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1 year ago

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Which of these shows unbalanced forces at work on an object? A. an ice skater turning as he skates around an ice rink B. a bicyc

le chained to a bike rack at school C. a car driving at a constant 25 miles per hour on a straight street D. a water skier pulled by a boat at a constant 30 mph

2 answers:

Serggg [28]1 year ago

5 0

The answer is A people think ok..................................................!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

rosijanka [135]1 year ago

4 0

I’m pretty sure the answer is a

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If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with spee

Alexus [3.1K]

Answer:

457.81 Hz

Explanation:

From the question, it is stated that it is a question under Doppler effect.

As a result, we use this form

fo = (c + vo) / (c - vs) × fs

fo = observed frequency by observer =?

c = speed of sound = 332 m/s

vo = velocity of observer relative to source = 45 m/s

vs = velocity of source relative to observer = - 46 m/s ( it is taking a negative sign because the velocity of the source is in opposite direction to the observer).

fs = frequency of sound wave by source = 459 Hz

By substituting the the values to the equation, we have

fo = (332 + 45) / (332 - (-46)) × 459

fo = (377/ 332 + 46) × 459

fo = (377/ 378) × 459

fo = 0.9974 × 459

fo = 457.81 Hz

7 0

2 years ago

A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

<u>The relation between the change in pressure with height is given as:</u>

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

<em>Now putting the respective values:</em>

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0

1 year ago

A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer

Mekhanik [1.2K]

The initial volume of the gas is
$V_i = 600 cm^3$
while its final volume is
$V_f = 200 cm^3$
so its variation of volume is
$\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3$

The pressure is constant, and it is
$p=400 kPa = 400 \cdot 10^3 Pa$

Therefore the work done by the gas is
$W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J$
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
$Q=+100 J$

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
$\Delta U = Q-W = 100 J - (-160 J)=+260 J$
where the positive sign means the internal energy of the gas has increased.

7 0

1 year ago

An alpha particle is identical to a(n) _____.

Alexxx [7]

Helium atom, in other words, it consistis of a particle having four protons and two neutrons.

3 0

2 years ago

Read 2 more answers

A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a

Yanka [14]

Answer:

Diameter of the cylinder will be $d=2.998\times 10^4m$

Explanation:

We have given young's modulus of steel $E=207GPa=207\times 10^9Pa$

Change in length $\Delta l=0.38mm$

Length of rod $l=500mm$

Load F = 11100 KN

Strain is given by $strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}$

We know that young's modulus $E=\frac{stress}{strain}$

So $207\times 10^9=\frac{stress}{7.6\times 10^{-4}}$

$stress=1573.2\times 10^{-5}N/m^2$

We know that stress $=\frac{force}{artea }$

So $1573.2\times 10^{-5}=\frac{11100\times 1000}{area}$

$area=7.055\times 10^{8}m^2$

So $\frac{\pi }{4}d^2=7.055\times 10^{8}$

$d=2.998\times 10^4m$

6 0

2 years ago