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3 years ago

5

calculate the work done by a girl of mass 40 kg when she climbs a stair case of 20 steps each of height 10 cm and acceleration i

s 10Nkg^-1

2 answers:

Volgvan3 years ago

5 0

Answer:

i think its a 800x² hope you like it

TEA [102]3 years ago

3 0

Answer:

8964

Explanation:

465

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a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u

n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

$a_{c}$ = $\frac{v^{2} }{r}$

$a_{c}$ = 23.04/10.3

$a_{c}$ = 2.23 m/s²

It continues to fly but now with some tangential acceleration

$a_{t}$ = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

$a_{net}$ = $\sqrt{a_{c} ^{2} +a_{t} ^{2} }$

$a_{net}$ = $\sqrt{4.97 + 0.396}$

$a_{net}$ = 2.316 m/s²

So the magnitude of net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

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8 0

1 year ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

Suppose multiple stations are connected in wired network where CSMA/CD implemented. Among them, station A wants to communicate w

anastassius [24]

Answer:

bro i think ur from the uni of arid agriculture rawalpindi and today is your networking paper

Explanation:

6 0

3 years ago

A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an

AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0

3 years ago

The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m 828 m (2716.5 ft) and has more than 160 storie

Natasha_Volkova [10]

Answer:

h = height of the hotel room from the ground floor = 237.4m

Explanation:

Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh

PE1 is the potential energy of tourist at the ground floor

PE1 is the potential energy of tourist at the top (hotel room)

Given

PE1 = − 2.01 × 10⁵ J

PE2 = 0J

PE2 – PE1 = mgh

0 – (− 2.01 × 10⁵ J) = mgh

2.01 × 10⁵ J = 86.4×9.8×h

h = 2.01 × 10⁵/(86.4×9.8) = 237.4m

8 0

3 years ago