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3 years ago

A 2100 lb (1 kg=2.2 lbs) car is using 4500 N of traction in order to move with a constant

Physics

1 answer:

vredina [299]3 years ago

4 0

Answer:

$a = 0\ m/s^2$

Explanation:

The acceleration of a body is defined as the change in velocity of an object, with respect to time. So, in order to produce and acceleration in any object, the velocity of that object must change. And, if the velocity of the object is not changing with respect to time, then the object is said to have a uniform motion. The acceleration of the object in such uniform motion is always zero. Because the change in velocity is zero.

$acceleration = change in velocity/time\\acceleration = 0 / time\\acceleration = 0 m/s^2$

Since, the car in this case is moving with a constant speed. Therefore, its acceleration will also be zero.

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50 mL of soda in a soda can exerts ____ 50 mL of soda in a 1L bottle.

lutik1710 [3]

More pressure than..

4 0

3 years ago

Giving brainiest to correct answer.

mixas84 [53]

Answer:

$5.33\ m/s$

Explanation:

$We\ know\ that,\\Momentum=Mass*Velocity\\p=mv\\Hence,\\Lets\ first\ consider\ the\ case\ of\ the\ two\ balls\ 'Before\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Initial\ Velocity\ of\ the\ green\ ball=5\ m/s\\Initial\ Momentum\ of\ the\ green\ ball=5*0.2=1\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Initial\ Velocity\ of\ the\ pink\ ball=2\ m/s\\Initial\ Momentum\ of\ the\ pink\ ball=0.3*2=0.6\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'Before\ Collision'=1+0.6=1.6\ kg\ m/s$

$Hence,\\Lets\ now\ consider\ the\ case\ of\ the\ two\ balls\ 'After\ Collision':\\\\Mass\ of\ the\ green\ ball=0.2\ kg\\Final\ Velocity\ of\ the\ green\ ball=0\ m/s\\Final\ Momentum\ of\ the\ green\ ball=0\ kg\ m/s\\\\Mass\ of\ the\ pink\ ball=0.3\ kg\\Final\ Velocity\ of\ the\ pink\ ball=v\ m/s\\Final\ Momentum\ of\ the\ pink\ ball=0.3*v=0.3v\ kg\ m/s\\\\Total\ momentum\ of\ both\ the\ balls\ 'After\ Collision'=0+0.3v=0.3v\ kg\ m/s$

$As\ we\ know\ that,\\Through\ the\ law\ of\ conservation\ of\ momentum,\\In\ an\ isolated\ system:\\Total\ Momentum\ Before\ Collision=Total\ Momentum\ After\ Collision\\Hence,\\1.6=0.3v\\v=\frac{1.6}{0.3}=5.33\ m/s$

5 0

3 years ago

A type of energy embodied in oscillating electric and magnetic fields is called

yaroslaw [1]

Answer: Electromagnetic radiation

Explanation:

Electromagnetic radiation is a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.

To understand it better:

This radiation is spread thanks to the electromagnetic fields produced by moving electric charges and their sources can be natural or man-made.

It should be noted that the energy of electromagnetic radiation can vary and depending on its frequency it can be useful for various situations.

4 0

3 years ago

A bicycle rider traveling east at 10 km/hr sees a blue car pass her appearing to travel west at 50 km/hr. What is the blue car's

BigorU [14]

Answer:

$v_{bR} = 25 km/h$ towards west

Explanation:

As we know that the speed of the blue car as appear to the bicycle rider is given as

$v_{bc} = 50 km/h$ towards west

also it is given that bicycle is moving at speed of 10 km/h towards East

so here we have

$v_{bc} = v_b - v_c$

so we have

$v_b = v_{bc} + v_c$

$v_b = -50 + 10 = 40 km/h$ towards west

now speed of the red car is given as 15 km/h towards west

so here the relative speed of blue car with respect to red car is given as

$v_{bR} = v_b - v_R$

$v_{bR} = 40 - 15 = 25 km/h$ towards west

8 0

3 years ago

Water in a tank is pressurized by air and pressure measured using a multi-fluid manometer. Determine the gage pressure of air in

sattari [20]

Answer:

The gauge pressure of air is 110 kpa

Explanation:

Atmospheric pressure, $P_{atm}$ = 101 Kpa

$P_{gauge} + \rho_w gh_1 + \rho_o gh_2 -\rho_{Hg} gh_3 =P_{atm}$

$P_{gauge} = P_{atm} - \rho_w gh_1 - \rho_o gh_2 +\rho_{Hg} gh_3$

where;

ρw is the density of water = 1000 kg/m³

ρo is the density of oil = 800 kg/m³

ρHg is the density of mercury = 13,600 kg/m³

g is acceleration due to gravity = 9.8 m/s²

$P_{gauge} = 101,000 - (1000* 9.8*0.2) - (800* 9.8*0.3) +(13,600* 9.8*0.46)\\\\P_{gauge} = 101,000 - 1960 - 2352 + 13610.26\\\\P_{gauge} = 110,298.26 pa$

Therefore, the gauge pressure of air is 110 kpa

4 0

3 years ago