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Eduardwww
12 days ago
11

Which factors could be potential sources of error in the experiment? check all that apply.

Physics
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The cluster of decisions that managers make to assist the organization to achieve its goals is known as:
Softa [3030]
The answer is option C. SWOT analysis. The compilation of decisions made by managers to aid the organization in reaching its objectives is referred to as SWOT Analysis.
Here are the options. A. Strategy B. Scenario planning C. SWOT analysis D. Diversification E. Related diversification
8 0
3 months ago
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is sudde
Maru [3345]

Explanation:

Data provided:

Area A = 10 cm×2 cm = 20×10⁻⁴ m²

Separation distance d between the plates = 1 mm = 1×10⁻³ m

Battery voltage, or emf = 100 V

Resistance = 1025 ohm

Solution:

In an RC circuit, the voltage across the plates varies with time t. At the outset, the voltage matches that of the battery, V₀ = emf = 100V. However, after a certain time t, both the resistance and capacitance alter this, leading to a final voltage V expressed as

V = V_{0}(1-e^{\frac{-t}{RC} } )\\\frac{V}{V_{0} } = 1-e(^{\frac{-t}{RC} }) \\e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} }

Applying the natural logarithm to both sides,

e^{\frac{-t}{RC} } = 1- \frac{V}{V_{0} } \\\frac{-t}{RC} = ln(1-\frac{V}{V_{0} } )\\t = -RCln(1 - \frac{V}{V_{0} })

t = -RC ln (1-\frac{V}{V_{0} })        (1)

Next, we can determine the capacitance using the plates' area.

C = ε₀A/d

  = \frac{(8.85*10^{-12))} (20*10^{-4}) }{1*10^{-3} }

  = 18×10⁻¹²F

We can now find the time it takes for the voltage to drop from 100 to 55 V by substituting C, V₀, V, and R values into equation (1)

t = -RC ln (1-\frac{V}{V_{0} })

 = -(1025Ω)(18×10⁻¹² F) ln( 1 - 55/100)

 = 15×10⁻⁹s

= 15 ns

5 0
1 month ago
Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t
Softa [3030]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; \sum fy = ma_y

For the body with mass m;

T - mg = ma... (1)

For the body with mass M;

Mg - T = Ma... (2)

Combining equations 1 and 2 gives;

+Mg -mg = ma + Ma

Ma-Mg = -mg-ma

M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

3/4 g-g = -m(7/4 g)

-g/4 = -m(7/4 g)

1/4 = 7m/4

Multiplying gives: 28m = 4

m = 1/7 kg

Hence, the mass of the other box is 1/7 kg

3 0
3 months ago
A box of mass 3.1kg slides down a rough vertical wall. The gravitational force on the box is 30N . When the box reaches a speed
Ostrovityanka [3204]

Answer:

Explanation:

a) La fuerza neta que actúa sobre la caja en la dirección vertical es:

Fnet=Fg−f−Fp *sin45 °

aquí Fg representa la fuerza gravitacional, f es la fuerza de fricción, y Fp es la fuerza de empuje.

Fnet=ma

ma=Fg−f−Fp *sin45 °

​a=\frac{30-13-23*sin(45)}{3.1}

=0.24 m/s²

Vf =Vi +at

=0.48+0.24*2

Vf=2.98 m/s

b)

Fnet=Fg−f−Fp *sin45 °

=Fg−0.516Fp−Fp *sin45 °

=30-1.273Fp

Fnet=0 (Ya que la velocidad es constante)

Fp=30/1.273

=23.56 N

5 0
3 months ago
Read 2 more answers
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