348.34 m/s. When Superman reaches the train, his final velocity will be 348.34 m/s.
To solve this problem, we are going to use the kinematics equations for constant aceleration. The key for this problem are the equations and where is distance, is the initial velocity, is the final velocity, is time, and is aceleration.
Superman's initial velocity is , and he will have to cover a distance d = 850m in a time t = 4.22s. Since we know , and , we have to find the aceleration in order to find .
From the equation we have to clear , getting the equation as follows: .
Substituting the values:
To find we use the equation .
Substituting the values:
The answer is ultra violet radiation. From the air
Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
The mass of lead required to make a 1.00 cm3 fishing sinker is 11.3g.
What is mass?
Mass is a metric used in physics to express inertia, a fundamental characteristic of all matter. A mass of matter's resistance to altering its direction or speed in response to the application of a force is what it essentially is. The change that an applied force produces is smaller the more mass a body has.
Given :
Density of lead = 11.3 g/cm3
Volume of sinker = 1.00 cm3
One of a substance's attributes is density, which is calculated by dividing the mass by the volume. Mathematically:
Density : Mass / volume
therefore after putting the values,
mass= 11.3g
To learn more about density click on the link below:
brainly.com/question/18939565
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Image formed by a plane mirror is always virtual which means that the light rays do not actually come from the image but upright and these of the same shape and size are the object it is<span> reflecting.</span>