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*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer 1
Answer:

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

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A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is at rest just after the separation. How much work was done by the internal forces that caused the separation

Answers

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_(f) = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_(f)

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_(f) = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

Final answer:

The principle of conservation of momentum implies that no work is performed by the internal forces during the separation of the space vehicle. This is granted that external forces are ignored and the total momentum and kinetic energy of the closed system remain constant.

Explanation:

The subject you're asking about centers around the principle of conservation of momentum. In the case of this space vehicle, before separation, the momentum of the whole system is given by the product of the mass and velocity, mv. After separation, one piece is at rest, leaving the other piece with momentum mv. As there is no external force, the total momentum does not change, so no work is performed by the internal forces causing the separation.

In more detail, the principle of conservation of momentum states that the total linear momentum of a closed system remains constant, regardless of any interactions happening within the system. The system is 'closed' meaning that no external forces are acting upon it. In this case, the space vehicle and the two smaller pieces it separates into form a closed system. This is consistent with your question's stipulation to ignore external forces, such as gravitational forces.

This can also be understood from the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. If we consider the vehicle before and after the separation, the kinetic energy of the system remains the same: initially all the energy is concentrated in the moving vehicle, and after the separation, all the kinetic energy is transferred to the moving piece while the at-rest piece has none. Therefore, the work done by the internal forces - which would change the kinetic energy - must be zero.

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A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but the plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?

Answers

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

(V_1)/(T_1)=(V_2)/(T_2)\n\Rightarrow {V_2}=(V_1)/(T_1)* T_2\n\Rightarrow {V_2}=(12)/(298.15)* 373.15\n\Rightarrow {V_2}=15.01 L

∴ Final volume at 100°C is 15.01 Liters.

A white blood cell has a diameter of approximately 12 micrometers or 0.012 um a model represents its diameter as 24 um what ratio of model size

Answers

Answer:

The ratio of the model size is 1 : 2000

Explanation:

Given

Real Diameter =  0.012 um

Scale Diameter =  24 um

Required

Determine the scale ratio

The scale ratio is calculated as follows;

Scale = (Real\ Measurement)/(Scale\ Measurement)

Substitute values for real and scale measurements

Scale = (0.012\ um)/(24\ um)

Divide the numerator and the denominator by 0012um

Scale = (1)/(2000)

Represent as ratio

Scale = 1 : 2000

Hence, the ratio of the model size is 1 : 2000

The ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

The ratio of the model size to the actual size can be calculated using the given measurements:

Actual Diameter = 0.012 um

Model Diameter = 24 um

Ratio = Model Diameter / Actual Diameter

Ratio = 24 um / 0.012 um

Ratio = 2000

So, the ratio of the model size to the actual size is 1 : 2000. This means the model represents the white blood cell's diameter 2000 times larger than its actual size.

Learn more about white blood from the link given below.

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Can you guys please help me out

Answers

The answer is A I’m pretty sure it is

The International Space Station (ISS) is a space station orbiting the earth above the ground. If the radius of the earth is 3,958.8 miles, mass of earth is 5.972 x 10 24 kg, the period of the ISS at the orbit around the earth is 7.84 hours, can you calculate what is the distance from the ISS to the surface of the earth, in unit of miles

Answers

Answer:

8488 miles

Explanation:

The orbital period around an earth is given as:

T=2\pi \sqrt{(r^3)/(Gm) }

Where G = constant = 6.67 x 10ˉ¹¹ N m² kgˉ², m = mass of object, T = period taken to round the earth, r = distance from the center of the earth to the orbiting object = radius of earth + orbital altitude.

Given that T = 7.84 hours = 28224 seconds, m = 5.972 x 10²⁴ kg, radius of earth = 3,958.8 miles = 6371071 m

T=2\pi \sqrt{(r^3)/(Gm) }\n\nsquaring:\n\nT^2=4\pi^2 ((r^3)/(Gm) )\n\nr^3=(GmT^2)/(4\pi^2) \n\nr=\sqrt[3]{(GmT^2)/(4\pi^2) } \n\nr=\sqrt[3]{(6.67*10^(-11)*5.972*10^(24)*(28224)^2)/(4\pi^2) } \n\nr=20031232.62\ meters

r = radius of earth + distance from the ISS to the surface of the earth

distance from the ISS to the surface of the earth = r - radius of earth

distance from the ISS to the surface of the earth = 20031232.62 meters -  6371071 meters = 13660161.62 meters

distance from the ISS to the surface of the earth = 13660161.62 meters =  8488 miles

The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?

Answers

(a) The magnitude of the force F acting on the knot is 5.54 N.

(b) The angle α of the force F is 54.4⁰.

The given parameters:

  • F force at α
  • 5.7 N force at 50⁰
  • 6.2 N force at 44⁰
  • 6.7 N force at 43⁰

The net vertical force on the knot is calculated as follows;

F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\n\nF_y = F sin(\alpha) -4.51\n\nFsin(\alpha) = 4.51

The net horizontal force on the knot is calculated as follows;

F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\n\nF_x = -Fcos(\alpha) + 3.22\n\nFcos(\alpha) = 3.22

From the trig identity;

sin^2 \theta + cos^ 2 \theta = 1\n\n

(Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\n\nF^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\n\nF^2(1) = 30.71\n\nF = √(30.71) \n\nF = 5.54 \ N

The angle α of the force F is calculated as follows;

Fsin(\alpha) = 4.51\n\nsin(\alpha) = (4.51)/(F) \n\nsin(\alpha ) = (4.51)/(5.54) \n\nsin(\alpha ) = 0.814\n\n\alpha = sin^(-1)(0.814)\n\n\alpha = 54.5 \ ^0

Find the image uploaded for the complete question.

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The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have

F₁ + F₂ + F₃ + F₄ = 0

Decomposing each force into horizontal and vertical components, we have

F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0

F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0

Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to

F cos(α) ≈ - 3.22 N

F sin(α) ≈ 4.51 N

(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :

(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N²   →   F ≈ 5.5 N

(2) Use the definition of tangent to solve for α :

tan(α) = sin(α) / cos(α) ≈ 1.399   →   α ≈ 126º

or about 54º from the horizontal from above on the left of the knot.
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