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anygoal [31]
3 years ago
10

. Energy can neither be created nor be destroyed, but it can be changed from one form to another", this law is known as kinetic

energy potential energy conservation of energy conservation principle
Physics
1 answer:
Fynjy0 [20]3 years ago
3 0

Answer:

The law of conservation of energy

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A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
2 years ago
Which of the following is an example of work being done on an object?
tensa zangetsu [6.8K]

Answer:

d. All of these

Explanation:

work is said to be done when a force is applied to an object through a certain distance. the SI unit of workdone is joules or newton per meter

mathematically

workdone = force x distance.

from the answers,  work is being done because there is force applied in a certain distance.

  • from wagon is used to carry vegetables from a garden.
  • pulley is used to get water from a well.
  • hammer is used to remove a nail from a wall.

4 0
3 years ago
Read 2 more answers
What is the difference between a cannabinoid and a trichome?
Sedbober [7]
<span>Trichome density and type and cannabinoid content of leaves and bracts were quantitated during organ ontogeny for three clones of Cannabis sativa L. Trichome initiation and development were found to occur throughout leaf and bract ontogeny. On leaves, bulbous glands were more abundant than capitate-sessile glands for all clones, although differences in density for each gland type were evident between clones. On pistillate bracts, capitate-sessile glands were more abundant than the bulbous form on all clones, and both types decreased in relative density during bract ontogeny for each clone. The capitate-stalked gland, present on bracts but absent from vegetative leaves, increased in density during bract ontogeny. The capitate-stalked gland appeared to be initiated later than bulbous or capitate-sessile glands during bract development and on one clone it was first found midway in bract ontogeny. Nonglandular trichomes decreased in density during organ ontogeny, but the densities differed between leaves and bracts and also between clones. Specific regulatory mechanisms appear to exist to control the development of each trichome type independently.</span>
3 0
3 years ago
How many infrared photons of frequency 2.57 x 1013 Hz would need to be absorbed simultaneously by a tightly bound molecule to br
victus00 [196]

Answer:

94

Explanation:

f = 2.57 x 10^13 Hz

E = 10 eV = 10 x 1.6 x 10^-19 J = 1.6 x 10^-18 J

Energy of each photon = h f

Where, h is Plank's constant

Energy of each photon = 6.63 x 10^-34 x 2.57 x 10^13 = 1.7 x 10^-20 J

Number of photons = Total energy / energy of one photon

N = (1.6 x 10^-18) / (1.7 x 10^-20) = 94.11 = 94

6 0
2 years ago
In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac
Elan Coil [88]

Answer:

a=0.284\ m/s^2

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2

So, the acceleration experienced by the occupants of the spaceship is 0.284\ m/s^2.

5 0
3 years ago
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