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Vinil7 [7]
2 years ago
6

To get a feeling for inertial forces discuss the familiar cases of accelerating in a car in a straight line while increasing or

decreasing speed and turning the wheel to change direction. What direction do you feel a force in these scenarios and how does the strength of that force change if you either hit gas/brake harder or turn sharper?
Physics
1 answer:
inn [45]2 years ago
7 0

Answer:

Explanation:

When we accelerate in a car on a straight path we tend to lean backward because our lower body part which is directly in contact with the seat of the car gets accelerated along with it but the upper the upper body experiences this force  later on due to its own inertia. This force is accordance with Newton's second law of motion and is proportional to the rate of change of momentum of the upper body part.

Conversely we lean forward while the speed decreases and the same phenomenon happens in the opposite direction.

While changing direction in car the upper body remains in its position due to inertia but the lower body being firmly in contact with the car gets along in the direction of the car, seems that it makes the upper body lean in the opposite direction of the turn.

On abrupt change in the state of motion the force experienced is also intense in accordance with the Newton's second law of motion.

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Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston on which a number of small weights are
Amiraneli [1.4K]

Answer:

Explanation:

a ) At constant pressure , work done = P x Δ V

= 200 x 10³ x ( .1 - .04 )

= 12 x 10³ J .

b )

At constant temperature work done

= n RT ln v₂ / v₁

= PV ln v₂ / v₁

= 200 x 10³ x .04 ln .1 / .04

8 x 10³ x .916

= 7.33 x 10³ J .

5 0
1 year ago
In the swing carousel amusement park ride, riders sit in chairs that are attached by a chain to a large rotating drum as shown i
irinina [24]

Answer:\theta =44.068^{\circ}

Explanation:

Given

time taken to complete the circle=7.9 s

radius of circle(r)=15 m

velocity of rider is given by =\frac{2\pi r}{t}

v=\frac{2\pi 15}{7.9}=11.93 m/s

Let us suppose T is the tension in the chain and \thetais the angle which chain makes with vertical

Therefore T\sin \theta =\frac{mv^2}{r}-1

T\cos \theta=mg --2

Divide 1 & 2 we get

tan\theta =\frac{v^2}{rg}

tan\theta =0.968

\theta =44.068^{\circ}

8 0
2 years ago
The seismic activity density of a region is the ratio of the number of earthquakes during a given time span to the land area aff
Natalija [7]

Answer:

0.0059

Explanation:

According to the question the seismic activity density is given by

\text{Seismic activity density}=\frac{\text{Number of Earthquakes over a given time span}}{\text{The land area affected}}

Here,

Number of Earthquakes over a given time span = 424

The land area affected = 71300 mi²

So,

\text{Seismic activity density}=\frac{424}{71300}\\\Rightarrow \text{Seismic activity density}=0.0059

The seismic activity density is 0.0059

8 0
1 year ago
Read 2 more answers
Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa
Katarina [22]

Answer:

A 93%

Explanation:

P_1=P_2 = Pressure will be equal at inlet and outlet

\rho = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

v_1 = Velocity at inlet = 1.2 m/s

v_2 = Velocity at outlet

r_1 = Radius of inlet = \dfrac{1.5}{2}=0.75\ cm

r_2 = Radius of outlet

From Bernoulli's relation

P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s

From continuity equation

A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m

The fraction would be

\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%

The fraction is 93.0304%

8 0
1 year ago
--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t
REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

                                  <em> V = v cos Ф </em>

Substituting the values in the above equation

                                   v = 300 x cos 90°

                                      = 300 x 0

                                      = 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

5 0
2 years ago
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