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Leviafan [203]
2 years ago
10

Which of the following are elements in both civil and criminal trials. (select all that apply)

Physics
1 answer:
Ganezh [65]2 years ago
6 0

Answer:

a an b

Explanation:

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A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks on i
Maurinko [17]

Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

5 0
1 year ago
A 500 N force accelerates an object at 20 m s-2. What is its mass?
avanturin [10]

<u>Answer</u>: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

<u>Explanation:</u> Given force, F=500N

                                 acceleration, a=20 m/s^{2}

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

                         ⇒m=500/20=25

                 ∴ Mass of the object is 25 kg .

<u> </u><u>Reference Link: </u>brainly.com/question/1141170

#SPJ2

8 0
1 year ago
If an asteroid were going to impact Earth, would you prefer to know in advance or let the NEO remain a mystery until impact? Why
ivanzaharov [21]

I would like to know in advance
6 0
2 years ago
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
in 1859 a hunter brought 24 rabbits from England to Australia and release them to establish a population for sport hunting rabbi
slavikrds [6]
I would say the rabbits were breeding.
4 0
3 years ago
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