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Artist 52 [7]
3 years ago
9

Who said "give me a place to stand and a long enough lever and I can move the world"?

Physics
1 answer:
elena-s [515]3 years ago
6 0
I searched the quote and everyone says it was Archimedes
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A 1500 kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force
MakcuM [25]

The correct answer to the question is : 9375 N.

CALCULATION:

As per the question, the mass of the car  m = 1500 Kg.

The diametre of the circular track D = 200 m.

Hence, the radius of the circular path R = \frac{D}{2}

                                                                  = \frac{200}{2}\ m

                                                                  = 100 m.

The velocity of the truck v = 25 m/s.

When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.

Hence, the force acting on the car is centripetal force.

The magnitude of the centripetal force is calculated as -

                              Force F = \frac{mv^2}{R}

                                            =  \frac{1500\times (25)^2}{100}\ N

                                            = 9375 N.           [ANS}        

The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.


6 0
3 years ago
Find the coefficient of kinetic friction μk. express your answer in terms of some or all of the variables d1, d2, and θ.
Andre45 [30]
<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
6 0
3 years ago
Read 2 more answers
According to peacemaking criminology, toward which option should a country redirect the money it currently spends on imprisoning
ratelena [41]

Answer:

Explanation:

It is C. Environmental campaigns

8 0
3 years ago
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NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU
daser333 [38]

Acceleration = (change in speed)/(time for the change)

Change in speed = (end speed) - (start speed)

Change in speed = (10 m/s) - (20 m/s) = -10 m/s

Time for the change = 5.00 seconds

Acceleration = (-10 m/s) / (5 sec)

<em>Acceleration = -2 m/s²</em>

That's choice-A .

8 0
3 years ago
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4. A 1,000 kg truck moving at 10 m/s runs into a concrete wall. It takes 0.5 seconds for the truck to conipietery
Maru [420]

Answer:

\large \boxed{\text{h. 20 000 N}}  

Explanation:

Force is the change in momentum over time

F = Δp/Δt

1. Calculate the change in momentum

p₁ = mv₁ = 1000 kg × 10 m/s = 10 000 kg·m·s⁻¹

p₂ = 0

Δp = p₂ - p₁= (0 - 10 000) kg·m·s⁻¹ = -10 000 kg·m·s⁻¹

2. Calculate the force

\begin{array}{rcl}F & = & \dfrac{\Delta p}{\Delta t}\\\\& = & \dfrac{-10 000 \text{ kg$\cdot$m$\cdot$ s}^{-1}}{\text{ 0.5 s}}\\\\& = & \textbf{-20 000 N}\\\end{array}\\\text{The negative sign shows that the force is exerted opposite to the direction of motion.}\\\text{The magnitude of the force is $\large \boxed{\textbf{20 000 N}}$}

3 0
3 years ago
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