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2 years ago

Which of these is the most effective way for Leanna to cool down after an intense bike ride

Physics

1 answer:

Sonja [21]2 years ago

6 0

I am pretty sure the answer would be too stretch

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A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular vel

KengaRu [80]

Answer:

The net torque is 0.0372 N m.

Explanation:

A rotational body with constant angular acceleration satisfies the kinematic equation:

$\omega^{2}=\omega_{0}^{2}+2\alpha\Delta\theta$ (1)

with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

$\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha$

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

$\alpha=-\frac{-(15.1)^{2}}{2(114.3)}=1.00\frac{rad}{s^{2}}$

The negative sign indicates the sphere is slowing down as we expected.

Now with the angular acceleration we can use Newton's second law:

$\sum\overrightarrow{\tau}=I\overrightarrow{\alpha}$ (2)

with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:

$I = \frac{2MR^{2}}{5}$

With M the mass of the sphere an R its radius, then:

$I = \frac{2(1.85)(\frac{0.45}{2})^{2}}{5}=0.037 kg*m^2$

Then (2) is:

$\sum\overrightarrow{\tau}=0.037(-1.00)=0.037 Nm$

7 0

2 years ago

Read 2 more answers

a rod of some material 0.20 m long elongates 0.20 mm on heating from 21 to 120°c. determine the value of the linear coefficient

Rufina [12.5K]

Answer:

The value of the linear coefficient of thermal expansion is : α=1.01 *10⁻⁵ (ºC)⁻¹

Explanation:

Li = 0.2m

ΔL = 0.2 mm = 0.0002m

T1 = 21ºC

T2 = 120ºC

ΔT =99ºC

α =ΔL/(Li*ΔT)

α =0.0002m /(0.2m * 99ºC)

α = 1.01 *10⁻⁵ (ºC)⁻¹

4 0

2 years ago

A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

evablogger [386]

Answer:

$\text{heat loss} = 24864.05 \ W/m^2$

Explanation:

$T_1$ , $T_2$ are temperatures of gasses and liquid in Kelvins,

$t_1$ and $t_2$ are thicknesses of gas layer and steel slab in meters,
$h_1$ , $h_2$ are convection coefficients gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ is the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ are thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

using known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Using the rate equation :

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Similarly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature distribution is shown in the attached image

3 0

2 years ago

Last year a baseball player made 63 errors. This year he made 42. What percent decrease was there in the number of errors commit

Crank

Answer:

33.33 %

Explanation:

given,

error last year = 63

error this year = 42

percent decrease in the error = ?

to find the percentage difference in the error formula used is

= $\dfrac{difference}{original}\times 100$

= $\dfrac{63-42}{63}\times 100$

= $\dfrac{21}{63}\times 100$

= 33.33 %

Percentage decrease in the number of error is equal to 33.33%.

7 0

2 years ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$