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pishuonlain [190]
3 years ago
8

Which of the following is best described as an inclined plane twisted into a spiral?

Physics
1 answer:
melamori03 [73]3 years ago
6 0
It is a screw. Hope this helps!
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P1: Explain in detail - How can the motion of an object that is already moving change? (What are the different ways a moving obj
Tomtit [17]

Inertia: tendency of an object to resist changes in its velocity. An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

5 0
4 years ago
The critical angle for a certain air-liquid surface is 47.7°. What is the index of refraction of the liquid? Round to the neares
KengaRu [80]

Answer:

The index of refraction of the liquid is 1.35.

Explanation:

It is given that,

Critical angle for a certain air-liquid surface, \theta_1=47.7^{\circ}

Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1

Using Snell's law for air liquid interface as :

n_1\ sin\theta_1=n_2\ sin(90)

n_1\ sin(47.7)=1

n_1=\dfrac{1}{sin(47.7)}

n_1=1.35

So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.

5 0
3 years ago
Difference between object distance and image distance within 4 points .(class 10)​
Sergio039 [100]

The relationship between object distance, image distance, and focal length of a spherical mirror is given by

1/f=1/v+1/u

Where

f= focal length of a spherical mirror (distance between the pole and the principal focus of the mirror)

u= object distance (distance between pole and object)

v= image distance (the distance between pole and image)

6 0
3 years ago
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
4 years ago
What two things must a measurement include?
astraxan [27]
The number and units. An example would be 3cm or 4kg.
7 0
3 years ago
Read 2 more answers
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