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NeX [460]
3 years ago
13

27. The traffic officer issued violation tickets to traffic

Physics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

27: 85

28:75%

Explanation:

27:68=80

?=100 hence (68×100)÷80

=85

28:<em>1</em><em>8</em><em>/</em><em>2</em><em>4</em><em>×</em><em> </em><em>1</em><em>0</em><em>0</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>5</em><em>%</em>

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The "steam" above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot
KiRa [710]

Answer:

T_{f} = 85.7 ° C

Explanation:

For this exercise we will use the calorimetry heat ratios, let's start with the heat lost by the evaporation of coffee, since it changes from liquid to vapor state

      Q₁ = m L

Where m is the evaporated mass (m = 2.00 103-3kg) and L is 2.26 106 J / kg, where we use the latent heat of the water

     Q₁ = 2.00 10⁻³ 2.26 10⁶

     Q1 = 4.52 10³ J

Now the heat of coffee in the cup, which does not change state is

     Q coffee = M c_{e} ( T_{f} -T_{i})

Since the only form of energy transfer is terminated, the heat transferred is equal to the evaporated heat

    Qc = - Q₁

    M ce (T_{f} -T_{i}) = - Q₁

The coffee dough left in the cup after evaporation is

    M = 250 -2 = 248 g = 0.248 kg

   T_{f} -Ti = -Q1 / M c_{e}

   T_{f} = Ti - Q1 / M c_{e}

Since coffee is essentially water, let's use the specific heat of water,

    c_{e}= 4186 J / kg ºC

Let's calculate

     T_{f} = 90.0 - 4.52 103 / (0.248 4.186 103)

     T_{f} = 90- 4.35

     T_{f} = 85.65 ° C

     T_{f} = 85.7 ° C

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3 years ago
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Genrish500 [490]
Can i have more information?
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4 years ago
Charging a ballon and rubbing it on wool is an example of ___?
Travka [436]
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8 0
3 years ago
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A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.
pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package.  Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

\sum F_y=m_1*a_m_i_n = T-m_1*g

If the package is barely lifted, that means that T=m_2*g; then:

\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g

Solving the equation for a_mín, we have:

a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: \sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a

For the package: \sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: \sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a

We have two unknowns and two equations, so we can proceed. We can match both tensions and have:

m_1*a+m_1*g=m_2*g-m_2*a

Solving a, we have

(m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2

We can then replace this value of a in one for the sums of force and find the tension T:

T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N

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4 years ago
What controls whether a lava flow is aa or pahoehoe
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Explanation:

Basaltic lava

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