A baseball is thrown a distance of 20 m what is its speed if it takes 0.5 seconds to cover the distance

Answers

Answer 1

Answer:

40m/s

Explanation:


Related Questions

A projectile is thrown with a velocity of 40 m/s, 45 degrees above the +x-axis. Determine its initial x-velocity and its initial y-velocity.

Answers

x-component is V·cos(angle)

y-component is V·sin(angle)

When the angle is 45°, its sin and cos are both 1/2·√2 .

Vx = 20·√2 = 28.28 m/s

Vy = 20·√2 = 28.28 m/s

Why do we use copper wiring?

Answers

Explanation:

The lower the level of resistivity the more electrical conductivity a metal has. Copper has low resistivity, and therefore is an excellent conductor. Copper is also less oxidative than other metals

A young parent is dragging a 65 kg (640 N) sled (this includes the mass of two kids) across some snow on flat ground, by means of a rope attached to the sled. The rope is at an angle of 30 degrees with respect to the ground and the tension in the rope is 160 N. The sled is moving at a constant velocity of 1.5 m/s.
(a) Draw and label all forces acting on the kids + sled system. Indicate the relative size of each force by scaling the length of each force arrow appropriately
(b) Calculate the normal force acting on the system
(c) Calculate the force of friction acting on the system.
(d) Calculate the coefficient of friction between the sled and the snow.

Answers

Answer:

b) N = 560 N, c)  fr = 138.56 N, d)  μ = 0.247

Explanation:

a) In the attachment we can see the free body diagram of the system

b) Let's write Newton's second law on the y-axis

              N + T_y -W = 0

              N = W -T_y

let's use trigonometry for tension

             sin θ = T_y / T

             cos θ = Tₓ / T

             T_y = T sin θ

             Tₓ = T cos θ

we substitute

              N = W - T sin 30

we calculate

              N = 640 - 160 sin 30

              N = 560 N

c) as the system goes at constant speed the acceleration is zero

X axis

              Tₓ - fr = 0

               Tₓ = fr

we substitute and calculate

              fr = 160 cos 30

              fr = 138.56 N

d) the friction force has the formula

             fr = μ N

             μ = fr / N

we calculate

             μ = 138.56 / 560

             μ = 0.247

A uniform 140 g rod with length 57 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 30 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 11 cm on each sides of the center, at which time the system rotates at an angular speed of 23 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer:

The correct answer is "12 rad/s"

Explanation:

The given values are,

Mass of rod,

M = 140 g

i.e.,

   = 0.14 kg

Length,

L = 57 cm

i.e.,

  = 0.57 m

Mass of beads,

M = 30 g

i.e.,

   = 0.03 kg

Angular speed,

r = 11 cm

i.e.,

 = 0.11 m

Now,

The inertia of rods will be:

=  [tex]\frac{1}{12}ML ^2[/tex]

On substituting the values, we get

=  [tex]\frac{1}{12}\times 0.14\times (0.57)^2[/tex]

=  [tex]0.0037905 \ kg-m^2[/tex]

The inertia of beads will be:

=  [tex]mr^2[/tex]

On substituting the values, we get

=  [tex]0.03\times (0.11)^2[/tex]

=  [tex]0.000726 \ kg-m^2[/tex]

The total inertia will be:

=  [tex]Inertia \ of \ rods+Inertia \ of \ beads[/tex]

=  [tex]0.0037905 + 0.000726[/tex]

=  [tex]0.0045165 \ kg-m^2[/tex]

now,

The angular speed of the system will be:

⇒ [tex]L_1w_1=L_2w_2[/tex]

On substituting the values in the above equation, we get

⇒ [tex]0.0045165\times 23 = (0.0037905 + (0.03\times 0.285^2)\times 2 )\times w_2[/tex]

⇒ [tex]0.1038795 = 0.0037905 + (0.00243675\times 2 )\times w_2[/tex]

⇒             [tex]w_2 = 12 \ rad/s[/tex]

A 1.6 kg ball is attached to the end of a 0.40 m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.80 kg block that is at rest on a horizontal frictionless surface. The speed of the block just after the collision is 3 m/s. What is the speed of the ball just after the collision

Answers

Answer:

the speed of the ball just after the collision is 1.5 m/s.

Explanation:

Given;

mass of the ball, m₁ = 1.6 kg

initial velocity of the ball, u₁ = 0

mass of the block, m₂ = 0.8 kg

initial velocity of the block, u₂ = 0

final velocity of the block, v₂ = 3 m/s

let the final velocity of the ball after collision = v₁

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.6 x 0   +    0.8 x 0       =   1.6 x v₁     +  0.8 x 3

0 = 1.6v₁  + 2.4

-1.6v₁ = 2.4

v₁  = -2.4 / 1.6

v₁ = - 1.5 m/s

v₁ = 1.5 m/s (in opposite direction of the block)

Therefore, the speed of the ball just after the collision is 1.5 m/s.

Which types of reactions are also redox reactions?​

Answers

Answer:

Your ans is B)

Explanation:

Hope its right

if wrong firgive me

Which of the following is an example of the transformation of gravitational potential energy into motion energy (kinetic energy)?

