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Tatiana [17]
3 years ago
6

Question 11 of 11

Physics
2 answers:
patriot [66]3 years ago
7 0

Answer:

d

Explanation:

VladimirAG [237]3 years ago
6 0

Answer:

D.

Explanation:

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A sixth grader moves down the hall at 1.2 m/s. When he sees a bunch of eighth graders coming, he
balandron [24]

Answer:

0.75m/s^2

Explanation:

3.6-1.2=2.4 m/s (Change in velocity)

2.4/3.2=0.75 m/s/s or m/s^2

5 0
3 years ago
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
the two forces f1 and f2 acting at A have a resultant force of Fr= -100lb. determine the magnitude and coordinate direction angl
faust18 [17]
Two forces F<span>1 and </span>F<span>2 act on the screw eye. The resultant force </span>FR<span> has a magnitude of 125 lb and the coordinate direction angles shown in (Figure 1) . Determine the magnitude of </span>F<span>2. Determine the coordinate direction angle </span>α<span>2 of </span>F<span>2. Determine the coordinate direction angle </span>β<span>2 of </span>F<span>2. Determine the coordinate direction angle </span>γ<span>2 of </span>F<span>2.</span>
8 0
3 years ago
if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?
KonstantinChe [14]
Since the stone was dropped from height, its initial velocity = 0 m/s

Using  v² = u² + 2gs.

Where g ≈ 10 m/s²,  u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s 
6 0
3 years ago
What does the area under the velocity-time graph represent.
V125BC [204]

Answer:

The distance traveled!

Explanation:

This is a velocity time graph of an object moving in a straight line due North.

8 0
3 years ago
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