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nexus9112 [7]
3 years ago
7

A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 190m. Find the the acceleration of t

he car.
Physics
1 answer:
Mama L [17]3 years ago
4 0

Answer:

a = 7.75 [m/²]

Explanation:

To solve this problem we must use the following equation of kinematics.

x=x_{0} +v_{o} *t + (\frac{1}{2})*a*t^{2}

where:

x = final distance = 190 [m]

Xo =  initial distance = 0

Vo = initial velocity = 0 (car starts from the rest)

a = acceleration [m/s²]

t = time = 7 [s]

190 = 0 + (0*7) + 0.5*a*(7²)

190 = 0.5*49*a

a = 7.75 [m/²]

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When looking at the chemical symbol, the charge of the ion is displayed as the
Black_prince [1.1K]

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

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6 0
3 years ago
A tray of electronic components contains 15 components, 4 of which are defective. If 4 components are selected, what is the poss
dlinn [17]

Answer:

a) 0.0007326

b) 0.03223

c) 0.2418

d) 0.2418

Explanation:

To find different probabilities for the selection of components among eleven good and four defective components, we will use the Combination.

a) C(4,4) = 1; C(15,4) = 1365

P = \frac{C(4,4)}{C(15,4)} = \frac{1}{1365} = 0.0007326

b) C(4,3) = 4; C(11,1) = 11

P = \frac{C(4,3)*C(11,1)}{C(15,4)} = \frac{4*11}{1365} = 0.03223

c) C(4,2) = 6; C(11,2) = 55

P = \frac{C(4,2)*C(11,2)}{C(15,4)} = \frac{6*55}{1365} = 0.2418

d) C(11,4) = 330

P = \frac{C(11,4)}{C(15,4)} = \frac{330}{1365} = 0.2418

8 0
3 years ago
A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s
mafiozo [28]

Answer:

block velocity   v = 0.09186 = 9.18 10⁻² m/s  and speed bollet   v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

     p₀ = m v₀ + 0

After the crash

   p_{f} = (m + M) v

    p₀ = p_{f}

    m v₀ = (m + M) v                    (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

    Em₀ = K = ½ m v2

Final

    E m_{f}= Ke = ½ k x2

   Emo = E m_{f}

   ½ m v² = ½ k x²

   v² = k/m  x²

Let's look for the spring constant (k), with Hook's law

   F = -k x

   k = -F / x

   k = - 0.75 / -0.25

   k = 3 N / m

Let's calculate the speed

  v = √(k/m)   x

  v = √ (3/8.00)   0.15

  v = 0.09186 = 9.18 10⁻² m/s

This is the spped of  the  block  plus bullet rsystem right after the crash

We substitute calculate in equation  (1)

   m v₀ = (m + M) v

  v₀ = v (m + M) / m

  v₀ = 0.09186 (0.008 + 0.992) /0.008

  v₀ = 11.5 m / s

6 0
3 years ago
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