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IrinaVladis [17]
3 years ago
14

according to newton's third law, when a horse pulls on a cart, the cart pulls back on the horse with an equal force on the horse

. if in fact the cart pulls back on the horse as hard as the horse pulls forward on the cart, how is it possible for the horse to move the cart?
Physics
1 answer:
Debora [2.8K]3 years ago
8 0
Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.

When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
 
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>


</span>
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A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
3 years ago
"the number of transistors per square inch on an integrated chip doubles every 18 months." this observation is known as ________
creativ13 [48]
That is called Moore's Law
5 0
3 years ago
calculate the change in potential energy of a body of mass 2kg that is lifted through a height of 10m above the ground level. if
LuckyWell [14K]

Answer:

Use the potential energy formula

Explanation:

PE=mgh

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4 0
3 years ago
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 51.5 A. The resistanc
Sliva [168]

Answer:

a).Jcu= 436.44x10^{3} \frac{A}{m^{2}}

b).λcu=1.05728 \frac{kg}{m}

c).Jal= 274.66 x10^{3} \frac{A}{m^{2}}

d).λcu=0.487 \frac{kg}{m}

Explanation:

a).

ζcu=1.7x10^{-8}Ωm

ζal=2.76x10^{-8}Ωm

A=\frac{w}{R}

wcu=ζcu*l

Acu=\frac{1.7x10^{-8}*1000}{0.144} =1.18x10^{-4} m^{2}

J=\frac{I}{Acu}=\frac{51.5A}{1.18x10^{-4}m^{2}} \\J=436.44x10^{3} \frac{A}{m^{2}}

b).

mass per unit Copper

λcu=Dcu*Acu

λcu=8960 \frac{kg}{m^{3}}*1.18x10^{-4} m^{2}

λcu=1.05728 \frac{kg}{m}

c).

wal=ζal*l

Aal=\frac{2.7x10^{-8}*1000}{0.144} =0.187x10^{-3} m^{2}

J=\frac{I}{Aal}=\frac{51.5A}{0.187x10^{-3}m^{2}} \\J=274.66x10^{3} \frac{A}{m^{2}}

d).

mass per unit Aluminum

λal=Dal*Aal

λal=2600 \frac{kg}{m^{3}}*0.1875x10^{-3} m^{2}

λcu=0.487 \frac{kg}{m}

3 0
3 years ago
A 275-g sample of nickel at 100.0°C is placed in 100.0 mL of water at 22.0°C. What is the final temperature of the water? Assume
Mademuasel [1]

Answer:

39.6138 °C

Explanation:

Heat gain by water = Heat lost by nickel

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{nickel}\times C_{nickel}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{nickel}\times C_{nickel}\times (T_i-T_f)

For water:

Volume = 100.0 mL

Density of water= 1 g/mL

So, mass of the water:  

Mass\ of\ water=Density \times {Volume\ of\ water}  

Mass\ of\ water=1 g/mL \times {100.0\ mL}  

Mass of water  = 100 g

Initial temperature = 22.0 °C

Specific heat of water = 4.186 J/g°C

For nickel:

Mass = 275 g

Initial temperature = 100 °C

Specific heat of nickel = 0.444 J/gK = 0.444 J/g°C

So,  

100\times 4.186\times (T_f-22.0)=275\times 0.444\times (100-T_f)

418.6\times (T_f-22.0)=122.1\times (100-T_f)

418.6\times T_f+122.1\times T_f=21419.2

T_f=39.6138\ ^{0}C

Thus,  

The final temperature of the combined metals is 39.6138 °C

4 0
3 years ago
Read 2 more answers
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