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4 years ago

13

From the edge of a roof you throw a snowball downward that strikes the ground with 100J of kinetic energy. then you throw a seco

nd snowball upward with the same initial speed, and this too falls to the ground. neglecting air resistance, the second snowball hits the ground with a kinetic energy of:
a. 100J
b. 200J
c. more than 200J
d. less than 200J
e. none of the above

1 answer:

Vanyuwa [196]4 years ago

7 0

Answer:

The second snowball hits the ground with a kinetic energy of 100 Joules

Explanation:

Given that,

From the edge of a roof you throw a snowball downward that strikes the ground with 100 J of kinetic energy. It is a case of conservation of energy.

At the highest point,

$mgh+\dfrac{1}{2}mu^2=mgh'+0$

$100=mgh'$

At lowest point,

$mgh'=K$

From above two equation, we get :

Kinetic energy, K = 100 J

So, the second snowball hits the ground with a kinetic energy of 100 Joules. So, the correct option is (A).

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If a puddle of water is frozen, do the particles in the ice have kinetic energy? Explain.

TEA [102]

They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though

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4 years ago

People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th

fiasKO [112]

Answer:

636.619772368 A

Explanation:

$\tau$ = Torque = $1\times 10^{-3}\ N/m$

B = Magnetic field of Earth = $5\times 10^{-5}\ T$

A = Area

d = Diameter = 20 cm

Current is given by

$I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A$

The current is 636.619772368 A

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3 years ago

A planet exerts a gravitational force of magnitude 9e22 N on a star. If the planet were 2 times closer to the star (that is, if

Dmitrij [34]

To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.

Planet gravitational force

$F_p = 6*10^{22}N$

$F_p = \frac{GMm}{R^2}$

$F_p = 9*10^{22}N$

Distance between planet and star

$r = \frac{R}{2}$

Gravitational force is

$F = \frac{GMm}{r^2}$

Applying the new distance,

$F = \frac{GMm}{(\frac{R}{2})^2}$

$F = 4\frac{GMm}{R^2}$

Replacing with the previous force,

$F = 4F_p$

Replacing our values

$F= 4(9*10^{22}N)$

$F = 36*10^{22}N$

Therefore the magnitude of the force on the star due to the planet is $36*10^{22}N$

5 0

3 years ago

A person weighing 645 N climbs up a ladder to a height of 4.55 m what is the increase in gravitational energy

Lemur [1.5K]

Answer: 2934.75 Joules

Explanation:

Potential energy can be defined as energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

<em>P.E = mgh</em>

Where P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per second square.

h represents the height measured in meters.

Given the following data;

Weight =645

Height = 4.55

<em>P.E = mgh</em>

But we know that weight = mg = 645N

Substituting into the equation, we have;

<em>P.E = 645 • 4.55</em>

<em>P.E = 2934.75J</em>

Potential energy, P.E = 2934.75 Joules.

3 0

3 years ago

a ball rolls horizontally of the edge of the cliff at 4 m/s, if the ball lands at a distance of 30 m from the base of the vertic

algol13

Answer:

Approximately $281.25\; \rm m$ . (Assuming that the drag on this ball is negligible, and that $g = 10\; \rm m \cdot s^{-2}$ .)

Explanation:

Assume that the drag (air friction) on this ball is negligible. Motion of this ball during the descent:

Horizontal: no acceleration, velocity is constant (at $v(\text{horizontal})$ is constant throughout the descent.)
Vertical: constant downward acceleration at $g = 10\; \rm m \cdot s^{-2}$ , starting at $0\; \rm m \cdot s^{-1}$ .

The horizontal velocity of this ball is constant during the descent. The horizontal distance that the ball has travelled during the descent is also given: $x(\text{horizontal}) = 30\; \rm m$ . Combine these two quantities to find the duration of this descent:

$\begin{aligned}t &= \frac{x(\text{horizontal})}{v(\text{horizontal})} \\ &= \frac{30\; \rm m}{4\; \rm m \cdot s^{-1}} = 7.5\; \rm s\end{aligned}$ .

In other words, the ball in this question start at a vertical velocity of $u = 0\; \rm m \cdot s^{-1}$ , accelerated downwards at $g = 10\; \rm m \cdot s^{-2}$ , and reached the ground after $t = 7.5\; \rm s$ .

Apply the SUVAT equation $\displaystyle x(\text{vertical}) = -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t$ to find the vertical displacement of this ball.

$\begin{aligned}& x(\text{vertical}) \\[0.5em] &= -\frac{1}{2}\, g \cdot t^{2} + v_0\cdot t\\[0.5em] &= - \frac{1}{2} \times 10\; \rm m \cdot s^{-2} \times (7.5\; \rm s)^{2} \\ & \quad \quad + 0\; \rm m \cdot s^{-1} \times 7.5\; s \\[0.5em] &= -281.25\; \rm m\end{aligned}$ .

In other words, the ball is $281.25\; \rm m$ below where it was before the descent (hence the negative sign in front of the number.) The height of this cliff would be $281.25\; \rm m\!$ .

5 0

3 years ago