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Gala2k [10]
3 years ago
15

If the work done to stretch an ideal spring by 4.0 cm is 6.0 J, what is the spring constant (force constant) of this spring? A)

300 N/m B) 3000 N/m C) 3500 N/m D) 7500 N/m E) 6000 N/m
Physics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

D) 7500 N/m

Explanation:

Given :

Work done by the spring, W = 6 J

Stretch in the spring, x = 4 cm = 0.04 m

Consider k be the spring constant.

The relation between work done and stretch of the spring is:

W=\frac{1}{2}kx^{2}

Substitute the suitable values in the above equation.

6=\frac{1}{2}k(0.04)^{2}

k=\frac{12}{1.6\times10^{-3} }

k = 7500 N/m

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A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?
Mkey [24]

The car's mass is 1600 kg.

Its weight is (mass) x (gravity).  

On Earth, that's (1600 kg) x (9.8 m/s²)  =  15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

So the minimum force required to lift the car is <em>15,680 Newtons</em>, directed vertically upward.

5 0
3 years ago
Find the net charge of a system consisting of 6.25×10^6 electrons and 7.75×10^6 protons. Express your answer using three signifi
jok3333 [9.3K]

Answer:

2.40 x 10⁻¹³ C

Explanation:

n_{e} = number of electrons = 6.25 x 10⁶

q_{e} = charge on electron = - 1.6 x 10⁻¹⁹ C

n_{p} = number of protons = 7.75 x 10⁶

q_{p} = charge on proton =  1.6 x 10⁻¹⁹ C

Net charge is given as

Q = n_{e} q_{e} + n_{p} q_{p}

Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)

Q = 2.40 x 10⁻¹³ C

6 0
3 years ago
1. What is the average speed of a runner that<br> travels 100 meters in 9.7 seconds?
alexdok [17]

Answer:

10.30928 m/s

Explanation:

6 0
3 years ago
What is the net force on a 50-newton weight hanging on a string tied to the ceiling?
Lady bird [3.3K]
The net force on the hanging object is zero. If it were not zero, then the object would be accelerating in some direction.
4 0
4 years ago
12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

v = 30\cdot t (2)

Free fall (v_{o} = 900\,\frac{m}{s}, a = -9.807\,\frac{m}{s}, t_{o} = 30\,s, 30\,s \le t \le 120\,s)

v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

5 0
3 years ago
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