the body and D the distance from this point to the axis of interest

Let's apply these relationships to our problem, the center of mass of a uniform rod coincides with its geometric center, in this case the rod is 1 m long, so the center of mass is in

L = 100.00 cm (1m / 100 cm) = 1.0000 m

$x_{cm}$ = 50 cm = 0.50 m

Let's calculate the moment of inertia for this point, suppose the rod is on the x-axis and use the concept of linear density

λ = M / L = dm / dx

dm = λ dx

Let's replace in the moment of inertia equation

I = ∫ x² ( λ dx)

We integrate

I = λ x³ / 3

We evaluate between the lower limits x = -L/2 to the upper limit x = L/2

I = λ/3 [(L/2)³ - (-L/2)³] = lam/3 [L³/8 - (-L³ / 8)]

I = λ/3 L³/4

I = 1/12 λ L³

Let's replace the linear density with its value

I = 1/12 (M/L) L³

I = 1/12 M L²

Let's calculate with the given values

I = 1/12 M 1²

I = 1/12 M

This point is the center of mass of the rod

Icm = I = 1/12 M = 8.333 10-2 M

Now let's use the parallel axis theorem to calculate the moment of resection of the new axis, which is 0.30 m from one end, in this case the distance is

D = $x_{cm}$ - x

D = 0.50 - 0.30

D = 0.20 m

Let's calculate

$I_{30}$ = $I_{cm}$ + M D²

$I_{30}$ = 1/12 M + M 0.202

$I_{30}$ = M (1/12 + 0.04)

$I_{30}$ = M 0.123

To find the relationship between the two moments of inertia, divide the quantities

$I_{30}$ / $I_{50}$ = M 0.123 / (M 8.3 10-2)

$I_{30}$ / $I_{50}$ = 1.48

The correct answer is d 1.5

6 0

2 years ago

29. 2072 Set C Q.No. 10c

Annette [7]

Answer:

$90.2^{\circ}C$

Explanation:

Considering the thermal conductivity of aluminium and brass as $k_{al}=205 W/mK$ and $k_{br}=109 W/mk$ respectively

The temperature at the end of aluminium and brass are given as $T_{al}=150^{\circ}C$ and $T_{br}=20^{\circ}C$ respectively with length of rod L=1.3 m , Length of aluminium $L_{al}=0.8 m$ , length of brass $L_{br}=0.5 m$ and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

$H_{br}=H_{al}$

$k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}$

Upon re-arranging

$T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}$

$(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}$

$T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}$

$T=90.2^{\circ}C$

Therefore, the temperatures at which the metals are joined is $90.2^{\circ}C$

6 0

2 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0

2 years ago

Read 2 more answers

A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

ikadub [295]

In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

KE1 = KE2

The kinetic energy of the system before the collision is solved below.

KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s.

6 0

2 years ago

hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward

dimulka [17.4K]

For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

t = 5 m / 6.65 m/s

t = 0.75 s

Result:

The time will be <u>0.75 seconds.</u>

7 0

2 years ago

Read 2 more answers