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matrenka [14]
3 years ago
5

An inclined plane is 5.00 m long and 3.00 m high. What is the ideal mechanical advantage of this machine?

Physics
1 answer:
Dvinal [7]3 years ago
4 0
 <span>inclined plane formula is length/hight 

so 5/2= 2.5 

and work= f x d so  
work= 5 x 104 x 10= 5200 W 
</span>
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90 percent a day to keep things running smoothly
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3 years ago
A parallel-plate capacitor has square plates that have edge length equal to 1.20×102cm and are separated by 1.00 mm. It is conne
lora16 [44]

Answer:

Energy stored in the capacitor will be 1.835\times 10^{-6}J

Explanation:

We have given edge length of capacitor = 1.2\times 10^2cm=1.2m

So area A = 1.2×1.2 = 1.44 m^2

Separation is given as d  1 mm = 10^{-3}m

We know that capacitance is given by C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 1.44}{10^{-3}}=12.744\times 10^{-9}F

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3 0
3 years ago
Read 2 more answers
I need an answer for this plzz!!<br>number 2 <br>anybody can help ??
Anuta_ua [19.1K]
2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
8 0
3 years ago
A 2300 kg truck has put its front bumper against the rear bumper of a 2500 kg suv to give it a push. with the engine at full pow
valentina_108 [34]
F=ma
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a=18000/4800
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Final answer
7 0
3 years ago
Read 2 more answers
A 0.10-kg ball, traveling horizontally at 25 m/s, strikes a wall and rebounds at 19 m/s. What is the 7) magnitude of the change
IRISSAK [1]

Answer:

Change in momentum will be -4.4 kgm/sec

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Mass of the ball is given m = 0.10 kg

Initial velocity of ball v_1=25m/sec

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So change in momentum is equal to =m(v_2-v_1)=0.1\times (-19-25)=-4.4kgm/sec ( here negative sign shows only direction )

So option (A) will be correct answer

5 0
2 years ago
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