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alina1380 [7]
4 years ago
7

What is Hooke's law? Also, what does it say about how elastic materials respond to force? Explain.

Physics
1 answer:
Eva8 [605]4 years ago
3 0
It <span>states that the force F needed to extend or compress a spring by some distance X is proportional to that distance.

For elastic materials, they extend more in same amount of force, (as they are directly proportional), due to it's elastic nature (presence of large deforming force)

Hope this helps!</span>
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9) An electrical appliance has a resistance of 25 N. When this electrical ap-
Neko [114]

(C)

Explanation:

From Ohm's law,

V = IR

Solving for I,

I = V/R

= (230 V)/(25 ohms)

= 9.2 A

6 0
3 years ago
If the Universe started as a singularity, it would mean that _____ started out _____. all mass and energy, spread out across the
timurjin [86]

Answer: All mass and energy; in a point.

Explanation: All mass and energy is expressed to have emerged from a point, if the universe is established in solitary.

8 0
4 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
2 years ago
Read 2 more answers
Please help on questions D AND E and please show working out as well so I can understand better thank you only QUESTIONS D AND E
tester [92]

Answer:

d. 87,500 J

e. 49,600 J

Explanation:

The total energy is the heat absorbed by the copper plus the heat absorbed by the water.

d)

E = m₁C₁ΔT + m₂C₂ΔT

E = (1 kg) (390 J/kg/°C) (10 °C) + (2 kg) (4180 J/kg/°C) (10 °C)

E = 87,500 J

e)

E = m₁C₁ΔT + m₂C₂ΔT

E = (2 kg) (390 J/kg/°C) (10 °C) + (1 kg) (4180 J/kg/°C) (10 °C)

E = 49,600 J

5 0
3 years ago
In a cathode ray tube, the number of electrons that reach the fluorescent screen is controlled by the:
EleoNora [17]

Answer:

Grid

Explanation:

The number of electrons attaining the fluorescent light is regulated by the grid in a cathode ray tube.When the heater is supplied with current, it allows the electrons to emit from the cathode. Before approaching the anode the electrons pass through grid opening. The number of electrons attempting to reach the anode may also be regulated by managing the number of electrons passing through the grid.

6 0
3 years ago
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