Can someone please help me with this problem?

Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a

masha68 [24]

Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
$F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N$
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
$\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m$
And by using Hook's law, we can find the constant of the spring:
$k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m$

4 0

3 years ago

Which forces are shown on a free body diagram?

notsponge [240]

Answer:

What is a Free Body Diagram?

The free body diagram helps you understand and solve static and dynamic problem involving forces. It is a diagram including all forces acting on a given object without the other object in the system. You need to first understand all the forces acting on the object and then represent these force by arrows in the direction of the force to be drawn.

Explanation:

5 0

2 years ago

The freight cars a and b have a mass of 20 mg and 15 mg, respectively. if the cars collide and couple together, what is the velo

igomit [66]

Suppose car A is moving with a velocity Va, and car b with a velocity Vb,

According the principle of conservation of momentum:

Va x Ma + Vb x Mb = (Ma + Mb) V

V = (Va x Ma + Vb x Mb)/(Ma +Mb)

V = speed of cars after coupling

V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)

Put in the values of Va and Vb, and get the V

7 0

2 years ago

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Particle q₁ has a charge of 2.7 μC and a velocity of 773 m/s. If it experiences a magnetic force of 5.75 × 10⁻³ N, what is the s

Ne4ueva [31]

The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
$F=qvB \sin \theta$
where $\theta$ is the angle between the directions of v and B.

1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is $q=2.7 \mu C=2.7 \cdot 10^{-6}C$ . In this case, v and B are perpendicular, so $\theta=90^{\circ}$ , therefore we have:
$B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T$

2) In this second case, the angle between v and B is $\theta=55^{\circ}$ . The charge is now $q=42.0 \mu C=42.0 \cdot 10^{-6}C$ , and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
$F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N$

5 0

2 years ago

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1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande

kvv77 [185]

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

5 0

2 years ago