Ask question

Ask question

All categories

3 years ago

6

A car is driving at a velocity of 24 m/s. If its breaks can supply an acceleration of -5.0 m/s/s, how much time will be required

to bring the car to a stop?

1 answer:

DedPeter [7]3 years ago

4 0

Around 5 seconds i think

You might be interested in

A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will

gtnhenbr [62]

1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
$x(t)=v_0t$
$y(t)=h- \frac{1}{2}gt^2$
where the motion on the x-axis is a uniform motion with constant speed $v_0=18.1 m/s$ , while the motion on the y-axis is a uniformly accelerated motion with constant acceleration $g=9.81 m/s^2$ , and initial height $h=1.85 m$ .

First of all, we can find the time t at which the ball reaches the ground by requiring $y(t)=0$ :
$0=h- \frac{1}{2}gt^2$
$t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }= 0.61 s$

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
$x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m$

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one:
$v_x=18.1 m/s$
On the y-axis, the velocity is given by
$v_y(t)=gt$
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
$v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s$
And the speed of the ball is the magnitude of the resultant of the two components:
$v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s$

8 0

3 years ago

Read 2 more answers

Fragments of molecules that have at least one electron available for bonding are called

Luda [366]

They are called radicals.................

5 0

3 years ago

A building that is 105.50 ft. feet tall is casting a shadow that is 37.50 ft. feet. If the building next to it, is casting a sha

Arlecino [84]

115.35 ft

Set the proportion up 37.50/105.50 = 41/x and solve for x

7 0

3 years ago

Which was the most important effect of the bessemer process

koban [17]

Answer: The effect of Bessemer process is its reduction in cost for steel production.

Explanation:

In the manufacturing of steel, Bessemer process was the first method discovered for mass production of steel. This was discovered by Sir Henry Bessemer and Williams Kelly both from United States. This method aids in the removal of impurities from iron and converts it to steel in few minutes (this usually takes a full day to achieve). The economy of the country improved as steel was made faster and cheaper causing companies to build thousands of new railroads.

6 0

3 years ago

Derive the formula for the moment of inertia of a uniform, flat, rectangular plate of dimensions l and w, about an axis through

Ad libitum [116K]

Answer:

A uniform thin rod with an axis through the center

Consider a uniform (density and shape) thin rod of mass M and length L as shown in (Figure). We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line. In this example, the axis of rotation is perpendicular to the rod and passes through the midpoint for simplicity. Our task is to calculate the moment of inertia about this axis. We orient the axes so that the z-axis is the axis of rotation and the x-axis passes through the length of the rod, as shown in the figure. This is a convenient choice because we can then integrate along the x-axis.

We define dm to be a small element of mass making up the rod. The moment of inertia integral is an integral over the mass distribution. However, we know how to integrate over space, not over mass. We therefore need to find a way to relate mass to spatial variables. We do this using the linear mass density of the object, which is the mass per unit length. Since the mass density of this object is uniform, we can write

λ = m/l (orm) = λl

If we take the differential of each side of this equation, we find

d m = d ( λ l ) = λ ( d l )

since

λ

is constant. We chose to orient the rod along the x-axis for convenience—this is where that choice becomes very helpful. Note that a piece of the rod dl lies completely along the x-axis and has a length dx; in fact,

d l = d x

in this situation. We can therefore write

d m = λ ( d x )

, giving us an integration variable that we know how to deal with. The distance of each piece of mass dm from the axis is given by the variable x, as shown in the figure. Putting this all together, we obtain

I=∫r2dm=∫x2dm=∫x2λdx.

The last step is to be careful about our limits of integration. The rod extends from x=−L/2x=−L/2 to x=L/2x=L/2, since the axis is in the middle of the rod at x=0x=0. This gives us

I=L/2∫−L/2x2λdx=λx33|L/2−L/2=λ(13)[(L2)3−(−L2)3]=λ(13)L38(2)=ML(13)L38(2)=112ML2.

4 0

3 years ago