Answer:
To find the acceleration of the object we have to apply Newton second law of motion that is F = mass × acceleration.
Explanation:
Given ,
F = 130N
M = 24kg
A = ?
F = m× a
then ,
130N = 24kg ×a
a = 130/24 = 5 m/s.
Answer:
C - higher volume
Explanation:
The pitch or frequency of sound that an object can produce depends upon its size and configuration . The shape of hand of all are same so the frequency of sound produced by hands of all will be almost same . Hence frequency of sound produced by the hands of Anand and Kumar would have been almost the same .
But the intensity of sound produced by them would have been different . Intensity represents energy a sound carries . Hard hitting clap will produce sound of higher intensity . Intensity of sound is also called high volume sound . So Kumar's clap will carry greater energy and hence greater volume of sound .
Answer:
about 19.6° and 73.2°
Explanation:
The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...
y = -4.9(x/s·sec(α))² +x·tan(α)
where s is the launch speed in meters per second.
We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):
-13.6111·tan(α)² +50·tan(α) -16.0511 = 0
This has solutions ...
tan(α) = 0.355408 or 3.31806
The corresponding angles are ...
α = 19.5656° or 73.2282°
The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.
_____
I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.
Answer:
7.78 * 10³ m/s
Explanation:
Orbital velocity is given as:
v = √(GM/R)
G = 6.67 * 10^(-11) Nm/kg²
M = 5.98 * 10^(24) kg
R = radius of earth + distance of the satellite from the surface of the earth
R = 2.15 * 10^(5) + 6.38 * 10^(6)
R = 6.595 * 10^(6) m
v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))
v = √(6.048 * 10^7)
v = 7.78 * 10³ m/s
