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katrin [286]
3 years ago
9

A fish swims 12.0 m in 5.0 s. It swims the first 4.0 m in 2.0 s, the next 3.0 m in 1.2 s, and the last 5.0 m in 1.8 s. What is t

he average velocity of the fish during the time interval between 0.0 s and 3.2 s?
Physics
1 answer:
Ivahew [28]3 years ago
3 0

Answer:

B.) 2.2 m/s

Explanation:

It was on quizlet and I got it right on the test.

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A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed
Shkiper50 [21]

Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

3 0
3 years ago
An iron wire has a length of 1.50 m and a cross sectional area of 0.290 mm2. If the resistivity of iron is 10.0 ✕ 10−8 Ω · m and
lapo4ka [179]

Answer:

1.35 A

Explanation:

Applying,

V = IR

I = V/R..................... Equation 1

I = Current, V = Voltage, R = Resistance.

But,

R = Lρ/A............... Equation 2

Where L = Length of the wire, ρ = resistivity, A = Cross-sectional area of the wire.

Sustitute equation 2 into equation 1

V = AV/Lρ............... Equation 3

From the question,

Given: V = 0.7 V, A = 0.290 mm² = 2.9×10⁻⁷ m², L = 1.5 m, ρ = 10×10⁻⁸ Ω.m

Substitute these values into equation 3

I = (0.7× 2.9×10⁻⁷)/(1.5× 10×10⁻⁸ )

I = (2.03×10⁻⁷)/(15×10⁻⁸)

I = 1.35 A

5 0
2 years ago
No two electrons in an atom can have the same set of four quantum numbers' is
miv72 [106K]

Answer:

D. Pauli's exclusion principle

Explanation:

<em>A. Newton's laws</em> are related to the motion, they state that "Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it", "  Force equals mass times acceleration." and "  For every action there is an equal and opposite reaction"  

<em>B. Bohr's law </em>depicts an atom as a small, positively charged nucleus surrounded by electrons. These electrons travel in circular orbits around the nucleus.  

<em>C. Aufbau principle</em>, also called the building-up principle or the aufbau rule, states that in the ground state of an atom or ion, electrons fill atomic orbitals of the lowest available energy levels before occupying higher levels  

<em>D. Pauli's exclusion principle</em> states that <em>no two fermions (e.g., electrons) in an atom can have the same set of quantum numbers,</em> hence they have to "pile up" or "build up" into higher energy levels.

I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

f = \frac{v}{\lambda}

Where,

v = Velocity \rightarrow 20m/s

\lambda = Wavelength \rightarrow 35*10^{-2}m

Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

The frequency is directly proportional to the angular velocity therefore

\omega = 2\pi f

\omega = 2\pi *57.14

\omega = 359.03rad/s

Now the maximum speed from the simple harmonic movement is given by

V_{max} = A\omega

Where

A = Amplitude

Then replacing,

V_{max} = (1*10^{-2})(359.03)

V_{max} = 3.59m/s

Therefore the maximum speed of a point on the string is 3.59m/s

8 0
3 years ago
half life worksheet answer key 3. What percent of a sample As-81 remains un-decayed after 43.2 seconds
Ivahew [28]

The final mass after decay can be obtained by using under given relation:

half life period of As-81 = 33 seconds

mf = mi x (1/2^n)

= 100 x ( 1/2^(43.2/33))

= 40.4 %


3 0
3 years ago
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