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Mama L [17]
3 years ago
11

What are the quantities and units of the formula of distance in accelerated motion?

Physics
1 answer:
creativ13 [48]3 years ago
8 0

Answer:The equation is

s = ut + 1/2 at²

Where s is the distance, u is an initial velocity a the acceleration and t is time. If starting from rest the first term can be ignored giving

s = 1/2 at²

Explanation: :b

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<span>U could compare them using the intensity technique when bending waves are negligible in comparison with quasi-longitudinal waves.</span>
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3 years ago
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A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple
jok3333 [9.3K]

Answer:

C) 1 s

Explanation:

The period of a mass-spring system is given by the formula:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass hanging on the spring

k is the spring constant

As we can see from the equation above, the period of the system does NOT depend on the initial amplitude of the oscillation. Therefore, even if the initial amplitude is changed from 5 cm to 10 cm, the period of the system will remain the same, 1 s.

4 0
3 years ago
i need help please. this is for physics but everything i search for related to this comes up as chemistry
Annette [7]

The car tyre contains air initially at a pressure of 195 kPa after travelling several km the temperature of the air inside a car tyre rises from 30 to 70°C if the tyre is rigid and does not expand then the new pressure inside the tyre would be 220.74 kPa.

<h3>What is pressure?</h3>

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

By using the Charles law for gases which states that the volume of the gas remains constant then the pressure of the gas is directly proportional to the temperature.

As given in the problem the tyre is rigid and does not expand this means the volume of the tyre remains constant.

The mathematical expression for Charles's law is as follows

P₁/P₂ = T₁/T₂

First, we have to change the temperature from degree Celcius to the kelvin temperature scale

K = 273 + C

where k is the temperature in kelvin and the C is degrees of Celcius

Initially, the temperature was 30° C

T₁ = 273 + 30

T₁ = 303 K

Then after travelling the temperature of the air inside a car tyre rises from 30 to 70°C

T₂= 273+ 70

T₂ =343 K

The car tyre contains air initially at a pressure of 195 KPa

P₁ = 195 kPa

Lets us take the final pressure of the air would be P₂

By substituting the values in the formula

P₁/P₂ = T₁/T₂

195/P₂ = 303/343

P₂ = 220.74 kPa

Thus, the new pressure inside the tyre would be  220.74 kPa.

Learn more about pressure learn more

brainly.com/question/28012687

#SPJ1

7 0
1 year ago
A car of mass 200 kg, moving with a forward acceleration of 3 m/s-ıs acted upon by
Shkiper50 [21]

Answer:

600 N

Explanation:

5 0
2 years ago
The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when
mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, a=-8\ m/s^2 (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

x_1=ut_1+\dfrac{1}{2}at^2

20=4u+\dfrac{1}{2}(-8)\times 4^2

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

v=u+at

0=21+(-8)t

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

v=21+(-8)\times 11

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|

d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.    

5 0
3 years ago
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