(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim
e between , when the sphere is launched and , when the sphere hits the target. Label for the horizontal component of the sphere’s velocity and the vertical component of the sphere’s velocity.
1 answer:
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Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
=
[ (805 N)( (3-1) m) + ( 395 N )(
) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
=
[ 1610 + 592.5 + 500 + 410 ]
=
[ 3112.5 ]
= 1037.5 N
= 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
= ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
= 2520 N - 1037.5 N
= 1482.5 N
= 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
Answer:
The angular magnification is
Explanation:
From the question we are told
The focal length is
The near point is
The angular magnification is mathematically represented as
Substituting values
Answer:
Explanation:
Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies where r is the distance
Since intensity is given then when at 3 m
Since we have the constant then at 4m
Intensity,
Answer:
The frequency of radiation is
Explanation:
Given:
Wavelength m
Speed of light
For finding the frequency of radiation,
Therefore, the frequency of radiation is