A baseball player (hitter) hits the ball with her bat. The ball flies north at an angle a above ground and with an initial speed v towards a fielder whose job is to catch the ball before it hits the ground. The distance between the fielder and the hitter is D. At the moment when the bat hits the ball, the fielder starts running to the south towards the place where he believes the ball will fall. Assume the fielder runs with a constant speed all the time the ball is in the air. Assume the hand of the fielder catches the ball just before it hits the ground and that the ball is hit by the bat at ground level.
How fast must the fielder run to be able to catch the ball?

Answer:

microsecond = 1 × 10-6 seconds

1 second = 1 × 100 seconds

1 microsecond = (1 / 1) × 10-6 × 10-0 seconds

1 microsecond = (1) × 10-6-0 seconds

1 microsecond = (1) × 10-6 seconds

1 microsecond = 1 × 1.0E-6 seconds

1 microsecond = 1.0E-6 seconds,just do like it

Answer:

6.3 x [tex]10^{-3}[/tex]

Explanation:

1 millisecond = 1000 microseconds

0.0063 millisecond = 6.3 microseconds

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

Answer:

The frictional force between two bodies depends mainly on three factors: (I) the adhesion between body surfaces (ii) roughness of the surface (iii) deformation of bodies

Answer:

The scientific notation of 738 is 7.38 x 1002.

Answer:

     a = 6.1 m / s²

Explanation:

For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle

Let's find the relative initial velocity of the two vehicles

          v₀ = v₀₂ - v₀₁

          v₀ = 25.4 - 13.6

          v₀ = 11.8 m / s

the fastest vehicle

          x = v₀ t + ½ a t²

The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is

          x = 11.4 m

let's use the expression

           v² = v₀² - 2 a x

how the vehicle stops v = 0

          a = v₀² / 2x

          a = [tex]\frac{11.8^2}{2 \ 11.4}[/tex]

          a = 6.1 m / s²

this velocity is directed to the left

Answer:

a)  T = -22796.5 N,  b)  F = 3000 N

Explanation:

a) For this part we use Newton's second law

Let's set a reference frame with the x-axis in the direction of motion and the y-axis in the vertical direction.

We make a free-body diagram for each body,

the tractor has the bar force (T) and the push force (F) and the friction force (fr₁)

Y axis

        N₁ -W₁ = 0

        N₁ = M₁ g

X axis

        F - T - fr₁ = M₁ a

the Trailer has the bar force (T) and the friction force (fr₂)

Y axis  

      N₂ - W₂ = 0

      N₂ = m₂ g

X axis

     T - fr₂ = m₂ a

let's write the system of equations

      F - T - fr₁ = M₁ a          (1)

           T - fr₂ = m₂ a

we solve

       F - (fr₁ + fr₂) = (M₁ + m₂) a

indicate that the total friction forces are fr = 3000N

       fr = fr₁ + fr₂

        F =[tex]\frac{(M_1+m_2) a}{fr}[/tex]

let's calculate

        F =[tex]\frac{(2000+1500) \ 3}{3000}[/tex]

        F = 3.5 N

The friction force is

       fr = μ N

the norm of the system is N = N₁ + N₂

       μ = [tex]\frac{fr}{N_1 + N_2}[/tex]

       μ = [tex]\frac{3000}{2000+1500}[/tex]

       μ = 0.858

with this value we can find the friction force 1 and substitute in equation 1

       F - T - μ N₁ = M₁ a

       T = F - M₁ (a + μ g)

        T = 3.5 - 2000 (3 + 0.858  9.8)

        T = -22796.5 N

b) when the system moves with constant velocity the acceleration is zero

            F - T - fr₁ = 0

                  T - fr₂ = 0

we solve

           F + (fr₁ + fr₂) = 0

           F = fr₁ + fr₂

           F = 3000 N

Answer:

Explanation:

Let the stretch produced in first spring and second spring be x₁ and x₂ respectively. 300 N is the restoring force created in them .

k₁x₁ = k₂x₂ = 300 N .

k₁ = 11000 N/m

11000 x₁ = 300N

x₁ = 300 N / 11000

= 0.027 m

= 2.7 cm .

