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Furkat
3 months ago
10

Write a hypothesis about the effect of increasing voltage on the current in the circuit. Use the "if . . . then . . . because .

. ." format and be sure to answer the lesson question: "How do changes in voltage or resistance affect current in an electric circuit?"
Physics
2 answers:
kicyunya [3.2K]3 months ago
4 0

Sample Response: If the voltage in a circuit is raised while resistance stays constant, then the current will inevitably rise since Ohm's law shows that voltage and current are directly proportional.

Yuliya22 [3.3K]3 months ago
3 0

Hypothesis: An increase in voltage should result in a corresponding rise in current because according to Ohm's Law,

I \propto V

I=\frac{V}{R}

Ohm's Law indicates that current is proportional to voltage when resistance remains constant. Hence, if resistance stays the same, elevating the voltage will lead to an increase in current. Conversely, if voltage remains unchanged and resistance increases, current will decrease.

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A parachutist, after opening her parachute, finds herself gently floating downward, no longer gaining speed. She feels the upwar
Sav [3153]

As the parachutist is descending at a steady rate

we can conclude that

a = \frac{dv}{dt}

Acceleration indicates the change in velocity

given the constant velocity in this scenario

a = 0

Thus, in this situation, we find the acceleration to be zero

It’s understood from Newton's second law

F_{net} = ma

where a is equal to 0

F_{net} = 0

F_{net} = F_g - F_b

Here, the force due to gravity

equals the force due to buoyancy

F_gHence, we can deduce

F_b

therefore

F_g - F_b = 0

as such the upward force is counteracted by the downward force.

5 0
2 months ago
A rocket takes off vertically from the launch pad with no initial velocity but a constant upward acceleration of 2.25 m/s^2. At
kicyunya [3294]

Answer:

A) 328 m

B) 80.22 m/s

C) 8.18 sec

Explanation:

A)

The rocket's initial acceleration is 2.25 m/s²

Time until engine failure is 15.4 s

Initial velocity u during takeoff = 0 m/s

Distance covered while the engine is functional =?

We apply Newton's laws for this calculation

S = ut + \frac{1}{2}at^{2}

Here, S represents the distance traveled under the rocket's thrust

S = (0 x 15.4) + \frac{1}{2}(2.25 x 15.4^{2})

S = 0 + 266.81 m = 266.81 m

Before the engine fails, the final velocity can be found using:

v = u + at

v = 0 + (2.25 x 15.4) = 34.65 m/s

Once the engine fails, the rocket decelerates under gravitational pull at g = -9.81 m/s²  (acting downwards)

The upward initial velocity when freefall begins is v = 34.65 m/s

The final velocity is reached at peak height, where the rocket halts, therefore:

u = 0 m/s

The distance covered during this freefall will be s =?

Utilizing the equation

v^{2} = u^{2} + 2gs

0^{2} = 34.65^{2} + 2(-9.81 x s)

0 = 1200.6 - 19.62s

-1200.6 = -19.62s

s = -1200.6/-19.62 = 61.19 m

Thus,

Maximum Height = 266.81 m  + 61.19 m =  328 m

B)

At maximum height, the rocket’s initial upward velocity drops to 0 m/s (the rocket completely stops)

The descent occurs freely under g = 9.81 m/s² (acting downwards)

The distance covered during the fall will be 328 m

Final velocity v just prior to impact =?

Applying v^{2} = u^{2} + 2gs

v^{2} = 0^{2} + 2(9.81 x 328)

v^{2} = 0 + 6435.36

v = \sqrt{6435.36} = 80.22 m/s

The time taken before reaching the pad is found as follows

v = u + gt

80.22 = 0 + 9.81t

t = 80.22/9.81 = 8.18 sec

7 0
3 months ago
A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What
kicyunya [3294]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

6 0
3 months ago
Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi
ValentinkaMS [3465]
1) The buoyant force acting on an object submerged in a fluid can be described as:
B=d_f V_d g
where d_f indicates the fluid's density, V_d represents the volume of the fluid displaced, and g=9.81~m/s^2 signifies the gravitational acceleration.

2) To determine the volume of the displaced fluid, we note that the titanium object is entirely submerged in the fluid (air), thus this volume matches the volume of 1 Kg of titanium, which has a density of d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3. Using the correlation between density, volume, and mass, we derive
V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3

3) We can now revisit the equation in step 1) to compute the buoyant force. Given that the air density is d_f = 1~Kg/m^3, this provides us with
B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N

4) The weight of 1 Kg of titanium is:
W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N
Therefore, the buoyant force is negligible when compared to the weight.
7 0
3 months ago
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