How man significant figures does the following value gave 43.023

A roller coaster accelerates from an initial speed of 6.0 m/s to a final of 70 m/s over 4 seconds. What is the rigth formula to

Anton [14]

Answer:

a = 16 m/s²

Explanation:

Given that,

Initial speed of a roler coaster, v₁ = 6 m/s

Final speed of a roller coaster, v₂ = 70 m/s

Time, t = 4s

We know that, acceleration of an object is equal to the rate of change of velocity. Let the acceleration be a. Its formula is given by :

$a=\dfrac{v_2-v_1}{t}$

or

$v_2=v_1+at$

The correct option is (1). Its value is equal to :

$a=\dfrac{70-6}{4}\\\\a=16\ m/s^2$

Its value is equal to 16 m/s².

6 0

3 years ago

A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take

Maslowich

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$x_{o}$ , $y_{o}$ - Initial horizontal and vertical position of the projectile, measured in meters.

$v_{o,x}$ , $v_{o,y}$ - Initial horizontal and vertical speed of the projectile, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

$x$ , $y$ - Current horizontal and vertical position of the projectile, measured in meters.

Given that $x_{o} = 0\,m$ , $y_{o} = 80\,m$ , $v_{o,x} = 360\,\frac{m}{s}$ , $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the kinematic equations are, respectively:

$x = 360\cdot t$

$y = 80-4.094\cdot t^{2}$

(a) If $y = 0\,m$ , the time taken for the projectile to reach the water is:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If $t \approx 4.420\,s$ , the horizontal scope of the projectile is:

$x = 360\cdot (4.420)$

$x = 1591.2\,m$

The horizontal scope of the projectile is 1591.2 meters.

(c) If $t = 2\,s$ , the height that remains to descend is:

$y = 80-4.094\cdot (2)^{2}$

$y = 63.624\,m$

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

6 0

3 years ago

The distance from a crest to a trough is called the wave length

V125BC [204]

Answer:

Explanation:

The highest surface part of a wave is called the crest, and the lowest part is the trough. The vertical distance between the crest and the trough is the wave height. The horizontal distance between two adjacent crests or troughs is known as the wavelength.

I hope it's helpful!

5 0

3 years ago

Calculate the kinetic energy in joules of a 1500 kg automobile moving at 19 m/s

Svetlanka [38]

Using
KE = ½mv² = ½×1500×19×19 = 270750 joules

5 0

3 years ago

It takes 0.5 s to complete a 3 m wave. What is the speed of the wave?

Mademuasel [1]

Answer:

30

Explanation:

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i think that is right

4 0

3 years ago