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solong [7]
2 years ago
15

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to

an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly move at a constant velocity of 256m/s
Physics
1 answer:
Salsk061 [2.6K]2 years ago
7 0

Explanation:

1.2 \mathrm{N} ; 2 \mathrm{N}

2.200 \mathrm{N} ; 200 \mathrm{N}

4.2 \mathrm{N} ; 2 \mathrm{N} ; 4 \mathrm{N}

5.2 \mathrm{N} ; 2 \mathrm{N} ; 2 \mathrm{N}

6.2 \mathrm{N} ; 2 \mathrm{N} ; 3 \mathrm{N}

8.200 \mathrm{N} ; 200 \mathrm{N} ; 5 \mathrm{N}

In only the above cases (i.e 1,2,4,5,6,8 ) the object possibly moves at a constant velocity of 256 \mathrm{m} / \mathrm{s}

You should have noticed that the sets of forces applied to the object are the same asthe ones in the prevous question. Newton's 1st law (and the 2nd law, too) makes nodistinction between the state of re st and the state of moving at a constant velocity(even a high velocity).

In both cases, the net force applied to the object must equal zero.

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A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac
maks197457 [2]
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ =  (r₁/ r₂)² · ω<span>₁
     
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ =  (4)² · ω<span>₁
     = 16 </span>· ω<span>₁

Hence, the protostar will be rotating 16 </span><span>times faster.</span>
5 0
2 years ago
Which statement is true?
iogann1982 [59]
B 
Think of inertia of getting into a car accident without a seat belt although the car stops you will not you would likely fly out the window
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1 year ago
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Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box, what is the magnit
vlabodo [156]
Newton's third law tells us that for every force there is an equal and opposite force.  This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.
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2 years ago
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For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s




Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




7 0
2 years ago
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