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3 years ago

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Malcolm and Ravi raced each other. The average of their maximum speeds was 260 km/h If doubled, Malcolm's maximum speed would be

80km/h end text more than Ravi's maximum speed. What were Malcolm's and Ravi's maximum speeds

1 answer:

AfilCa [17]3 years ago

8 0

Answer

Given,

Average speed of Malcolm and Ravi = 260 km/h

Let speed of the Malcolm be X and speed of the Ravi Y.

From the given statement

$\dfrac{X+Y}{2}=260$

$X + Y = 520$ ....(i)

$2X - Y = 80$ ....(ii)

Adding both the equations

3 X = 600

X = 200 km/h

Putting value in equation (i)

Y = 520 - 200

Y = 320 Km/h

Speed of Malcolm = 200 Km/h

Speed of Ravi = 320 Km/h

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3 years ago

Which element does not have the same number of electrons in its outermost shell as the other elements in its group?(1 point)

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3 years ago

Read 2 more answers

In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?

Serjik [45]

Answer:

1

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2 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

5 0

3 years ago

A ball is thrown vertically upwards from the edge of the cliff and hits the ground at the base of the cliff with a speed of 30 m

olya-2409 [2.1K]

To solve this problem we will apply the linear motion kinematic equations. From the definition of the final velocity, as the sum between the initial velocity and the product between the acceleration (gravity) by time, we will find the final velocity. From the second law of kinematics, we will find the vertical position traveled.

$v = v_0 -gt$

Here,

v = Final velocity

$v_0$ = Initial velocity

g = Acceleration due to gravity

t = Time

At t = 4s, v = -30m/s (Downward)

Therefore the initial velocity will be

$-30 = v_0 -9.8(4)$

$v_0 = 9.2m/s$

Now the position can be calculated as,

$y = h +v_0t -\frac{1}{2}gt^2$

When it has the ground, y=0 and the time is t=4s,

$0 = h+(9.2)(4)-\frac{1}{2} (9.8)(4)^2$

$h = 41.6m$

Therefore the cliff was initially to 41.6m from the ground

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2 years ago