Setting goals that match your lifestyle helps improve chances that they will be reached.

Nat2105 [25]

Answer:

The slope of the position time graph gives the velocity.

Explanation:

The slope of the position time graph gives the value of velocity.

In first graph,

The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts. and more than the second part, so the initial velocity is more than the final velocity.

In the second graph,

The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.

6 0

3 years ago

An object is traveling at 15 m/s and then after 5 seconds ends up traveling at 10 m/s. Calculate the speeds acceleration rate. W

SSSSS [86.1K]

Answer:

1) a = -1 m/s²

2) v = 12 m/s

Explanation:

Given,

The initial velocity of the object, u = 15 m/s

The final velocity of the object, v = 10 m/s

The time taken by the object to travel is, t = 5 s

Using the first equation of motion

v = u + at

a = (v - u) / t

Substituting the values

a = (10 - 15) / 5

= -1 m/s²

The negative sign indicates the body is decelerating

The acceleration of the object is, a = -1 m/s²

The speed of the object after 2 seconds

From the above equations of motion

v = 15 + (-1) 2

= 12 m/s

Hence, the speed of the object after 2 seconds is, v = 12 m/s

7 0

4 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

(c) 2.86s

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

the answer to (b) is 6.86m

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

the answer to (c) is 2.86s

6 0

3 years ago

Determine the average velocity and average speed for the following situation. A dog runs west for 35 meters then turns east for

mart [117]

Answer:

0.882 m/s average velocity and 1.71 m/s average speed

Explanation:

The dog travels a total of 35 m west and 110 m east.

110-35 = 75 m east of the starting position. Since velocity is a vector you must consider its first and final position and not the total distance traveled.

75 m / 85 s = 0.882 m/s average velocity

Speed is not concerned with direction so we instead add the total distance traveled which is 35+110 = 145 m. We then perform the same operation as before and divide by the time it took to run this distance.

145 m / 85 s = 1.71 m/s average speed

7 0

3 years ago

When the amplitudes of waves are equal, which frequency waves have the most energy?

Molodets [167]

Answer:

higher frequency waves will have more energy

Explanation:

5 0

2 years ago