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matrenka [14]
3 years ago
11

Please help

Physics
1 answer:
Aneli [31]3 years ago
5 0

Answer:

170 energy

Explanation:

I don't know the answer

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A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 74.3 ◦ with the floor. The ladder has a mass
olganol [36]

Answer: µ=0.205

Explanation:

The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,

f=Fw

The sum of the moments about the base of the ladder Is 0

ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²

Note that it doesn't matter WHAT the length of the ladder is -- it cancels.

Solve this for Fw.

0= 0.9637FwL - (67.91L)2.652

Fw=180.1/0.9637

Fw=186.87N

f=186.81N

Since Fw=f

We know Fw, so we know f.

But f = µ*Fn

where Fn is the normal force at the floor --

Fn = (25.8 + 67.08)kg * 9.8m/s² =

910.22N

so

µ = f / Fn

186.81/910.22

µ= 0.205

4 0
3 years ago
What is the process called when a solid changes to a<br> liquid?<br><br> ???20 points?ASAP please
svp [43]

Answer:

melting

Explanation:

6 0
3 years ago
Read 2 more answers
How do helmets protect your head? a. By creating a large area to spread out the pressure of a blow to your head. b. By focusing
Andru [333]
Your answer is A.

I hope this helps
5 0
3 years ago
Read 2 more answers
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
How could you test the hypothesis that elephants interpreted the ground signal as being farther away than the air signal?
bekas [8.4K]

Answer:

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

Explanation:

This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.

The speed of sound in solid soil is an average of the speed of its constituent media, giving values ​​between

 wood      3900 m / s

 concrete 4000 m / s

 fabrics     1540 m / s

 earth       5000 m / s wave S

 ground    7000 m / s P wave

 

we can see that the speed on solid earth is an order of magnitude greater than in air.

One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.

From the initial information, the wave going through the ground should arrive first.

3 0
4 years ago
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