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wolverine [178]
3 years ago
5

The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P

= 181 N, and (c) P = 352 N. The acceleration is positive if up the slope, negative if down the slope.
Physics
1 answer:
bogdanovich [222]3 years ago
5 0

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

a=\dfrac{P}{m}

a=\dfrac{181\ N}{68\ kg}

a=2.67\ m/s^2

(c) P = 352 N

a=\dfrac{P}{m}

a=\dfrac{352\ N}{68\ kg}

a=5.17\ m/s^2

Hence, this is the required solution.

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An 89.0 kg fullback moving east with a speed of 5.6 m/s is tackled by an 85.0 kg opponent running west at 2.84 m/s, and the coll
Korolek [52]

Answer:

a. v_f=1.477m/s

b. ΔK=1558.3J

c. E_k=1034.7 J

Explanation:

a).

Momentum conserved

p_{ix}=p_{fx}

m_1*v_1+m_2*v_2=(m_1+m_2)*v_f

v_f=\frac{m_1*v_1+m_2*v_2}{m_1+m_2}

v_f=\frac{89.0kg*5.6m/s+85.0kg*-2.84m/s}{(89.0+85.0)kg}

v_f=1.477m/s

b).

ΔK=K_i-K_f

\frac{1}{2}*m_1*v_1^2+\frac{1}{2}*m_2*v_2^2=\frac{1}{2}*(m_1+m_2)*v_f^2

\frac{1}{2}*89.0kg*(5.6m/s)^2+\frac{1}{2}*85.0kg*(2.84m/s)^2=\frac{1}{2}*(89.0+85.0)kg*(1.447m/s)^2

ΔK=1558.3J

c).

E_k=\frac{1}{2}*89kg*(5.8m/s)^2-\frac{1}{2}*(85+89)kg*(1.44m/s)^2

E_k=1034.7 J

d).

All of which has been lost as mechanical energy, and is now thermal energy warmer football players, noise a loud crunch for example.

8 0
3 years ago
What is the equation linking pressure difference, depth, density and g
Leokris [45]
Pressure difference = depth x density x gravity
7 0
3 years ago
If a 100-n net force acts upon a 50g car, what will the acceleration of the car be
Brrunno [24]

Answer:

2m/s

Explanation:

The formula for acceleration can be found by [ a = f/m ]

We are given the force [ 100N ] and the mass [ 50g ].

We can use these values to solve for the acceleration.

a = 100/50

a = 2m/s

Best of Luck!

3 0
3 years ago
28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp at a sp
Ad libitum [116K]

Answer:

a) # buses = 7  

Explanation:

For this exercise we use the kinematic equations, let's find the time it takes to reach the same height

   

     y =v_{oy}  t - ½ g t²

Let's decompose the speed, with trigonometry

      v₀ₓ = v₀ cos θ

      v_{oy} = v₀ sin  θ

      v₀ₓ = 40 cos 32

      v₀ₓ = 33.9 m / s

      v_{oy} = 40 sin32

      v_{oy} = 21.2 m / s

When it arrives it is at the same initial height y = 0

         0 = (v_{oy} - ½ gt) t

That has two solutions

       t = 0                    when it comes out

       t = 2 v_{oy} / g       when it arrives

       t = 2 21.2 /9.8

       t = 4,326 s

We use the horizontal displacement equation

       x = vox t

       x = 33.9   4.326

       x = 146.7 m

To find the number of buses we can use a direct proportions rule

    # buses = 146.7 / 20

    # buses = 7.3

    # buses = 7

The distance of the seven buses is

     L = 20 * 7 = 140 m

b) let's look for the scope for this jump

     R = vo2 sin2T / g

     R = 40 2 without 2 32 /9.8

     R = 146.7 m

As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)

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3 years ago
There are no risks when taking medication to treat anxiety.
lidiya [134]

Answer:

False

Explanation:

5 0
3 years ago
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