Correct question is;

1/0.12 = (1/0.05) + (1/d')

Answer:

d' = -1/700

Explanation:

1/0.12 = (1/0.05) + (1/d')

Let's rearrange to get;

(1/d') = (1/0.12) - (1/0.05)

(1/d') = (1/(12/100)) - (1/(5/100))

(1/d') = 100/12 - 100/5

Let's multiply through by 60 to get rid of the denominators on the right side;

> (1/d') = 500 - 1200

> (1/d') = -700

> d' = -1/700

Answer:

Explanation:

so opposite and equal , right?    forces are. soooo..

528+52= 580 N    is the force that is being exerted up on the scale

Answer:

The correct option is (e) "none of the above".

Explanation:

Given that,

A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.

If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.

We know that the time period can be calculated as :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

Where

l is the length

g is acceleration due to gravity

It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.

Answer:

The spring constant of the spring is 10.3 N/m.

Explanation:

Given that,

Mass of a bungee jumper, m = 59 kg

The period of oscillation, T = 0.25 min = 15 sec

We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]

Where

k is the spring constant

[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]

So, the spring constant of the spring is 10.3 N/m.

[tex]\implies F_1 < F_2[/tex]

[tex] \implies F_{net} = F_2 - F1[/tex]

[tex]\implies F_{net} = 42 -15[/tex]

[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]

The net force on the 4.0 kg box is 27 N towards EAST.

Answer:

The correct answer is Option A (decrease).

Explanation:

Other given choices are not connected to the given query. Thus the above is the right answer.

Answer:

C Quincy

Explanation:

Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.

I hope this helps!

Answer:

The correct answer is C. Quincy is 16 years old and lives in a dormitory

Explanation:

Quincy being 16 may be able to get a job and hold a membership at a gym even if not at a gym he still has the most freedom period with Lucinda having children and Frieda and Vladimir being to young to make the choices of exercising on their own.

Please tell me if I'm wrong so I may give you the correct answer!

Happy Holidays!!  

Solution-1:-

[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]

Solution:-2

[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]

Solution:-3

[tex]\boxed{\sf 320m}[/tex]

Solution:-4

[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]

Solution:-5

[tex]\boxed{\sf Mg_3N_2}[/tex]

Solution:-6

[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]

Solution:-7

[tex]\boxed{\sf {}^{}_{}N}[/tex]

Solution:-8

[tex]\boxed{\sf Galactuse}[/tex]

Solution:-9

[tex]\boxed{\sf Y-X}[/tex]

Solution:-10

[tex]\boxed{\sf Cell\:wall}[/tex]

Answer:

I hope this will help you. i tried my best

Wave-D (the red one) has the smallest amplitude of the 4 waves on this graph.

Answer: [tex]P=5573.43\ W[/tex]

Explanation:

Given

Mass of the elevator is [tex]M=650\ kg\\\[/tex]

Time period of ascension [tex]t=3\ s[/tex]

cruising speed [tex]v=1.75\ m/s[/tex]

Distance moved by elevator during this time

Suppose Elevator starts from rest

[tex]\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s[/tex]

Distance moved

[tex]\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m[/tex]

Gain in Potential Energy is

[tex]\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N[/tex]

Average power during this period is

[tex]\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W[/tex]

Answer:

The power is 331.7 W.

Explanation:

mass, m = 650 kg

time, t= 3 s

initial  velocity,  u = 0 m/s

final velocity, v = 1.75 m/s

(a) The power is defined as the rate of doing work.

Work is given by the change in kinetic energy.

W = 0.5 m (V^2 - u^2)

W = 0.5 x 650 x 1.75 x 1.75 =  995.3 J

The power is given by

P = W/t = 995.3/3 = 331.7 W

Answer:

[tex]K=0.023J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.15[/tex]

Velocity [tex]v=0.5m/s[/tex]

Angular Velocity [tex]\omega=8.4rad/s[/tex]

Generally the equation for Kinetic Energy is mathematically given by

[tex]K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)[/tex]

[tex]K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))[/tex]

[tex]K=0.023J[/tex]

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

       Em₀ = U = m g h

Final point. In the highest part of the loop

       Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

      I = m R²

      v = w R

we substitute

       Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

       em_f = m v² + 2 m g R

Energy is conserved

       Emo = Em_f

       mgh = m v² + 2m g R

       h = v² / g + 2R

The lowest velocity that the ball can have at the top of the loop is v> 0

      h> 2R

The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration  ( i.e. statement in the question is False )

Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.