Answers

Answer:

c

Explanation:

because a drop of water is falling and that is gravitational potential energy into motion energy

A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 11.5 pC. The inner cylinder has a radius of 0.550 mm, the outer one has a radius of 7.20 mm, and the length of each cylinder is 13.0 cm.
(a) What is the capacitance?
(b) What applied potential difference is necessary to produce these charges on the cylinders?

Answers

Answer:

A)2.811 × 10^-12 F

B)4.09V

Explanation:

(a) What is the capacitance?

Capacitance of coaxial cylinder can be determined using below expresion

C= (2π ε0 L)/ Ln[ rb/ra]

Where

ε0= permittivity of free space= 8.85×10^−12 Fm^-1

L= length of each cylinder =13.0 cm= 13×10^-2m

rb= radius of outer cylinder= 7.20 mm= 7.20×10^-3m

ra=radius of inner cylinder=0.550 mm= 0.550×10^-3 m

If we substitute the values we have,

C= (2π ε0 L)/ Ln[ rb/ra]

C= ( 2 × π × 8.85×10^−12 ×13×10^-2) / Ln[

7.20×10^-3/0.550×10^-3]

C=( 7.2288×10^-12 )/2.5729

C=2.811 × 10^-12 F

B) (b) What applied potential difference is necessary to produce these charges on the cylinders?

Vba= Q/C

Where Vba=potential difference

Q= charge on each = 11.5 pC. = 11.5×10^-12

C= capacitance= 2.811 × 10^-12 F

If we substitute the values we have

Vba=(11.5×10^-12)/2.811 × 10^-12

= 4.09V

What is Gravity?
PLEASE ANSWER ​

Answers

Answer:

Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun.

Please mark as brainliest

Have a great day, be safe and healthy

Thank u  

XD

Answer:

GRAVITY:

In physics, gravity is the force that attracts a body towards the centre of the earth, or towards any other physical body having mass.

Direction of Wave Travel
С
Which letter
correctly
identifies the
wavelength of
this wave?
А
B
A. A
B. B
C. C

Answers

Answer:

I THINK it's A but I'm not completely sure

A Is the answer! Good luck

PLEASE HELP ME WITH ONE QUESTION
If 1000 J of heat energy is lost from a piece of hot metal which is placed into a cup of cool water, how much energy is gained by the water?

Answers

Answer:

1000J

Explanation:

According to the law of calorimetry;

Heat lost by cold substance = Heat gained by hot substance

If 1000 J of heat energy is lost from a piece of hot metal,

heat lost by hot substance = 1000J

Since we are to determine how much energy is gained by the water, using the law

Heat lost by hot metal = heat gained by the water

Hence heat gained by the water = 1000J

A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a hill of unknown height. It then goes down that first hill and up another circularly shaped hill of height 10 meters and radius 15 meters. The ride is designed so that the cart just barely loses contact with the track at the top of the second hill (10 meters above ground level). What is the height of the starting hill relative to ground

Answers

Answer:

hinit = 17.5 m

Explanation:

Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

       [tex]E_{o} = U_{o} = m*g*h_{init} (1)[/tex]

When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.We can express this final energy as follows:

       [tex]E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)[/tex]

In order to find hinit, we need to make (1) equal to (2), and solve for it.In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).We know also that there is a force that keeps the car along the circular track, which is the centripetal force.This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.So, the centripetal force must be equal to the weight.The centripetal force can be expressed as follows:

       [tex]F_{c} = m*\frac{v_{f} ^{2}}{R} (3)[/tex]

We have just said that (3) must be equal to the weight:

       [tex]F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)[/tex]

Simplifying, and rearranging, we can solve for vf², as follows:

       [tex]v_{f}^{2} = R*g (5)[/tex]  

Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

      [tex]h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)[/tex]

A student sets up a standing wave of wavelength 6 meters in a coiled spring by moving his hand up and down twice each second. What is the velocity of the wave?

Answers

Answer:

Velocity = 12 m/s

Explanation:

Given the following data;

Wavelength = 6 meters

Period = 0.5 seconds. This is due to the fact that the student is moving his hand up and down twice each second.

To find the velocity;

Velocity = wavelength * frequency

But, frequency = 1/period

Frequency = 1/0.5

Frequency = 2

Substituting the values into the velocity formula, we have;

Velocity = 6 * 2

Velocity = 12 m/s

Therefore, the velocity of the wave is 12 meters per seconds.

An elevator motor in a high-rise building can do 3500 kJ of work in 5 min. Find the power developed by the motor. Explain if you can plz

Answers

Answer:

P = 11666.6 W

Explanation:

Given that,

Work done by the motor, W = 3500 kJ

Time, t = 5 min = 300 s

We need to find the power developed by the motor. Power developed is given by :

[tex]P=\dfrac{E}{t}\\\\P=\dfrac{3500\times 10^3}{300}\\\\P=11666.7\ W[/tex]

So, the required power is 11666.6 W.

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