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kondor19780726 [428]
2 years ago
9

A light spring obeys Hooke's law. The spring's unstretched length is 34.0 cm. One end of the spring is attached to the top of a

doorframe and a weight with mass 7.00 kg is hung from the other end. The final length of the spring is 44.5 cm. (a) Find its spring constant (in N/m).
Physics
1 answer:
sleet_krkn [62]2 years ago
5 0

When the spring is extended by 44.5 cm - 34.0 cm = 10.5 cm = 0.105 m, it exerts a restoring force with magnitude R such that the net force on the mass is

∑ F = R - mg = 0

where mg = weight of the mass = (7.00 kg) g = 68.6 N.

It follows that R = 68.6 N, and by Hooke's law, the spring constant is k such that

k (0.105 m) = 68.6 N   ⇒   k = (68.6 N) / (0.105 m) ≈ 653 N/m

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PLEASEEE HELPPP!!!
Alja [10]

Given :

A mover slides a refrigerator weighing 650 N at a constant velocity across the floor a distance of 8.1 m.

The force of friction between the refrigerator and the floor is 230 N.

To Find :

How much work has been performed by the mover on the refrigerator.

Solution :

Since, refrigerator is moving with constant velocity.

So, force applied by the mover is also 230 N ( equal to force of friction ).

Now, work done in order to move the refrigerator is :

W = Force\times distance\\\\W = 230 \times 8.1\ N\ m\\\\W = 1863\ N\ m

Hence, this is the required solution.

3 0
3 years ago
Acceleration of a free-falling object in a frictionless environment increases as a function of time.
Digiron [165]

A free-falling object is an object moving under the effect of gravitational forces alone

The correct option to select for the True or False question is False

The reason the above selected option is correct is as follows:

According to Newton's second law of motion, we have;

Force = Mass × Acceleration

The force of gravity is F_{g} =G \cdot \dfrac{M \cdot m}{r^{2}}

Where;

G \cdot \dfrac{M }{r^{2}} = Acceleration \ due \ to \ gravity , \ g \approx 9.81 m/s^2

m  = The mass of the object

∴ The force acting on an object in free fall, F_g = m × g

Therefore the acceleration of an object in free fall is the constant acceleration due to gravity, and it therefore, does not change with time

The correct option for the question, acceleration of a free-falling object in a frictionless environment increases as a function of time is <u>False</u>

<u></u>

Learn more about object in free fall here:

brainly.com/question/13712424

brainly.com/question/11698474

6 0
2 years ago
1. Define weight
andrezito [222]

1. Weight is the gravitational pull with which the earth attracts the body towards the center of the earth. The S.I unit is Newton (N)

2. The weight of an object is related to its mass with the below equation.

W = mg

Where W = weight, m = mass and g = acceleration due to gravity

3. The mass m = 50 kg and g = 9.8 m/s^{2}

Substitute the parameters in the equation above.

W = 50 x 9.8

W = 490 N

4. An object is accelerating when it speeds up. If the object slows down, it means it is decelerating. The correct answer is option A

5. Newton's second law of motion is:

F = ma

Where F = force applied, m = mass and a = acceleration

Therefore, Newton's second law of motion relates an object's acceleration to its net force acting on it. The correct answer is option C

6. According to Newton's second law of motion which relates an object's acceleration to its mass, doubling the net force acting on an object a doubles its acceleration. Because mass is always constant.

7. Since the weight of an object is related to its mass with the equation.

W = mg,  If the mass of an object doubles, its weight will also doubles. Option A is the correct answer.

8. If you know the mass of an object, you can calculate its weight with the formula F = mX when X = 9.8m/s^{2}

9. Force is expressed in unit as Newton (N)

10. The parameters given are :

mass m = 20kg

Force F = 40N

To calculate acceleration, use the formula F = ma

Substitute all the parameters into the equation.

40 = 20a

a = 40/20

a = 2m/s^{2}

The correct option is D

Learn more here : brainly.com/question/18835375

8 0
3 years ago
The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f
Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb,  stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

W =k\int_{6}^{10.1} x

W = \dfrac{k}{2}[x^2]_6^{10.1}

W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

3 0
3 years ago
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven t
Bond [772]

Answer:

\frac{D_{jet}}{D_{prop}}=2.865

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

D=\frac{1}{2}CpAv^2\\

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\   \frac{D_{jet}}{D_{prop}}=2.865

4 0
3 years ago
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