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Veronika [31]
3 years ago
7

When you sneeze, the air in your lungs accelerates from rest to 155 km/h in approximately 0.70 s . what is the acceleration of t

he air in m/s2?
Physics
1 answer:
ollegr [7]3 years ago
6 0
<span>62 m/s^2 First, convert 155 km/h to m/s 155 km/h * 1000 m/km / 60 min/h / 60 s/h = 43.05556 m/s Now take the speed in m/s and divide by the number of seconds it took to accelerate it to that speed. 43.05556 m/s / 0.70 s = 61.5 m/s^2 Since we only have 2 significant figures, round the result to 2 figures. 61.5 m/s^2 = 62 m/s^2</span>
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A car accelerates for 10 seconds. During this time, the angular
bagirrra123 [75]

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, \omega_i = 10 rad/s

final angular velocity, \omega_f = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2

Therefore, the angular acceleration of the car is 1.5 rad/s²

4 0
3 years ago
The first law of thermodynamics can be given as ________.
MrRa [10]
The first law of thermodynamics can be written as
\Delta U = Q-W
where
\Delta U is the variation of internal energy of the system
Q is the amount of heat absorbed by the system
W is the work done by the system on the surrounding.

Using this form, the sign convention for Q and W becomes:
Q > 0 --> heat absorbed by the system (because it increases the internal energy)
Q < 0 --> heat released by the system (because it decreases the internal energy)
W > 0 --> work done by the system (for instance, an expansion: when the system expands, it does work on the surrounding, and so the internal energy decreases, this is why there is a negative sign in the formula Q-W)
W < 0 --> work done by the surrounding on the system (for instance, a compression: when the system is compressed, the surrounding is doing work on the system, and so the internal energy of the system increases)
7 0
3 years ago
A ball is dropped from rest at the top of a 6.10 m
natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

\frac{v_1}{v} = e ( coefficient of restitution ) = \frac{1}{\sqrt{10} }

and

\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

e = \sqrt{\frac{h_1}{6.1} }

e = \sqrt{\frac{h_2}{h_1} }

So on

e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }

(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}= .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

3 0
3 years ago
A bald eagle is flying to the left with a speed of 34 meters
Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U  + at

Here,

U = 34 m/s

a =  - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V  = 34 + (-8)3

 = 34 - 24

 V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0
2 years ago
A machine that operates a ride at the fair requires 2500 J to lift a 294 N child 5.0 m. What is the efficiency of this machine?
NeX [460]

Answer:

η = 58.8%

Explanation:

Work is defined as the force applied by the distance traveled by the body.

W =F*d

where:

W = work [J] (units of joules)

F = force = 294 [N]

d = distance = 5 [m]

W = 294*5\\W = 1470 [J]\\

Efficiency is defined as the energy required to perform an activity in relation to the energy actually added to perform some activity. This can be better understood by means of the following equation.

efficiency = W_{done}/W_{required}\\efficiency = 1470/2500\\efficiency = 0.588 = 58.8%

5 0
3 years ago
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