Ask question

Ask question

All categories

3 years ago

12

During a softball game, a shortstop catches a ground ball. The action force is the ball pushing on the glove. What is the reacti

on force?

2 answers:

Tanzania [10]3 years ago

4 0

Its the floor pushing up on the player, this one is correct

Andrei [34K]3 years ago

4 0

Answers to the entire test. . .

The answer to this question, is on the first picture!

You might be interested in

The radius of a small ball is around 3.79747 cm. The radius of a basketball is about 3.16 times larger. What is the ratio of the

Svetradugi [14.3K]

Explanation:

The ratio of the areas is the square of the ratio of the radii.

A/A = 3.16² = 9.99

The ratio of the volumes is the cube of the ratio of the radii.

V/V = 3.16³ = 31.6

3 0

4 years ago

A sprinter starts from rest and accelerates to her maximum speed of 10.5 m/s in a distance of 11.0 m. (a) What was her accelerat

Anvisha [2.4K]

Answer:

a)a=5.01m/s^2

b)t=11.26s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)

{Vf^{2}-Vo^2}/{2.a} =X(2)

X=Xo+ VoT+0.5at^{2} (3)

X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve

to solve the question a, we can use the ecuation number 2

Vo=0

Vf=10.5 m/s

x=11m

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{10.5^{2}-0^2}/{2x11} =a

a=5.01m/s^2

to find the time we can use the ecuation number 1

Vf=Vo+a.t

t=(Vf-Vo)/a

t=(10.5-0)/5.01=2.09s

part b

in this case the spees is constant, so the movement is defined by the following ecuation

X=VT

t=x/v

t=96.3/10.5=9.17s

to find the total time we sum the times when the speed is constant and when the acceleration is constan

t=9.17+2.09

t=11.26s

8 0

3 years ago

A uniform electric field contains a number of particles. All are experiencing forces in the same direction as the electric field

myrzilka [38]

Answer: D

All the particles must be uncharged

Explanation:

If all the particles are positively charged, then there will be force of repulsion between them which will give different directions away from each other. The same is applicable if they are all negatively charged.

If the particles are positively and negatively charged, their will be force of attraction between them which will give different directions towards each other.

For all to be experiencing forces in the same direction, We can conclude that

All the particles must be uncharged.

8 0

3 years ago

The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

6 0

3 years ago

An orienteer runs 400m directly east and then 500m to the northeast (at a 45 degree andle from due east and from due north). Pro

Aleks04 [339]

Answer:

Final displacement with respect to the starting position is 832.37 meter

Explanation:

Lets consider that the orienteer start to run on the point of (0,0) point of a coordinate system. When he runs towards to east side about 400m, he will be at the point A(400,0). After he runs to the northeast (at a 45 degree angle from due east and from due north), approximately he will make 353.55 meter to east and 353.55 meter to the north side and he will be at the point of B(753.55, 353.55)

From the starting point total displacement will be 832.37 meter. Please check the attached graphical solution.

6 0

3 years ago