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VARVARA [1.3K]
3 years ago
10

Simplify the expression [9+2(2-4)]/3

Physics
1 answer:
il63 [147K]3 years ago
4 0

Answer:

5/3

Explanation:

[9+2(2-4)]/3

BODMAS

[9+2(-2)]/3

[9-4]/3

5/3

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umka21 [38]
Location & Sunlight Availability.
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8 0
2 years ago
The 20-g bullet is travelling at 400 m/s when it becomes embedded in the 2-kg stationary block. The coefficient of kinetic frict
nikklg [1K]

Answer:

The distance the block will slide before it stops is 3.3343 m

Explanation:

Given;

mass of bullet, m₁ = 20-g = 0.02 kg

speed of the bullet, u₁ =  400 m/s

mass of block, m₂ = 2-kg

coefficient of kinetic friction,  μk = 0.24

Step 1:

Determine the speed of the bullet-block system:

From the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the speed of the bullet-block system after collision

(0.02 x 400) + (2 x 0) = v (0.02 + 2)

8 = v (2.02)

v = 8/2.02

v = 3.9604 m/s

Step 2:

Determine the time required for the bullet-block system to stop

Apply the principle of conservation momentum of the system

v(m_1+m_2) -F_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -N \mu_kt = v_f(m_1 +m_2)\\\\v(m_1+m_2) -g(m_1 +m_2) \mu_kt = v_f(m_1 +m_2)\\\\3.9604(2.02)-9.8(2.02)0.24t = v_f(2.02)\\\\8 - 4.751t = 2.02v_f\\\\3.9604 - 2.352t = v_f

when the system stops, vf = 0

3.9604 -2.352t = 0

2.352t = 3.9604

t = 3.9604/2.352

t = 1.684 s

Thus, time required for the system to stop is 1.684 s

Finally, determine the distance the block will slide before it stops

From kinematic, distance is the product of speed and time

S = \int\limits {v} \, dt \\\\S = \int\limits^t_0 {(3.9604-2.352t)} \, dt\\\\ S = 3.9604t - 1.176t^2

Now, recall that t = 1.684 s

S = 3.9604(1.684) - 1.176(1.684)²

S = 6.6693 - 3.3350

S = 3.3343 m

Thus, the distance the block will slide before it stops is 3.3343 m

3 0
3 years ago
Read 2 more answers
A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont
PIT_PIT [208]

Answer:

v' = 2.83 m/s

Explanation:

Velocity of wave in stretched string is given by the formula

v = \sqrt{\frac{T}{\mu}}

here we know that

T = 4 N

also we know that linear mass density is given as

\mu = 1 kg/m

so we have

v = \sqrt{\frac{4}{1}} = 2 m/s

now the tension in the string is double

so the velocity is given as

v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s

v' = 2.83 m/s

4 0
3 years ago
Find the speed of a long distance runner who runs 30km in 6 hours
slava [35]

30mi/6hrs is a speed of 5 mph, which converts to a pace of 12 min/mi.

7 0
2 years ago
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QUESTION 1 Linear Motion
Misha Larkins [42]

Answer:

48.51ms / 174.6 km/h

Explanation:

y = 1/2 x g x t^2         v = g x t          

when y = 120m

120 = 1/2 x 9.8 x t^2

t^2 = 24.49

t = 4.95s

when t = 4.95s

v = 9.8 x 4.95

v = 48.51 m/s = 174.6 km/h

I'd say its realistic. But I don't really know that sry

3 0
2 years ago
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