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professor190 [17]
3 years ago
12

A net force of 500N acts on a 100 kg cart what is the acceleration

Physics
1 answer:
Len [333]3 years ago
5 0

The formula is sum of the forces = Mass* Acceleration


If there is no friction (which i'm assuming not)

500N = 100 * A

A = 5 M/S/S (Or M/S^2)

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The free length of the spring that is attached to the 0.3-lb slider is 3 in. If the slider is released from rest when x = 6 in.,
hichkok12 [17]

Answer:

a = 64 ft / s²

Explanation:

The force in a spring is given by Hooke's law

          F = -k x

Let's use the initial data to calculate the spring constant

         k = F / x

Reduscate to the English system

          x = 3 in (1foot/12 in) =0.25 foot

         k = 0.3 / 0.25

         k = 1.2  lb / foot

Now we can use Newton's second law

        F = ma

        a = F / m

        a = -k x / m

        m = w / g

        m = 0.3 / 32 = 0.009375

         x= 6 in (1foot /12 in)= 0.5 foot

        a = - 1.2  0.5  / 0.009375

        a = 64 ft / s²

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3 years ago
Okay, so I've been stuck on this for a couple days. I've looked online and whatnot for answers or some help and nothing. So I ca
Lina20 [59]
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Which is an example of an abiotic factor in a pond environment?
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3 0
3 years ago
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2 years ago
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A 325 N box moves with a constant velocity of 3.75 m/s when pushed with a force of 425 N exerted at a 35.2 angle below the horiz
ANTONII [103]

Answer: 0.6

Explanation:

If we draw a free body diagram of the box we will have the following:

Net force in the x-axis:

N-W-Fsin \theta=0 (1)

Net force in the y-axis:

-F_{r}+Fcos \theta=0 (2)

Where:

N is the normal force

W=325 N is the weight of the box

F=425 N is the force exerted on the box

\theta=35.2\° is the angle below the horizontal

F_{r}=\mu_{k}N is the friction force, being \mu_{k} the coefficient of kinetic friction

Isolating N from (1):

N=W+Fsin \theta (3)

Substituting (3) in (2):

-\mu_{k}(W+Fsin \theta)+Fcos \theta=0 (4)

Finding \mu_{k}:

\mu_{k}=\frac{Fcos \theta}{W+Fsin \theta} (5)

\mu_{k}=\frac{425 N cos (35.2\°)}{325 N+425 N sin (35.2\°)} (6)

Finally:

\mu_{k}=0.6 (5)

6 0
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