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geniusboy [140]

2 years ago

5

A softball is fouled off with a vertical velocity of 20 m/s and a horizontal velocity of 15 m/s. what is the resultant velocity

of the ball

1 answer:

Kruka [31]2 years ago

6 0

25 m/s is the answer

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A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules

kobusy [5.1K]

Answer:2250J

Explanation:

mass(m)=20kg

velocity(v)=15m/s

Kinetic energy=(m x v^2)/2

Kinetic energy =(20 x 15^2)/2

Kinetic energy =(20x15x15)/2

Kinetic energy=4500/2

Kinetic energy=2250J

3 0

3 years ago

Explain Rutherford's experiment?

Ipatiy [6.2K]

Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.

7 0

3 years ago

Q 24, 25, 26 i dont get them and need answers for it

Alexandra [31]

Answer:

24) W = 75 [J]; 25) W = 1794[J]; 26) n = 8.8 (times) or 9 (times)

Explanation:

24) This problem can be solved by means of the following equation.

$DU = Q-W\\$

where:

DU = internal energy difference [J]

Q = Heat transfer [J]

W = work [J]

Since there are no temperature changes the internal energy change is equal to zero

DU = 0

therefore:

$Q = W\\$

The work is equal to the heat transfered, W = 75 [J].

25) The heat transfer can be calculated by means of the following equation.

$Q = m*c_{p}*DT\\where:\\m = mass = 0.4[kg]\\c_{p} = specific heat = 897[J/kg*K]\\DT= 5 [C]$

Q = 0.4*897*5 = 1794[J]

Work is equal to heat transfer, W = 1794[J]

26) Each time the bag falls the potential energy is transformed into heat energy, which is released into the environment. In this way the potential energy is equal to the developed heat.

$E_{p}=Q\\\\E_{p}=m*g*h$

where:

m = mass = 0.5[kg]

g = gravity = 9.81[m/s^2]

h = 1.5 [m]

$E_{p}=0.5*9.81*1.5\\E_{p}=7.36[J]$

The heat developed can be calculated by means of the following equation.

$Q=m*c_{p}*DT\\Q=0.5*130*1\\Q=65[J]$

The number of times will be calculated as follows

n = 65/7.36

n = 8.8 (times) or 9 (times)

7 0

3 years ago

In which type of bone does ossification occur in the membranes?

Afina-wow [57]

In the formation of flat bones such as the skull the mandibles and the clavicles

3 0

3 years ago

A 3-kg wheel with a radius of 35 cm is spinning in the horizontal plane about a vertical axis through its center at 800 rev/s. A

Kruka [31]

Answer:

$\omega_f = 585.37 \ rev/s$

Explanation:

given,

mass of wheel(M) = 3 Kg

radius(r) = 35 cm

revolution (ω_i)= 800 rev/s

mass (m)= 1.1 Kg

I_{wheel} = Mr²

when mass attached at the edge

I' = Mr² + mr²

using conservation of angular momentum

$I \omega_i = I' \omega_f$

$(Mr^2) \times 800 = ( M r^2 + m r^2) \omega_f$

$M\times 800 = ( M + m )\omega_f$

$3\times 800 = (3+1.1)\times \omega_f$

$2400 = (4.1)\times \omega_f$

$\omega_f = 585.37 \ rev/s$

3 0

3 years ago