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3 years ago

A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s. What is its acceleration?

Physics

2 answers:

frosja888 [35]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf -7.09 \ m/s^2}}$

Explanation:

We are asked to find the acceleration of a car. Acceleration is the change in velocity with respect to time. We will use the following formula:

$a= \frac {v_f-v_i}{t}$

In this formula, $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time. The car slows down from 27.7 meters per second to 10.9 meters per second in 2.37 seconds. Therefore:

$v_f$ = 10.9 m/s
$v_i$ = 27.7 m/s
$t$ = 2.37 s

Substitute the values into the formula.

$a= \frac{ 10.9 m/s-27.7 m/s}{2.37 \ s}$

Solve the numerator.

10.9 m/s - 27.7 m/s= -16.8 m/s

$a= \frac{-16.8 \ m/s}{2.37 \ s}$

Divide.

$a= -7.088607595 \ m/s/s$

$a= -7.088607595 \ m/s^2$

The original measurements of velocity and time have 3 significant figures, so our answer must have the same. For the number we found, that is the hundredth place. The 8 in the thousandth place tells us to round the 8 in the hundredth place up to a 9.

$a \approx - 7.09 \ m/s^2$

The acceleration of the car is approximately -7.09 meters per second squared. The acceleration is negative because the car is slowing down.

Sunny_sXe [5.5K]3 years ago

4 0

Answer:

- 7.088 m/s²

Explanation:

As we know that,

★ Acceleration = Change in velocity/Time

→ a = (v - u)/t

Here,

Initial velocity (u) = 27.7 m/s
Final velocity (v) = 10.9 m/s

→ a = (10.9 m/s - 27.7 m/s)/2.37 s

→ a = -16.8/2.37 m/s²

→ a = -7.088 m/s² [Answer]

Negative sign denotes that the velocity is decreasing.

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3 years ago

When viewed in yellow light, an object that reflects all the colors of light will appear_____.

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Answer:

Yellow.

Explanation:

An object that reflects all the incidents colors of light will surely reflect off the color yellow, and when the reflected ray hits our eyes, we will perceive the object as yellow. The object that reflects off all colors appears to us as white. So if you want to make sure, then just shine yellow light on a white fabric and see what happens!

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3 years ago

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

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$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

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baherus [9]

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3 years ago

Read 2 more answers

how fast will and in what direction will a 20kg object accelerate if one force pushes at a 30 degree angle and another pushes at

DiKsa [7]

Answer:

$|a|=2.83\ m/s^2$

$\theta=75^o$

Explanation:

Net Force And Acceleration

The Newton's second law relates the net force applied on an object of mass m and the acceleration it aquires by

$\vec F_n=m\vec a$

The net force is the vector sum of all forces. In this problem, we are not given the magnitude of each force, only their angles. For the sake of solving the problem and giving a good guide on how to proceed with similar problems, we'll assume both forces have equal magnitudes of F=40 N

The components of the first force are

$\vec F_1=$