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4 years ago

A moving object is in equilibrium. Which best describes the motion of the object if no forces change?

2 answers:

8 0

If the object is in equilibrium that means that the sum of the forces on it is zero and the net force is zero. If none of the forces changes then the object continues in constant uniform motion. That means constant speed in a straight line.

katrin2010 [14]4 years ago

5 0

Answer:

Uniform speed linear motion .

Explanation:

Any object can be in the state of equilibrium is sum of all the forces acting on it is zero means zero net force. If no forces change over it then the object will be uniform motion which means speed will remain constant. Best example for this situation is uniform motion along a straight line or uniform speed linear motion.

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Work is done if you carry a plant across a room at constant velocity. True False

BlackZzzverrR [31]

it is false hope this helps

5 0

4 years ago

Read 2 more answers

The acceleration due to gravity on the moon is about 5.4 ft/s2 . If your weight is 150 lbf on the earth:

Arada [10]

Answer:

4.662 slugs

Explanation:

Your mass on the moon should always be the same as any planet you are on (due to law of mass conservation), only your weight be different as gravitational acceleration is different on each planet.

If you weight 150 lbf on Earth, and gravitational acceleration on Earth is 32.174 ft/s2. The your mass on Earth is

m = W / g = 150 / 32.174 = 4.662 slugs

which is also your mass on the moon.

4 0

3 years ago

A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr

irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

$a = \dfrac{T}{m_2} - g$ .......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

$a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$ .........(2)

from equations 1 and 2

$\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}$

$\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

$T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)$

$T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}$

$T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}$

$T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}$

T = 90.61 N

from equation (1)

$a = \dfrac{90.61}{8} - 9.8$ .......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

$t = \sqrt{2.63}$

t = 1.62 s

5 0

3 years ago

The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

sattari [20]

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

8 0

4 years ago

Figure 23.9 shows a sliding mass on a spring. Assume there is no friction.

Tom [10]

<span>Without friction, there will be undamped simple harmonic motion. The force of the spring is proportional to the distance from the equilibrium point. The period of oscillation will be independent of the amplitude.

I hope my answer has come to your help. God bless and have a nice day ahead!</span>

4 0

3 years ago