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2 years ago

Starting from rest, a car undergoes a constant acceleration of 6 m/s^2. How far will the car travel in the two seconds?

1 answer:

4 0

If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.

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How can being scientifically literate help you in your own life?

stiv31 [10]

Answer: You can betterly understand what's around you and how it works

8 0

2 years ago

Consider the following list of numbers. 129, 685, 125, 511, 601, 52, 46 The height of a binary search tree is the maximum number

NemiM [27]

Answer:

The height of the tree is three (3) deep

Explanation:

It's 3 deep

Under 129, comes 125 and 685;

Under 125, comes 52 : Under 685, comes 511;

Under 52, comes 46 : Under 511, is 601.

This is illustrated below.

129

∧

125,685

|,|

52,511

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46,601

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4 years ago

Why can people float effortlessly on the dead sea?

ahrayia [7]

There water has a lot of salt in it, so it would be denser. People float better when the water is denser.

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3 years ago

An astronaut holds a rock 100m above the surface of Planet X. The rock is then thrown upward with a speed of 15m/s, as shown in

Harlamova29_29 [7]

The acceleration due to gravity of the planet X is 1 m/s².

The given parameters;

height above the ground, h = 100 m
initial velocity of the rock, u = 15 m/s
time of motion of the rock, t = 10 s

The acceleration due to gravity is calculated as follows;

$h = ut - \frac{1}{2} gt^2\\\\100 = 15(10) - (0.5\times 10^2)g\\\\100 = 150 - 50g\\\\50g = 150-100\\\\50g = 50\\\\g = 1 \ m/s^2$

Thus, the acceleration due to gravity of the planet X is 1 m/s²

Learn more here: brainly.com/question/24564606

7 0

2 years ago

A constant torque is applied to a rigid wheel whose moment of inertia is 2.0 kg · m2 around the axis of rotation. If the wheel s

Jlenok [28]

Answer:

The applied torque is 3.84 N-m.

Explanation:

Given that,

Moment of inertia of the wheel is $2\ kg-m^2$

Initial speed of the wheel is 0 (at rest)

Final angular speed is 25 rad/s

Time, t = 13 s

The relation between moment of inertia and torque is given by :

$\tau=I\alpha \\\\\tau=I\times \dfrac{\omega_f}{t}\\\\\tau=2\times \dfrac{25}{13}\\\\\tau=+3.84\ N-m$

So, the applied torque is 3.84 N-m.

4 0

3 years ago