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3 years ago

Give an example of an unbalanced forces acting on an object

Physics

2 answers:

MariettaO [177]3 years ago

7 0

An unbalanced foreces a group of people playing tug of war

777dan777 [17]3 years ago

4 0

Forces that are equal in size but opposite in direction are called balanced forces. Balanced forces do not cause a change in motion. When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal but opposite force

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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

6 0

3 years ago

A parked car's horn (belonging to a musician) emits a concert A of frequency 440 on a day when the speed of sound is 342 m/s. Yo

julsineya [31]

Answer:

19.08 m/s

Explanation:

f = actual frequency emitted by the parked car's horn = 440 Hz

V = speed of sound = 342 m/s

f' = frequency of the horn observed by you = 466 Hz

v = speed of your car moving towards the parked car = ?

frequency of the horn observed by you is given as

$f' = \frac{Vf}{V - v}$

$466 = \frac{(342)(440)}{342 - v}$

v = 19.08 m/s

3 0

3 years ago

Two subway stops are separated by 1210 m. If a subway train accelerates at 1.30 m/s2 from rest through the first half of the dis

solong [7]

Answer:

Part 1) Time of travel equals 61 seconds

Part 2) Maximum speed equals 39.66 m/s.

Explanation:

The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

$v^{2}=u^2+2as$

where

'v' is the final speed

'u' is initial speed

'a' is acceleration of the body

's' is the distance covered

Applying the given values we get

$v^2=0+2\times 1.30\times \frac{1210}{2}\\\\v^{2}=1573\\\\\therefore v=39.66m/s$

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

$v=u+at\\\\v=0+1.30\times t\\\\\therefore t=\frac{39.66}{1.30}=30.51seconds$

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance

Thus total time of journey equals $T=2\times 30.51\approx61seconds$

Part b)

the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals $39.66m/s$

4 0

3 years ago

Someone help me with these two questions ASAP!!!

Sonbull [250]

For the first question it is the fourth option. Cryosphere is a term for the portions of earth that are covered in water when the water is in solid form. this includes both snow and ice.

For the second question the answer is a delta is formed at the mouth of the river a sediment is carried down stream. The hydrosphere refers to all water on earth.

5 0

2 years ago

Read 2 more answers

Help please I don't understand this

BARSIC [14]

It's either staying there or is going at the same pace

4 0

3 years ago