What is the shorter electromagnetic wave called? A. more powerful. B. less powerful. C. hotter. D.colder
1 answer:
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Answer:
a)this graph is also a line b) in both cases we have a uniform movement
Explanation:
In this exercise we have a uniform movement
v = d / t
d = v t
in the table we give some values to make the graph
t (s) d (m)
1 10
2 20
3 30
In the attached we can see the graph that is a straight line
we have another vehicle at v = 50 me / S
t (s) d (m)
1 50
2 100
3 150
this graph is also a line
b) in both cases we have a uniform movement
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
<em>From the question;</em>
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt -
h = 26.54(1.7) -
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m
Answer:
speed of the mass is 3.546106 m / s
Explanation:
given data
mass = 77.3 g = 77.3 × kg
spring constant k = 12.5 N/m
amplitude A = 38.9 cm = 38.9 × m
to find out
the speed of the mass
solution
we will apply here conservation energy that is
K.E + P.E = Total energy ..................1
so that Total energy = K.E max = P.E max
we know amplitude so we find out first P.E max that is
PE max = K.E + P.E
(1/2)kA² = (1/2)mv² + (1/2)kx²
kA^² = mv²+ kx²
so here v² will be
v² = k(A² - x²) / m
v = √[(k/m)×(A² - x²)] ............2
here x = (1/2)A so from from 2 equation
v = √[(k/m)×(A² - (A/2)²)]
v = √[(k/m)×(3/4×A²)]
now put all value
v = √[(12.5/ 77.3 × )×(3/4×(38.9 ×
)²)]
v = 3.546106 m / s
speed of the mass is 3.546106 m / s