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3 years ago

9

In the qualifying round of the 50-yd freestyle in the sectional swimming championship, Dugan got an early lead by finishing the

first 25.00 yd in 10.01 seconds. Dugan finished the return leg (25.00 yd distance) in 10.22 seconds. Determine Dugan's average velocity for the entire race. Number only, round to the nearest hundredth.

1 answer:

Nata [24]3 years ago

6 0

Answer:

v = 0

Explanation:

Given that,

Total distance is 50 yards

Dugan got an early lead by finishing the first 25.00 yd in 10.01 seconds

Dugan finished the return leg (25.00 yd distance) in 10.22 seconds.

We need to find Dugan's average velocity for the entire race. As he returns at the initial position. As a result, the net displacement is equal to 0. So,

Average velocity = net displacement/time

v = 0

Hence, his average velocity for the entire race is 0.

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3. The soccer kicker kicks a 2 kg football with a force of 68 N. How fast will the ball accelerate down the field?

Katen [24]

Answer:

$\boxed {\boxed {\sf m= 2 \ kg , \\a= 34 \ m/s^2, \\\F= 68 \ N}}$

Explanation:

The formula for force is:

$F=m*a$

If we rearrange the formula to solve for a (acceleration), the formula becomes

$\frac{F}{m} =a$

The force is 68 Newtons. Let's convert the units to make the problem easier later on. 1 N is equal to 1 kg*m/s², so the force of 68 N is equal to 68 kg*m/s².

The mass is 2 kilograms.

$F=68 \ kg*m/s^2 \\m= 2\ kg$

Substitute the values into the formula.

$\frac{68 \ kg*m/s^2}{2 \ kg} =a$

Divide. Note that the kilograms will cancel each other out (hence why we changed the units).

$\frac{68 \ m/s^2}{2}=a$

$34 \ m/s^2=a$

The acceleration is<u> </u><u>34 meters per second squared.</u>

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3 years ago

It is proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. W

Anvisha [2.4K]

Answer:

962291.57928 m²

Explanation:

$P_r$ = Pressure = $2\dfrac{I}{c}$ (full reflection)

I = Intensity = $\dfrac{P}{A}=\dfrac{P}{4\pi r^2}$

P = Power = $3.9\times 10^{26}\ W$

c = Speed of light = $3\times 10^8\ m/s$

M = Mass of Sun = $1.99\times 10^{30}\ kg$

m = Mass of ship = 1500 kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Force of radiation is given by

$F_r=P_rA\\\Rightarrow F_r=2\dfrac{I}{c}\times A\\\Rightarrow F_r=2\dfrac{P}{4\pi r^2c} A$

This force will balance the gravitational force as stated in the question

$\dfrac{GMm}{r^2}=2\dfrac{P}{4\pi r^2c} A\\\Rightarrow A=\dfrac{4\pi cGMm}{2P}\\\Rightarrow A=\dfrac{4\times \pi\times 3\times 10^8\times 6.67\times 10^{-11}\times 1.99\times 10^{30}\times 1500}{2\times 3.9\times 10^{26}}\\\Rightarrow A=962291.57928\ m^2$

The area of the must be 962291.57928 m²

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3 years ago

Can you respond this two questions, please? :

Andrews [41]

Where's the diagram for question 1?

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3 years ago

You decide to roll a 0.11-kgkg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelen

Oksanka [162]

The de Broglie wavelength $\lambda = 4.0\times 10^{-30}$ m

We know that

de Broglie wavelength = $\lambda = \frac{h}{mv}\lambda = \frac{6.63\times 10^{-34}}{0.11\times 1.5 \times 10 ^{-3}}$

$\lambda = 4.0\times 10^{-30}$ m

<h3>What is de Broglie wavelength?</h3>

According to the de Broglie equation, matter can behave like waves, much like how light and radiation do, which are both waves and particles. A beam of electrons can be diffracted just like a beam of light, according to the equation. The de Broglie equation essentially clarifies the notion of matter having a wavelength.

Therefore, whether a particle is tiny or macroscopic, it will have a wavelength when examined.

The wave nature of matter can be seen or observed in the case of macroscopic objects.

To learn more about de Broglie wavelength with the given link

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1 year ago

A rope passes over a fixed sheave with both ends hanging straight down. The coefficient of friction between the rope and sheave

Oliga [24]

Answer:3.51

Explanation:

Given

Coefficient of Friction $\mu =0.4$

Consider a small element at an angle \theta having an angle of $d\theta$

Normal Force $=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}$

$N=T\cdot d\theta$

Friction $f=\mu \times Normal\ Reaction$

$f=\mu \cdot N$

and $T+dT-T=f=\mu Td\theta$

$dT=\mu Td\theta$

$\frac{dT}{T}=\mu d\theta$

$\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta$

$\frac{T_2}{T_1}=e^{\mu \pi}$

$\frac{T_2}{T_1}=e^{0.4\times \pi }$

$\frac{T_2}{T_1}==e^{1.256}$

$\frac{T_2}{T_1}=3.51$

7 0

3 years ago