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pshichka [43]
1 year ago
8

Two balls are dropped from rest and allowed to fall. If one ball is allowed to fall for 1 s and the other for 3 s compare the di

stances they have fallen.
Physics
1 answer:
kirill115 [55]1 year ago
3 0

The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

  • time of fall of the first ball, t = 1 s
  • time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m

The distance traveled by the second ball is calculated as follows;

h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m

Thus, the second ball traveled a greater distance because it spent more time in motion.

Learn more here:brainly.com/question/5868480

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Greater than

Explanation:

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When the cloud shrinks under the right conditions, a star may be formed.

Thus, Diameter of clouds are much higher than a star.

Moment of inertia of cloud is greater than the star's inertial.

so, angular velocity of the star would be greater than angular velocity of the rotating gas.

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When does a star reach equilibrium?
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1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.
Kisachek [45]

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] =  11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]

7 0
3 years ago
What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat
Amiraneli [1.4K]

Answer:633.8 KJ

Explanation:

Given

mass of water\left ( m\right )=250gm

Initial temperature\left ( T_i\right )=20^{\circ}C

Final temperature \left ( T_f\right )=100^{\circ}C

Specific heat of water \left ( c \right )=4190 J/kg-k

heat of vaporization\left ( L\right )=22.6\times 10^5 J/kg

Heat required for process\left ( Q\right )=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc\left ( T_f-T_i\right )+mL

Q=0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5

Q=\left ( 0.838+5.5\right )\times 10^5

Q=6.338\times 10^5J=633.8 KJ

4 0
3 years ago
A 1.2-kg mass is projected from ground level with a velocity of 30 m/s at some unknown angle above the horizontal. A short time
Dmitriy789 [7]

Answer:

K_f = 351.84 J

Explanation:

Using the conservation of energy K:

E_i = E_f

so:

\frac{1}{2}mv^2 = K_f + mgh

where m is the mass, v the initial velocity, K_f is the kinetic energy of the mass as it clears the fence, g the gravity and h the altitude.

Then, replacing values, we get:

\frac{1}{2}(1.2kg)(30m/s)^2 = K_f + (1.2kg)(9.8m/s^2)(16m)

solving for K_f:

K_f = 351.84 J

3 0
3 years ago
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