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Ilya [14]
3 years ago
8

A bug flies at a velocity of 0.75 m/s into an oncoming breeze blowing at 0.25 m/s. What is the resultant velocity of the bug?

Physics
1 answer:
MakcuM [25]3 years ago
4 0
0.5m/s.  Take the velocity of the bug and subtract from the velocity of the oncoming breeze because they are opposite.  If the breeze was in the same direction that the bug was flying, then your answer would be the sum of the two.
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A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)  

v₀ is the initial voltage on the cap  

v is the voltage after time t  

R is resistance in ohms,  

C is capacitance in farads  

t is time in seconds  

RC = τ = time constant  

τ = 20µ x 1.5k = 30 ms  

v = v₀e^(t/τ)  

0.5 = 1.5e^(t/30ms)  

e^(t/30ms) = 10/3  

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0
3 years ago
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

4 0
2 years ago
The steering wheel of a car has a radius of 0.19 m, and the steering wheel of a truck has a radius of 0.25 m. The same force is
kodGreya [7K]

Answer:

\frac{T_t}{T_c} = 1.32

Explanation:

The torque applied on an object can be calculated by the following formula:

T = Fr

where,

T = Torque

F = Applied Force

r = radius of the wheel

For car wheel:

T_c = Fr_c\\

For truck wheel:

T_t = Fr_t

Dividing both:

\frac{T_t}{T_c} = \frac{Fr_t}{Fr_c}

for the same force applied on both wheels:

\frac{T_t}{T_c} = \frac{r_t}{r_c} \\

where,

rt = radius of the truck steering wheel = 0.25 m

rc = radius of the car steering wheel = 0.19 m

Therefore,

\frac{T_t}{T_c} = \frac{0.25\ m}{0.19\ m} \\

\frac{T_t}{T_c} = 1.32

8 0
3 years ago
Make the curvature radius 0.6 m, the refractive index 1.5, and the diameter 0.6 m. place the lamp so that the source of light is
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We are given
r = 0.6 m
n = 1.5
D = 0.6 m, R1 = 30 cm
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We use the formula
1 / f = ( n - 1) (1/R1 - 1/R2)
Substituting and solving for f
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f = 80 cm

The focal length is 80 cm and the distance of the focal plane from the lesn is 80 cm - 30 cm = 50 cm.<span />
5 0
2 years ago
The valuation of business depends on​
Eddi Din [679]

Answer:

Growth Prospects. This factor looks at how much potential the business has to grow in the future.

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Location. As with real estate, business is all about location, location, location.

Concentration.

Staff and Management.

Reputation.

Explanation:

5 0
2 years ago
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