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2 years ago

a car accelerate at 9 m/s squared. Assuming the car starts from rest how far will it travel in 10 seconds

Physics

1 answer:

rusak2 [61]2 years ago

7 0

Answer:

450m

Explanation:

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 0 + t + \frac{1}{2} a {t}^{2} \\ \frac{1}{2 } \times 9 \times {10}^{2} = 450m$

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Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

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two charges + 2.6 uc and -5.4uc experience an attractive force of 6.5mN. What is the separation between the charges?

djyliett [7]

Answer:

...

Explanation:

....

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Drag the tiles to the correct boxes to complete the pairs.

marishachu [46]

Answer:

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Water pollution---> eutrophication

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Light pollution---> sky glow

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a ray of light incident on a mirror, at an angle of 45°. Another mirror is placed at an angle of 45° to the first ones as shown.

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Answer:

If the ray of light is deflected by 45 degrees by the first mirror its total deflection by mirror (I) is 90 deg. (incident = 45 and exit ray equals 45 deg)

The second mirror will cause a net deflection of 90 degrees and the total deflection will be 180 deg or in opposite direction to the incident ray.

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Answer:

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I know it's one of these, try getting a second opinion

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