Ask question

Ask question

All categories

4 years ago

14

Concept Simulation 2.3 provides some background for this problem. A ball is thrown vertically upward, which is the positive dire

ction. A little later it returns to its point of release. The ball is in the air for a total time of 7.96 s. What is its initial velocity? Neglect air resistance.

1 answer:

KATRIN_1 [288]4 years ago

8 0

Answer:

39 m/s

Explanation:

Hi!

The equation of motion of the ball is:

$y(t) = y_0 + v_{0y}t-(1/2)(9.8 m/s^2)t^2$

Where y_0 is the initial horizontal position, which we will take as zero.

v_0y is the initial velocity and t is the time

To find the initial velocity we will set y(7.96s) = 0

$0 = v_{0y} (7.96 s)-(4.9 m/s^2)(7.96 s)^2\\v_{0y} =(4.9 m/s^2) (7.96 s)\\v_{0y}=39.004 m/s$

You might be interested in

Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 15. What force is needed to acc

slavikrds [6]

Answer:

Force, F = 77 N

Explanation:

A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.

We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :

F = m × a

Where

m = mass = 55 kg

a = acceleration = 1.4 m/s²

$F=55\ kg\times 1.4\ m/s^2$

F = 77 N

So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.

3 0

3 years ago

An atom of a certain element has 36 protons, 36 electrons, and a mass number of 84. At room temperature, this element is a very

Grace [21]

Answer:

12 neutrons

Explanation:

The number of protons also shows the atomic number. Therefore the element in question is Krypton (Kr), which also is a noble gas.

Neutrons = Mass number - protons - electrons

Here neutrons = 84 - 36 - 36 = 12

5 0

3 years ago

Can someone please Help me with this? It’s Due today

kirza4 [7]

helium group no val elect

mg is reactive when activated. when burned, very intense

pot\asssium 1 valence elect ... KCl eg

theone with H and sodium in it

http://perendis.webs .com

6 0

3 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

3 years ago

What provides the force on the person on the passenger seat?

frutty [35]

When the car speeds up, slows down, or goes around a curve,
passengers need a force applied to them to make them do the
same thing, otherwise they won't keep up with the car.

The force on the passenger is applied by means of friction between
the upholstery and the seat of his pants, and also by the seat-back
or his seat-belt.

4 0

3 years ago

Read 2 more answers