In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.

Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration

Learn more : https://brainly.com/question/24658194

Answer:

resultant vector =0

Explanation:

because it is connected head to tail in a closed figure

Answer:

a. Total displacement = 140 km/h

b. Average velocity = 35 km/h

Explanation:

Given the following data;

Average velocity A = 80 km/h

Time A = 2.5 hours

Average velocity B = 40 km/h

Time B = 1.5 hours

a. To find the total displacement for the 4 h trip;

Total time = Time A + Time B

Total time = 2.5 + 1.5

Total time = 4 hours

Next, we would determine the displacement at each velocity.

Mathematically, displacement is given by the formula;

Displacement = velocity * time

Substituting into the formula, we have;

Displacement A = 80 * 2.5

Displacement A = 200 km/h

Displacement B = 40 * 1.5

Displacement B = 60 km/h

Total displacement = Displacement A - Displacement B

Total displacement = 200 - 60

Total displacement = 140 km/h

b. To find the average velocity for the total trip;

Mathematically, the average velocity of an object is given by the formula;

[tex] Average \; velocity = \frac {total \; displacement}{total \; time} [/tex]

Substituting into the formula, we have;

[tex] Average \; velocity = \frac {140}{4} [/tex]

Average velocity = 35 km/h

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

Answer:

0.028 M.

Explanation:

NOTE: This question is a chemistry question. However, the answer to the question can be obtained as shown below:

We'll begin by calculating the number of mole in 2.52 g of oxalic acid, C₂H₂O₄. This can be obtained as follow:

Mass of C₂H₂O₄ = 2.52 g

Molar mass of C₂H₂O₄ = (2×12) + (2×1) + (4×16)

= 24 + 2 + 64

= 90 g/mol

Mole of C₂H₂O₄ =?

Mole = mass / molar mass

Mole of C₂H₂O₄ = 2.52 / 90

Mole of C₂H₂O₄ = 0.028 mole

Finally, we shall determine the molarity of the solution. This can be obtained as follow:

Mole of C₂H₂O₄ = 0.028 mole

Volume = 1 L

Molarity =?

Molarity = mole / Volume

Molarity = 0.028 / 1

Molarity = 0.028 M

Therefore, the molarity of the solution is 0.028 M.

Answer: [tex]3.32\times 10^4\ kN/C[/tex]

Explanation:

Given

Charge at the center of the sphere is [tex]q=2\ nC[/tex]

Charge distributed over the entire sphere [tex]Q=1\ nC[/tex]

Radius of sphere [tex]r=4.8\ cm[/tex]

Using Guass law

[tex]\Rightarrow \oint \vec{E}\cdot \vec{dA}=\dfrac{q_{enc}}{\epsilon_o}\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+\dfrac{Q}{4\pi R^2}\times 4\pi r^2)\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+Q\dfrac{r^2}{R^2})\\\\\Rightarrow E\cdot (2.4\times 10^{-2})^2=9\times 10^9(2+1\cdot \dfrac{1}{4})\times 10^{-9}\\\\\Rightarrow E=3.32\times 10^4\ kN/C[/tex]

Direct relation is present between distance and time which means one has direct impact on the value of other.

There is direct relationship between distance and time. Direct relation means if one increases the other automatically increases while on the other hand, if one decreases the other will also decreases. If there is a large amount of distance so more time is required to pass it while on the other hand, if there is less distance so less time is required to complete it so we can conclude that there is direct relationship between distance and time.

https://brainly.com/question/24323600

Answer:

B. [tex]30\,s\,< t \le 40\,s[/tex].

Explanation:

The vehicle is travelling eastwards when its velocity is positive and westwards when velocity is negative. According to the graph, [tex]v > 0[/tex] for [tex]30\,s\,< t \le 40\,s[/tex]. Hence, correct answer is B.

In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.