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sveticcg [70]
4 years ago
14

Finding spring constant given displacement of spring and distance traveled by a cart

Physics
1 answer:
Alborosie4 years ago
5 0
 - (spring constant) (new length of spring - original length of spring) = Force applied to spring.
that is
-kx=F

Did you only have how far the cart traveled? No mass or acceleration or speed or time taken?
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A child is sliding down a slide at the playgound. is mechanicalenergy conserved
Flauer [41]

No. Mechanical energy is not conserved.  There's quite a bit of friction on the slide.  So some of the potential energy is lost to heat on the way down, and the child arrives at the bottom with hot pants and less kinetic energy than you might expect.

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3 years ago
A car drives for 30 km with a speed of 30m/s. How much time does it take the car to travel this distance?
Artist 52 [7]

                   Time  =  (distance)  /  (speed)

                           =  (30 km)  /  (30 m/s)

                           =  (30,000 m)  /  (30 m/s)

                           =  (30,000 / 30)  sec

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                           =      16 minutes  40 seconds  
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3 years ago
Describe mechanical energy in your own words
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3 0
3 years ago
Read 2 more answers
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
The drawing represents a mountain bike trail. A rider began at Point 1 on the trail and stopped at Point 3. At which point did h
Citrus2011 [14]

Answer:

Point 2.

Explanation:

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P.E = mgh

Where:

P.E is the potential energy.

m is the mass of the object.

g is acceleration due to gravity.

h is the height to which the object is located.

From the above equation, we can thus say that potential energy depends on the height of the object since the mass of the object is always constant i.e as the height of the object increase, the potential energy also increases and as the height of the object decrease, the potential energy also decreases.

Now, considering the diagram in the question given, we can see that point 2 is the lowest height to which the rider is located. At this point i.e point 2, the rider will have the least potential energy.

5 0
3 years ago
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