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goldfiish [28.3K]
2 years ago
9

A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at

her destination, the tire pressure has increased to 245 kPa. What is the temperature of the tires if we assume that the tires do not expand?
Physics
1 answer:
GuDViN [60]2 years ago
4 0

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K

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miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

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    = 1. 579 x 10 ^-22N

6 0
1 year ago
A car hits another and the two bumpers lock together during the collision. is this an elastic or inelastic collision?
valkas [14]
Inelastic.
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8 0
2 years ago
A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
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= 1016.31 N * cos(53) 
= 611.63 N 

The total normal force Fn then is: 
Fn = Fg + Fpn 
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= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
= (Fp * sin(θp)) + 0 
= 1016.31 N * sin(53) 
= 811.66 N</span>

 

> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0
2 years ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
horrorfan [7]

Answer:

The value is t_1 =  9 \  s

Explanation:

Generally the velocity attained by the sled after t = 3.10 s is mathematically evaluated using the kinematic equation as follows

v  =  u   +  at

Here u = 0 \ m/s

a = 13.5 m/s^2

So

v  =  0   +  13.5 *  3.10

=> v  =  41.85 \ m/s

The is distance it covers at this time is

s =  u *  t  +  \frac{1}{2} a *  t^2

=> s =  +  \frac{1}{2} * 13.5 *  3.10^2

=> s =64.87

Now when sled stops its the final velocity is v_f =  0 m/s while the initial velocity will be the velocity after its acceleration i.e v  =  41.85 \ m/s

So

v_f  =  v  +  a_1t_1

Here  a_1 =  - 4.65, the negative sign shows that it is deceleration

So

           0  =  41.85  - 4.65 *  t_1

=> t_1 =  9 \  s

3 0
1 year ago
The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's
Mice21 [21]

Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

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The frequency in the spring is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{L} is the linear mass density

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m = 0.0022 kg

or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.

3 0
2 years ago
Read 2 more answers
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