Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring compressed by a distance d = 0.21 m at an angle of θ = 32° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.

Answers

Answer 1

Answer:

Answer:

$v_0=17.3m/s$

Explanation:

In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:

$E_1=E_2=E_3\n\nU_e_1=U_g_2+K_2=U_g_3+K_3\n\n(1)/(2)kd^2=mg(d\sin\theta)+(1)/(2)mv_0^2=mgh+(1)/(2)m(v_0\cos\theta)^2$

Since we only care about the velocity $v_0$ , we can keep only the second and third parts of the equation and solve:

$mgd\sin\theta+(1)/(2)mv_0^2=mgh+(1)/(2)mv_0^2\cos^2\theta\n\n(1)/(2)mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\n\nv_0=\sqrt{(2g(h-d\sin\theta))/(1-\cos^2\theta)}\n\nv_0=\sqrt{(2(9.8m/s^2)(4.4m-(0.21m)\sin32\°))/(1-\cos^232\°)}\n\nv_0=17.3m/s$

So, the speed of the ball just after the launch is 17.3m/s.

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A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the internal energy of the system?

Answers

Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

$\Delta U=-12 J-(-28 J)=+16 J$

What is the fundamental frequency of a particular medium?. a.the lowest frequency at which a standing wave is possible. b. the highest frequency at which s standing wave is possible

Answers

The correct answer is definitely A) the lowest frequency at which a standing wave is possible. I choose this option because standing can occur only at a fundamental frequency and at odd harmonics --> the lowest frequence.

A rock at the top of a 30 meter cliff has a mass of 25 kg. Calculate the rock’s gravitational potential energy when dropped off the cliff. Assuming that energy is conserved and there is no air friction and gravity is 9.8 m/sec., at the bottom of the cliff the rock’s potential energy is completely converted to kinetic energy. Use the formula for kinetic energy to calculate the rock’s speed at the bottom of the cliff. Show all calculations.

Answers

Potential energy =

   (mass) x (gravity) x (height above the reference level) .

Relative to the bottom of the cliff, the potential energy
at the top of the cliff is

   (25kg) x (9.8 m/s²) x (30 meters)

   = (25 x 9.8 x 30) kg-m²/s²

   = 7,350 joules .

Kinetic energy = (1/2) x (mass) x (speed²)

The rock's kinetic energy at the bottom is
the same as its potential energy at the top.

7,350 joules = (1/2) x (25 kg) x (speed²)

Divide each side
by 12.5kg : 7,350 joules/12.5 kg = speed²

   7,350 kg-m²/s² / 12.5kg = speed²

   (7,350 / 12.5) m²/s² = speed²

      588 m²/s² = speed²
Take the square root
of each side:
   Speed = √(588 m²/s²)

   = 24.248... m/s (rounded)

Lets see:-

We have our formula for potential energy, which we are trying to solve for,

PE = mgh or potential energy = mass * gravity * height

So we know that PE all depends on these.

Height : 30 meters
Mass : 25 kg
Gravity (which is always constant) : 9.8 m/s/s

Now add into formula.

PE = 25*30*9.8
PE = 7350 Joules

Answer: PE = 7350 Joules

Density=2g/mL and volume=20mL what is mass

Answers

Answer:

40g

Explanation:

Mass = density x volume

= 2 x 20

= 40g

Answer:

m = V × ρ

= 20 milliliter × 2 gram/cubic meter

= 2.0E-5 cubic meter × 2 gram/cubic meter

= 4.0E-5 gram

= 4.0E-8 kilogram

Explanation:

The density of a material, typically denoted using the Greek symbol ρ, is defined as its mass per unit volume.

ρ =

where:

ρ is the density

m is the mass

V is the volume

The calculation of density is quite straightforward. However, it is important to pay special attention to the units used for density calculations. There are many different ways to express density, and not using or converting into the proper units will result in an incorrect value. It is useful to carefully write out whatever values are being worked with, including units, and perform dimensional analysis to ensure that the final result has units of mass volume. Note that density is also affected by pressure and temperature.

When Emma pushes a bag with a force of 27 newtons, the coefficient of kinetic friction between the bag and the floor is 0.23. What is the normal force acting on the bag by the floor?. A. 0.085 newtons B. 27 newtons C. 2.7 × 10^2 newtons . D. 1.2 × 10^2 newtons

Answers

The normal force acting on the bag by the floor is calculated using th data on the force used to push the bag and the coefficient of friction between the bag and the floor. The normal force is determined through the formula Force / kin.coeff. Substituting, 27 N / 0.23 is equal to 117. 39 N. The answer is D.

Answer:

the answer is D, hope this helps

Explanation:

When using a film camera, what must you do to bring an object into focus?a.move the lens toward or away from the film.
b.close the diaphragm.
c.adjust the speed of the shutter.
d.adjust the angle of the mirror.

Answers

Answer:

a. move the lens toward or away from the film.

Explanation:

Lets discuss the function of different components of the camera

a). Lens : Lens position is set to focus the image on the film as we know that

$(1)/(d_i) + (1)/(d_0) = (1)/(f)$

here we know that position of lens can adjusted by changing the value of distance from object and that will help to focus the image on the film.

b). diaphragm. = It is used to control the intensity of light to enter into the lens so that light is adjusted.

c). speed of the shutter = By changing the speed of shutter we can control the light entering into the lens which can control the brightness of picture

d) Angle of mirror : by adjusting the angle of mirror we can set the orientation of the image that is formed

Answer:

option (A) is correct

Explanation:

To bring the object in focus, you can change the distance between the film and the lens by moving the lens towards or away from the film.

As the lens is used to click the pictures of objects which are placed at different distances from the camera.

By using the lens formula

1 / f = 1 / v - 1 / u

Where, f is the focal length of the lens, u be the distance between lens and the object and v be the distance between image and the lens.

We can say that f is constant and u is constant so, we have to change the value of v that means the distance between the lens and the film to focus the images of the objects on the film.

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