8. A car travels at a constant velocity of 70 mph for one hour. By the end of the second hour, the car’s velocity was 60 mph. At
1 answer:
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<h2>The temperature of the air is 66.8° C</h2>
Explanation:
From the Newton's velocity of sound relationship , the velocity of sound is directly proportional to the square root of temperature .
In this case The velocity of sound = frequency x wavelength
= 798 x 0.48 = 383 m/sec
Suppose the temperature at this time = T K
Thus 383 ∝ I
The velocity of sound is 329 m/s at 273 K ( given )
Thus 329 ∝ II
Dividing I by II , we have
=
or = 1.25
and T = 339.8 K = 66.8° C
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and
The sum is represented as
For the the values given to us the sum is calculated as
Now the since the uncertainity inthe sum is
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals meters
Answer:
a) P = 44850 N
b)
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:
on substituting the values, we get
or
Load, P = 44850 N
Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>
b)The deformation () due to an axial load is given as:
on substituting the values, we get
or