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madam [21]
3 years ago
8

What is the effect on current if both the voltage and the resistance are halved?

Physics
1 answer:
bulgar [2K]3 years ago
3 0

The current through a resistor is

      <u> (voltage across the resistor) / (resistance of the resistor)</u> .

If the numerator and the denominator of that fraction are both
multiplied or divided by the same number, then the value of the
fraction <em>doesn't change</em>.


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A train moves from rest to a speed at 25m/s in 30.0 seconds. What is it’s acceleration?
sasho [114]

Answer:

25/30 = 5/6 m/s^2 5/6 meters per second squared

4 0
3 years ago
Which example has the least kinetic energy
Stells [14]

Answer:

D

Explanation:

Let’s calculate the kinetic energy for all of the choices.

a. (1/2)(100)(100)^2 = 50(10000)=500,000

b. (1/2)(100)(1)^2 = 50

c. (1/2)(10)(100)^2 = 5(10000) = 50,000

d. (1/2)(1)(1)^2 = 0.5

We can see that (d) has the least kinetic energy.

5 0
2 years ago
Magnesium chloride forms crystals. Which option is a description of this
shtirl [24]
The answer is B. This form of magnesium chloride is not a liquid but a solid that is white and colorless.
4 0
3 years ago
Read 2 more answers
The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o
LUCKY_DIMON [66]

Answer:

The average impact force is 12000 newtons.

Explanation:

By Impact Theorem we know that impact done by the sledge hammer on the chisel is equal to the change in the linear momentum of the former. The mathematical model that represents the situation is now described:

\bar F \cdot \Delta t = m \cdot  (v_{2}-v_{1}) (1)

Where:

\bar F - Average impact force, in newtons.

\Delta t - Duration of the impact, in seconds.

m - Mass of the sledge hammer, in kilograms.

v_{1}, v_{2} - Initial and final velocity, in meters per second.

If we know that \Delta t = 0.0020\,s, m = 4\,kg, v_{1} = -6\,\frac{m}{s} and v_{2} = 0\,\frac{m}{s}, then we estimate the average impact force is:

\bar F = \frac{m\cdot  (v_{2}-v_{1})}{\Delta t}

\bar F = 12000\,N

The average impact force is 12000 newtons.

5 0
3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
2 years ago
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