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SVEN [57.7K]
3 years ago
9

Describe a situation where an object has a changing velocity but constant speed.

Physics
1 answer:
ehidna [41]3 years ago
8 0
To summarize, an object moving in uniform circular motion is moving around the perimeter of the circle with a constant speed<span>. While the </span>speed<span> of the object is</span>constant<span>, its </span>velocity<span> is </span>changing<span>. </span>Velocity<span>, being a vector, has a </span>constant<span>magnitude </span>but<span> a </span>changing<span> direction.</span>
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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

7 0
3 years ago
A 0.500 kg bullet is fired from a gun at 25.0 m/s, how much kinetic energy does it have?
slamgirl [31]

Considering the definition of kinetic energy, the bullet has a kinetic energy of 156.25 J.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and in a rest position, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its rest state by applying a force to it.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Kinetic energy of a bullet</h3>

In this case, you know:

  • m= 0.500 kg
  • v= 25 m/s

Replacing in the definition of kinetic energy:

Ec= ½ ×0.500 kg× (25 m/s)²

Solving:

<u><em>Ec= 156.25 J</em></u>

Finally, the bullet has a kinetic energy of 156.25 J.

Learn more about kinetic energy:

brainly.com/question/25959744

brainly.com/question/14028892

#SPJ1

5 0
1 year ago
What is the maximum speed when the conditions are mass =450 kg, initial height= 30 m, and the roller coaster is initially at res
Zarrin [17]

Answer:

B. 24.2 m/s

Explanation:

Given;

mass of the roller coaster, m = 450 kg

height of the roller coaster, h = 30 m

The maximum potential energy of the roller coaster  due to its height is given by;

P.E_{max} = mgh\\\\PE_{max} = 450 *9.8*30\\\\PE_{max} = 132,300 \ J

P.E_{max} = K.E_{max} \ (law \ of \ conservation\ of \ energy)

K.E_{max} = \frac{1}{2}mv_{max}^2\\\\ v_{max}^2 = \frac{2K.E_{max}}{m}\\\\ v_{max}^2 = \frac{2*132300}{450}\\\\ v_{max}^2 =588\\\\v_{max} = \sqrt{588}\\\\  v_{max} = 24.2 \ m/s

Therefore, the maximum speed of the roller coaster is 24.2 m/s.

3 0
2 years ago
Even though you praise your dog for sitting inside on his bed and chewing the bone you gave him, your dog insists on going outsi
qwelly [4]

Answer:

dog

Explanation:

5 0
2 years ago
an empty propane tank dropped from a hot air balloon hits the ground with a speed of 143.8 m/s. from what height was the tank re
vodomira [7]
What you know:
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m
8 0
3 years ago
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