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DerKrebs [107]
3 years ago
12

Please can i have help with this question ​

Physics
1 answer:
kenny6666 [7]3 years ago
8 0
Opposites attract so north goes to south and Vice versa. When the same poles are facing each other they push away when they are opposites they attract.
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A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
Newtons third lawWhat action-reaction forces are involved when a rocket engine fires? Why doesnt a rocket need air to push on? A
dangina [55]

Answer: action forc roketorce

reaction force is engine fires

4 0
3 years ago
The tension of a guitar string is increased by 40%. By what factor odes the fundamental frequency of vibration change? a. 1.13 b
bogdanovich [222]

Answer:

<h3> b. 1.18</h3>

Explanation:

The fundamental frequency in string is expressed as;

F1 = 1/2L√T/m .... 1

L is the length of the string

T is the tension

m is the mass per unit length

If the tension is increased by 40%, the new tension will be;

T2 = T + 40%T

T2 = T + 0.4T

T2 = 1.4T

The new fundamental frequency will be;

F2 = 1/2L√1.4T/m ..... 2

Divide 1 by 2;

F2/F = (1/2L√1.4T/m)/1/2L√T/m)+

F2/F = √1.4T/m ÷ √T/m

F2/F = √1.4T/√m ×√m/√T

F2/F = √1.4T/√T

F2/F = 1.18√T/√T

F2/F = 1.18

F2 = 1.18F

Hence the fundamental frequency of vibration changes by a factor of 1.18

8 0
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.

5 0
3 years ago
Determine the kinetic energy of a 1000 kg roller coaster car that is moving with speed of 40.0 m/s​
ololo11 [35]

Answer:

KE=800,000

Explanation:

The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2

so 1000 is the mass and 40 is the velocity

KE=0.5*1000*40^2

KE=0.5*1,000*1,600

KE=800,000 Joules

8 0
3 years ago
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