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Llana [10]
3 years ago
8

What is a combination of two or more simple machines? a complicated machine, a compound machine ,a compiled machine, a converted

machine .
Physics
2 answers:
nadya68 [22]3 years ago
3 0
It's a compound machine
Nuetrik [128]3 years ago
3 0

Answer:

The correct answer is compound machine.

Explanation:

Compound machine are combination of two or more simple machines.Te example of compound machine is bicycle in which wheel, axle separately from different system.

The effort drawn by the compound machine are less than the efforts drawn by simple machines to move an object.They also perform more work in comparison to the work performed by the simple machines.

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10. A worker uses a pulley system to raise a 24.0 kg carton 16.5 m. A force of 129 N is exerted and the rope is pulled 33.0 m. a
lozanna [386]

Answer: Machanical advantage of the machine is 1.86

Explanation: Machanical advantage of a machine is the ratio of the Force to overcome which is the load in this case 24kg * 10= 240N to the force exerted(Effort) to overcome the load in this case 129N.

So, we have

MA = load/effort

= 240N/129N

= 1.86.

5 0
3 years ago
What does density have to do with heat?
joja [24]
If you take a fluid (i.e. air or water) and heat it, the portion that is heated usually expands. The same mass takes up more volume and as a consequence the heated portion becomes less dense than the portion that is<span><span> not heated.</span> </span>
8 0
3 years ago
Read 2 more answers
Jacki evaluated the expression below. 2 cubed (3 minus 1) + 4 (8 minus 12) = 2 cubed (2) + 4 (4) = 8 (2) + 16 = 16 + 16 = 32. Wh
Naddika [18.5K]

Answer:

Her error was that she did not subtract 12 from 8 correctly

Explanation:

Jackie did 8-12 instead of 12-8

8 0
3 years ago
If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass
mixas84 [53]

Answer:3.874 m/s^2

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

60mi/h \approx 26.8224m/s

34mi/h \approx 15.1994 m/s

we know acceleration is given by =\frac{velocity}{Time}

a=\frac{15.1994-26.8224}{3}

a=-3.874 m/s^2

negative indicates that it is stopping the car

Distance traveled

v^2-u^2=2as

\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s

s=\frac{488.419}{2\times 3.874}

s=63.038 m

7 0
3 years ago
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

5 0
3 years ago
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