Answer:
Time, t = 0.23 seconds
Explanation:
It is given that,
Initial speed of the ranger, u = 52 km/h = 14.44 m/s
Final speed of the ranger, v = 0 (as brakes are applied)
Acceleration of the ranger, 
Distance between deer and the vehicle, d = 87 m
Let d' is the distance covered by the deer so that it comes top rest. So,
d' = 26.06 m
Distance between the point where the deer stops and the vehicle is :
D=d-d'
D=87 - 26.06 = 60.94 m
Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :
t = 0.23 seconds
Hence, this is the required solution.
The full question is:
On a keyboard, you strike middle C, whose frequency is 256 Hz. What is the period of one vibration of this tone?
The period of a vibration is the time it takes for the particle to make one full oscillation. Frequency is by definition number of full oscillations per unit of time.
When the frequency is expressed in Hz that unit of time is one second.
So there is the following relation between frequency and period:
When we plug in the numbers we get:
Answer:
d) Decrease, increase, remain constant
Explanation:
If one tire has a slow leak, the air pressure within the tire will_DECREASE____with time due to outflow of air , the surface area between the tire and the road will__INCREASE__in time,due to flattening of tire.
The net force the tire exerts on the road will_REMAIN CONSTANT____in time. It is so because force does not depend upon area. It is pressure which depends upon area. As there is no change in the weight of the car , force on the road will remain constant.
There is no diagram below so I can't answer the question
The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
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