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3 years ago

If mass = 7.89 kg, weight on Earth =

Physics

1 answer:

marysya [2.9K]3 years ago

8 0

Answer:

77.322

Explanation:

you have to multiply the mass with gravity (which is 9.8).. which will give u that result

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In a horse cart system when would there be the smallest amount of acceleration

Assoli18 [71]

Answer:

well... when the horse stops/rests, or if it is blocked by a surface or anything of solid background.

Explanation:

If it is going up a hill or slope and it just starts to move that would also be considered the smallest amount of acceleration this can go for many things when it just starts to move. but I would go for when it rests amounting to your fitting of the question.

3 0

3 years ago

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t

postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^{\circ}$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4\times 0.001157$

$V_1=1.6198\times 10^{-3}\ m^3$

Final Volume $V_2=4V_1$

$V_2=4\times (1.6198\times 10^{-3})$

$V_2=6.4792\times 10^{-3}\ m^3$

Specific volume at this stage

$\nu _2=\frac{V_2}{m}$

$\nu _2=\frac{6.4792\times 10^{-3}}{1.4}$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})$

$T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)$

$T_2=370.7^{\circ} C$

4 0

3 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

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

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago

Name something that magnets has done for you in your life

kati45 [8]

Helped me learn how to spell words, held pictures or papers up, collected metal items that were lost

8 0

2 years ago

Read 2 more answers

Your new pet kitten is trapped in a ten foot deep hole. You need a contraption to safely rescue your poor animal. How can I solv

Amanda [17]

You could use a ladder or a rope

5 0

3 years ago