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3 years ago

400 N of load can be overcome by an effort of 50 N by using a lever. Calculate the mechanical advantage of the lever.

Physics

1 answer:

mars1129 [50]3 years ago

4 0

Answer:

load (l)=400N

Effort(E)=50N

mechanical advantage (MA)= load ÷Effort

(ma)=400÷50

(ma)=8

Explanation:

I copy pasted from the answer from the same question. Remember to first check if ur question is there

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If you ran 15 km/hr for 2.5 hours, how much distance would you cover?

SVETLANKA909090 [29]

Answer: 37.5 km

Explanation:

The question is that

If you ran 15 km/hr for 2.5 hours, how much distance would you cover ?

Where

Speed = 15 km/ hr

Time = 2.5 hours

Using the formula for speed.

Speed = distance/time

Substitute speed and time into the formula

15 = distance/ 2.5

Make distance the subject of formula by cross multiplying.

Distance = 15 × 2.5

Distance = 37.5 km.

6 0

3 years ago

The spring to launch a pinball in a pinball machine is compressed 25 cm and has a spring constant of 140 N/m.

mote1985 [20]

Answer:

I think it is 5.6. This is my answer

8 0

3 years ago

The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand

AlexFokin [52]

Answer:

$8.67791\times 10^{25}\ kg$

$0.34589\ m/s^2$

$0.07903\ m/s^2$

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

$r_m$ = Radius of Miranda = 236 km

m = Mass of Miranda = $6.6\times 10^{19}\ kg$

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg$

The mass of Uranus is $8.67791\times 10^{25}\ kg$

Acceleration is given by

$a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2$

Miranda's acceleration due to its orbital motion about Uranus is $0.34589\ m/s^2$

On Miranda

$g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2$

Acceleration due to Miranda's gravity at the surface of Miranda is $0.07903\ m/s^2$

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0

3 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

4 years ago

The internal energy of a system increases by 36 joules. The system does work using 17 joules. What amount of heat was added to t

Ray Of Light [21]

General internal energy expression:

Uf = (Ui-W) + Q

Where Uf = Final internal energy, Ui = Initial internal energy, W = Work done, Q = Heat added

But,
Uf -Ui = ΔU = 36 J, W = 17 J. Then,
ΔU = -W +Q => Q = ΔU+W = 36+17 = 53 J

Therefore, heat added to the system is 53 Joules

8 0

4 years ago

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