A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.

Answers

Answer 1

Answer:

Answer:

h = 16.9 m

Explanation:

When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:

Kinetic Energy Lost by Ball = Potential Energy Gained by Ball

(0.5)m(Vf² - Vi²) = mgh

h = (0.5)(Vf² - Vi²)/g

where,

Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)

Vi = Initial Speed of Ball = 18.2 m/s

g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)

h = maximum height the ball can reach = ?

Therefore, using values in the equation, we get:

h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)

h = 16.9 m

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While the block hovers in place, is the density of the block (top left) or the density of the liquid (bottom center) greater?

Answers

Answer:

for the body to float, the density of the body must be less than or equal to the density of the liquid.

Explanation:

For a block to float in a liquid, the thrust of the liquid must be greater than or equal to the weight of the block.

Weight is

W = mg

let's use the concept of density

ρ_body = m / V

m = ρ_body V

W = ρ_body V g

The thrust of the body is given by Archimedes' law

B = ρ_liquid g V_liquid

as the body floats the submerged volume of the liquid is less than or equal to the volume of the block

ρ_body V g = ρ_liquid g V_liquid

ρ_body = ρ liquid Vliquido / V_body

As we can see, for the body to float, the density of the body must be less than or equal to the density of the liquid.

The data listed below are for mechanical waves. Which wave has the greatest energy?

Answers

theres no data listed below

Answer:

amplitude = 14 cm; wavelength = 7 cm; period = 12 seconds

Explanation:

Sometimes, in an intense battle, gunfire is so intense that bullets from opposite sides collide in midair. Suppose that one (with mass M = 5.12 g moving to the right at a speed V = [08]____________________ m/s directed 21.3° above the horizontal) collides and fuses with another with mass m = 3.05 g moving to the left at a speed v = 282 m/s directed 15.4° above the horizontal. a. What is the magnitude of their common velocity (m/s) immediately after the collision? b. What is the direction of their common velocity immediately after the collision? (Measure this angle in degrees from the horizontal.) c. What fraction of the original kinetic energy was lost in the collision?

Answers

The magnitude of the speed is 83.0325 m\s, the direction is 62.7 degrees, and the fraction of kinetic energy lost is 0.895.

What is collision?

The collision is the phenomenon when two objects come in direct contact with each other. Then both the bodies exert forces on each other.

The mass, angle, and velocity of the first object are 5.12 g, 21.3°, and 239 m/s.

And the mass, angle, and velocity of the second object be 3.05 g, 15.4°, and 282 m/s.

The momentum (P₁) before a collision will be

$\rm P_1 = (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) \hat{x} + (m_1 u_1 sin \theta _1+ m_2 u_2 sin \theta _2) \hat{y}$

The momentum (P₂) after a collision will be

$\rm P_2 = (m_1 + m_2) u \ cos\ \theta \ \hat{x} \ + (m_1 + m_2) u \ sin \ \theta \ \hat{y}$

Applying momentum conservation, we have

$\rm (m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) = (m_1 + m_2) u \ cos\ \theta \ \n\n$ ...1

$\rm (m_1 u_1 sin \theta _1+m_2 u_2 sin \theta _2) \ =(m_1 + m_2) u \ sin \ \theta$ ...2

From equations 1 and 2, we have

$\rm \theta = tan \ ^(-1) ( (m_1 u_1 cos \theta _1 +m_2 u_2cos \theta _2))/( (m_1 u_1 sin \theta _1 - m_2 u_2 sin \theta _2))\n\n\n\theta = tan \ ^(-1) (5.12*239*cos21.3+3.05*282*cos15.4)/(5.12*239*sin21.3-3.05*282*sin15.4)\n\n\n\theta = 62.7^o$

From equation 1, we have

$\rm u = ((m_1 u_1 cos \theta _1 - m_2 u_2cos \theta _2) )/( (m_1 + m_2) \ cos\ \theta ) \n\n\nu = (5.12*239*cos21.3 - 3.05*282*cos15.4)/((5.12+3.05)cos62.2)\n\n\nu = 83.0325 m/s$

Then the change in kinetic energy, we have

$\rm \Delta KE = (1)/(2)m_1u_1^2+(1)/(2)m_2u_2^2-(1)/(2)(m_1+m_2)u^2\n\n\n\Delta KE = (1)/(2) * 5.12 * 239^2 + (1)/(2)*3.05*282^2 - (1)/(2)(5.12+3.05)*83.032^2\n\n\n\Delta KE = 239.34 \ J$

The fraction of kinetic energy lost will be

$\rm Energy \ lost = (239.34)/(267.5) = 0.895$

More about the collision link is given below.

brainly.com/question/13876829

Answer:

Detailed solution is given below

Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? Light from a lamp is shining on a surface. How can you increase the intensity of the light on the surface? A. Use a lens to focus the power into a smaller area. B. Increase the power output of the lamp. C. Either A or B.

Answers

Answer:

the correct option is C

Explanation:

The intensity of a lamp depends on the power of the lamp that is provided by the current flowing over it, therefore the intensity would increase if we raise the current.

Another way to increase the intensity is to decrease the area with a focusing lens, as the intensity is power over area, decreasing the area increases the power.

When we see the possibilities we see that the correct option is C

The electrons in the beam of a television tube have a kinetic energy of 2.20 10-15 j. initially, the electrons move horizontally from west to east. the vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of 3.00 10-5 t. (a) in what direction are the electrons deflected by this field component? due north due south due east due west (b) what is the magnitude of the acceleration of an electron in part (a)? m/s2

Answers

(a) The electrons move horizontally from west to east, while the magnetic field is directed downward, toward the surface. We can determine the direction of the force on the electron by using the right-hand rule:
- index finger: velocity --> due east
- middle finger: magnetic field --> downward
- thumb: force --> due north
However, we have to take into account that the electron has negative charge, therefore we have to take the opposite direction: so, the magnetic force is directed southwards, and the electrons are deflected due south.

b) From the kinetic energy of the electrons, we can find their velocity by using
$K= (1)/(2)mv^2$
where K is the kinetic energy, m the electron mass and v their velocity. Re-arranging the formula, we find
$v= \sqrt{ (2K)/(m) }= \sqrt{ (2 \cdot 2.20 \cdot 10^(-15) J)/(9.1 \cdot 10^(-31) kg) }=6.95 \cdot 10^7 m/s$

The Lorentz force due to the magnetic field provides the centripetal force that deflects the electrons:
$qvB = m (v^2)/(r)$
where
q is the electron charge
v is the speed
B is the magnetic field strength
m is the electron mass
r is the radius of the trajectory
By re-arranging the equation, we find the radius r:
$r= (mv)/(qB)= ((9.1 \cdot 10^(-31) kg)(6.95 \cdot 10^7 m/s))/((1.6 \cdot 10^(-19) C)(3.00 \cdot 10^(-5) T))=13.18 m$

And finally we can calculate the centripetal acceleration, given by:
$a_c = (v^2)/(r)= ((6.95 \cdot 10^7 m/s)^2)/(13.18 m)=3.66 \cdot 10^(14) m/s^2$

A skater has rotational inertia 4.2 kg-m2 with his fists held to his chest and 5.7 kg?m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they�re essentially on his rotation axis, how fast will he be spinning? Express your answer using two significant figures. ?f=

Answers

Answer: 38.5rad/s

Explanation: The calculations can be viewed on the image attached below. Thanks

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