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Nastasia [14]
3 years ago
11

A student places a block on a table and hangs one mass from the block. The student lets the block go and observes the block move

towards the end of the table where the mass was located. The student then places the block on the table and hangs a second, larger mass from the opposite end of the block. The block moves in the opposite direction as the first trial. What does this experiment demonstrate? Use information from the experiment to support your answer.​
Physics
1 answer:
NeTakaya3 years ago
7 0

Magnitude of acceleration

Explanation:

We know that acceleration can increase depending in the force applied on an object, any object with a greater mass will apply a greater force. F = M(a).

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A force of 450N accelerated a box across a frictionless surface at a rate of 15 m/s^2. What is the mass of the box
larisa [96]
Formula:
F = ma

m = F÷A
m: mass F:force A: accelerarion

m = 450N ÷ 15m/s^2
= 30kg

Hope this helps :)
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3 years ago
An object is in equilibrium when acted on by three coplanar forces.
andrey2020 [161]

Answer:

A is correct

Explanation:

The vertical force up is balanced by the vertical component of the largest force

The horizontal force to the right is balanced by the horizontal component of the largest force.

C and D will both tend to accelerate upward at some angle

B looks like the horizontal force right is larger than the sum of the horizontal components of the other two forces.

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3 years ago
What part of the brain is referred to as the seat to consciousness?
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8 0
4 years ago
How long does it take a bird to fly 2.5 km if a bird can fly 23km/h?
Lina20 [59]
Time = Distance / speed

Time =  2.5 km / 23 km/h

         = 0.1087 hours                      1 hour = 60 minutes
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6 0
4 years ago
An object of mass 0.50 kg is released from the top of a building of height 8 m. The object experiences a horizontal constant for
liberstina [14]

To solve this problem with the given elements we will apply the linear motion kinematic equations. We will start by calculating the time taken, with the vertical displacement data. Subsequently, with the components of the acceleration, we will obtain the magnitude of the total acceleration, to finally obtain the horizontal displacement with the data already found.

PART A) From vertical movement we know that the acceleration is equivalent to gravity and the displacement is 8m so the time taken to carry out the route would be

h = \frac{1}{2} gt^2

Here,

h = 8 m

g = 9.8 m/s^2

Replacing,

8 = 0.5 * 9.8 * t^2

t = 1.277 sec

PART B) Now, Magnitude of acceleration

a = \sqrt{a_x^2 + a_y^2}

a_x = \frac{1.9}{0.5} = 3.8 m/s^2

a_y = g = 9.8 m/s^2

Thus, magnitude of net acceleration

a = \sqrt{3.82^2  + 9.8^2}= 10.51 m/s^2

PART C) Finally the displacement along horizontal direction is:

s =v_0 t + \frac{1}{2} a t^2

s = 0 + \frac{1}{2} (3.8)(1.277)^2

s = 3.098 m

Therefore the distance traveled along the horizontal direction before it hits the ground is 3.098m

7 0
3 years ago
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