What factor determines an object's momentum?

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its te

Maksim231197 [3]

Answer:

Tt = 70 + 135e^-0.031t

13 minutes

Explanation:

Given that :

Initial temperature, Ti = 205°

Temperature after 2.5 minutes = 195°

Temperature of room, Ts= 70

Using the relation :

Tt = Ts + Ce^-kt

Temperature after time, t

When freshly poured, t = 0

205 = 70 + Ce^-0k

205 = 70 + C

C = 205 - 70 = 135°

T after 2.5 minutes to find proportionality constant, k

Tt = Ts + Ce^-kt

195 = 70 + 135e^-2.5k

125 = 135e^-2.5k

125 / 135 = e^-2.5k

0.9259 = e^-2.5k

Take In of both sides :

−0.076989 = - 2.5k

k = −0.076989 / - 2.5

k = 0.031

Equation becomes :

Tt = 70 + 135e^-0.031t

t when Tt = 160

160 = 70 + 135e^-0.031k

90 = 135e^-0.031t

90/135 = e^-0.031t

0.6667 = e^-0.031t

In(0.6667) = - 0.031t

−0.405465 = - 0.031t

t = 0.405465/ 0.031

t = 13.071

t = 13 minutes

8 0

2 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

Relation of Kilometre and mile

Alja [10]

Answer:

1km = o.621371 mile

Explanation:

1.609 kilometers equal 1 mile. The kilometer is a unit of measurement, as is the mille. However, a mile is longer than a kilometer.

5 0

2 years ago

Read 2 more answers

Real-world examples of power

Paha777 [63]

By definition, power is the amount of energy consumed (or produced) in a second. (or more precisely, it is the rate of change in energy).
so anything which uses energy in a known time period can be labeled with a power rating.

an example for power could be a nuclear plant; traditional nuclear plants produce somewhat close to 1 giga watts (which means 1 giga joules in a second)

3 0

2 years ago

An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b

lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

S = ut + 0.5at²

Here u = 0, and a = g

S = 0.5gt²

Distance traveled during the first second ( t =1 )

S = 0.5 x 9.81 x 1² = 4.905 m

Distance traveled during the first second = 4.905 m.

b) We have equation of motion

v² = u² + 2as

Here u = 0, s= 75 m and a = g

v² = 0² + 2 x g x 75 = 150 x 9.81

v = 38.36 m/s

Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

75 = 0.5 x 9.81 x t²

t = 3.91 s

We need to find distance traveled last second

That is

S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

Distance traveled during the last second of motion before hitting the ground = 33.45 m

3 0

3 years ago