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prohojiy [21]
3 years ago
11

A box rests on top of a flat bed truck. the box has a mass of m = 18 kg. the coefficient of static friction between the box and

truck is μs = 0.86 and the coefficient of kinetic friction between the box and truck is μk = 0.69. 1) the truck accelerates from rest to vf = 16 m/s in t = 16 s (which is slow enough that the box will not slide). what is the acceleration of the box?
Physics
1 answer:
konstantin123 [22]3 years ago
5 0
For this case, the first thing you should do is write the kinematic motion equation of the block.
 We have then:
 vf = vo + a * t
 Where,
 vf: Final speed.
 vo: Initial speed.
 a: acceleration.
 t: time.
 Substituting the values:
 (16) = (0) + a * (16)
 Clearing the acceleration:
 a = 16/16 = 1m / s ^ 2
 Note: the other data for this case are not used in this problem.
 answer:
 The acceleration of the box is 1m / s ^ 2
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Answer:

V = 156.13 [cm³]

Explanation:

El volumen de un solido con forma de paralepipedo se puede calcular por medio de la siguiente formula:

V = ancho*largo*alto

donde:

V = volumen [cm³]

ancho = 3.4 [cm]

largo = 11.2 [cm]

alto = 4.1 [cm]

Ahora reemplazando.

V = 3.4*11.2*4.1\\V = 156.13 [cm^{3}]

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3 years ago
The moon orbits the earth once every 27 days at a distance of 384400 km. The international space station orbits the earth at an
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Answer:

ive never done this befor but i think its 36 times aroud earth a day

Explanation:

384400/400=961  

961% of 27 days is 40minuets

1440/40=36

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3 years ago
A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
Yuliya22 [10]

Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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zavuch27 [327]
The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.

Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.

Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
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2 years ago
8750 J of heat are applied to a piece of aluminium causing a 56C increase in its
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Answer:146.8983 grams

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