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3 years ago

11

You drive a race car around a circular track of radius 300 m at a constant spe 108 km/h. Your acceleration will be: A) 39.9 m/s

B) 9.8 m/s C) 3.0 m/s D) 0.36 m/s

1 answer:

morpeh [17]3 years ago

8 0

Answer:

C

Explanation:

Centripetal acceleration is:

a = v² / r

First, convert km/h to m/s:

108 km/h × (1000 m / km) × (1 h / 3600 s) = 30 m/s

Therefore, the acceleration is:

a = (30 m/s)² / 300 m

a = 3 m/s²

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How does increased pressure result in the formation of a mineral?

drek231 [11]

The best answer is
A) <span>The atoms in the mineral get rearranged

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3 years ago

A 12,000kg. railroad car is traveling at +2m/s when it

ivann1987 [24]

<u>Answer:</u>

The final velocity of the two railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

$\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}$

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

$\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}$

Which is the final velocity of the two railroad cars

8 0

3 years ago

How many degrees are in each quadrant <br> A: 90°<br> B: 30°<br> C: 180°<br> D: 360°

Airida [17]

In one quadrant there are 90 degrees.

8 0

3 years ago

A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a

tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

$9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}$

W = 3.266 N

The mass of the meters stick is :

$m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg$

So, the mass of the meter stick is 0.333 kg.

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3 years ago

How do you find the oscillation period in seconds for different pendulum lengths?

GalinKa [24]

To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

Where L is the length of the pendulum and g is the acceleration due to gravity.

On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

$T=2\pi\sqrt{\frac{L}{g}}$

3 0

8 months ago