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Katarina [22]
3 years ago
9

Which wedge requires the least amount of force to do a job ?

Physics
2 answers:
yanalaym [24]3 years ago
6 0

The correct answer is:

C. Y

The wedge Y requires the least amount of force to do a job.

|Huntrw6|

Vlad [161]3 years ago
3 0

The answer is Y I'm sorry if i'm wrong guys

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Con lắc lò xo có độ cứng k = 100N/m được gắn vật có khối lượng m=0.1kg, kéo vật ra khỏi vị trí cân bằng 1 đoạn 5cm rồi buông tay
Delvig [45]

Answer:

The maximum velocity is 1.58 m/s.

Explanation:

A spring pendulum with stiffness k = 100N/m is attached to an object of mass m = 0.1kg, pulls the object out of the equilibrium position by a distance of 5cm, and then lets go of the hand for the oscillating object. Calculate the achievable vmax.

Spring constant, K = 100 N/m

mass, m = 0.1 kg

Amplitude, A = 5 cm = 0.05 m

Let the angular frequency is w.

w = \sqrt{K}{m}\\\\w = \sqrt{100}{0.1}\\\\w = 31.6 rad/s

The maximum velocity is

v_{max} = w A\\\\v_{max} = 31.6\times 0.05 = 1.58 m/s

8 0
3 years ago
Identify the forces acting on the cars and explain why the cars do not slide down the hill
Dvinal [7]
<span>Examples of outside forces acting on a car is gravity, wind, and other cars. Cars do not slide down hills because their weight, combined with the friction of their tires against the road, hold them in place. </span>
5 0
3 years ago
Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-hz sound waves into a room where the
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Given the following in the problem:

Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s

Get the wavelength of the sound by using the formula: 
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2 
Get the path length difference to the point from the two speakers 
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47

x = delta / lambda

If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.

Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference. 
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3 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

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3 years ago
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AysviL [449]
D. All of the above

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