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Debora [2.8K]
3 years ago
11

What is a vernier caliper used for?​

Physics
1 answer:
iris [78.8K]3 years ago
5 0

me ajudem por favor pra agora de noite

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A. If the object is not moving relative to the surface it's in contact with. the friction force is static friction. Draw a free-
Ksenya-84 [330]

Answer:

a. The free body diagram for this object has been attached. It shows all the forces acting on the body at rest, including the friction force in the opposite direction to sliding of the object (assume it's left to right).

b. Since the object is in contact with the surface, there is a normal force acting on both of them and is equal to the weight exerted by each. This perpendicular force is defined by Newton's second law of motion.

c. The force of friction always acts in a direction opposite to the direction of motion of the body. F = mg ('a' for acceleration is replaced by 'g' gravity because acceleration in this case is just gravity).

Hope that answers the question, have a great day!

6 0
3 years ago
A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I
Lorico [155]

Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

Required

Acceleration of the cart a

To get that, we will use the equation of motion

v = u+at

Substitute the given parameters

8.2 = 3.5+1.5a

1.5a = 8.2-3.5

1.5a = 4.7

a = 4.7/1.5

a = 3.13cm/s²

Hence the acceleration to the cart is 3.13cm/s²

3 0
3 years ago
In which item is energy stored in the form of gravitational potential energy? A. A slice of bread B. A compressed spring C. An a
marysya [2.9K]

The things that determine the amount of an object's gravitational potential energy are ...

-- mass of the object

-- gravitational acceleration in the place where the object is

-- height of the object above the ground or the floor

A). <em>a slice of bread; </em> No.  It's still a slice of bread even if it's on the ground.

B. <em>A compressed spring; </em> No.  It's still a compressed spring even if it's on the ground.

C. <em>An apple on a tree</em>; <em>Yes !</em>  It can't be an apple on a tree if it's on the ground.  

D. <em>A stretched bow string</em>; <em>No.</em>  It's still a stretched bowstring even if it's on the ground.

3 0
3 years ago
Read 2 more answers
A billiard ball moving at 5 m/s strikes a stationary ball of the same mass. After the collision, the original ball moves at a ve
SIZIF [17.4K]

90 degrees - 30 = 60 degrees

Velocity = (5m/s - 4.35m/s x cos(30)) / cos(60)

Velocity = 2.47 m/s

The answer is D) 2.47 m/s at 61.9 degrees

8 0
3 years ago
A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.
Daniel [21]

Answer:

a) a_{center} = 7.38 ~m/s^2

b) a_{end} = 14.77 ~m/s^2

c) v_{center} = 2.43~m/s

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

\tau =I\alpha

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2

So, the Newton's Second Law yields

\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

a = \alpha r = 7.38~m/s^2

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

a = \alpha L = 14.77~m/s^2

c) The conservation of energy can be used to find the velocity when the rod is vertical.

K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s

The linear velocity is v = ωr, so

v = 2.43 m/s.

4 0
3 years ago
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