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3 years ago

If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done

Physics

1 answer:

ioda3 years ago

6 0

Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

$Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J$

Therefore, the work done by the lawnmower is 236.14 J.

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A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do

soldi70 [24.7K]

Answer:

Specific heat at constant pressure is = 1.005 kJ/kg.K

Specific heat at constant volume is = 0.718 kJ/kg.K

Explanation:

given data

temperature T1 = 50°C

temperature T2 = 80°C

solution

we know energy require to heat the air is express as

for constant pressure and volume

Q = m × c × ΔT ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat is greater than the specific heat at constant volume,

so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is

Specific heat at constant pressure is = 1.005 kJ/kg.K

and

Specific heat at constant volume is = 0.718 kJ/kg.K

3 0

3 years ago

A boy notices that 45 complete waves travel past him in ninety seconds. What is the

san4es73 [151]

Answer:

0.5 Hz

Hope you find this helpful. Please mark me as brainliest!

5 0

3 years ago

What minimum speed does a 200 g puck need to make it to the top of a frictionless ramp that is 4.1 m long and inclined at 22 ∘?

myrzilka [38]

Answer:

5.5 m/ sec

Explanation:

Because the inclined surface is frictionless so we can assume that total change of energy is zero

i-e ΔE = 0

Or we can say that difference between final and initial energy is zero i-e

Ef- Ei =0

Where,

Ef= final energy at the top of the ramp= KEf+PEf

Ei= Initial energy at the bottom of the ramp=KEi+PEi

So we have

(KEf+PEf)-(KEi+PEi)=0

==>KEf-KEi+PEf-PEi=0 -------------(1)

KEf = mgh = 200×9.8×h

Where h= Sin 22 = h/d= h/4.1

0.375×4.1=h

or h= 1.54 m

So, PEf= 200×9.8×1.54=3018.4 j

and KEf= 1/2 m $Vf^{2}$ = 0.5×200×0=0 j

PEi= mgh = 200×9.8×0=0 j

KEi= 1/2 m $Vi^{2}$ =0.5×200× $Vi^{2}$ =100 $Vi^{2}$ j

Put these values in eq 1, we get;

0-100 $Vi^{2}$ +3018.4-0=0

-100 $Vi^{2}$ =-3018.4

==> $Vi^{2}= \frac{3018.4}{100}$ = 30.184

==> Vi = $\sqrt{30.184} = 5.5 m.sec$

7 0

3 years ago

What is the angle of reflection of the incident angle is 35°? 35 25° 55 90

Anestetic [448]

Answer:

i think its the third one or second one

4 0

3 years ago

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

molecular mass of water, $M_w=2+16=18\ g.mol^{-1}$

∵Density of water is $0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}$

∴mass of water in 316 mL, $m_w=316\times 0.992=313.5 g$

mass of glycerin, $m_g=154\ g$

pressure of mixture, $P_x=55.32\ torr= 55.32\ mm.Hg$

temperature of mixture, $T_x=40^{\circ}C$

Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.

moles of water in the given quantity:

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

moles of glycerin in the given quantity:

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

Now the mole fraction of water:

$X_w=\frac{n_w}{n_w+n_g}$

$X_w=\frac{17.42}{17.42+1.674}$

$X_w=0.912$

Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.

$\therefore P_{sol}=X_w\times P_x$

$\therefore P_{sol}=0.912\times 55.32$

$P_{sol}=50.4\ mm.Hg$

7 0

3 years ago