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crimeas [40]
3 years ago
11

On Mars gravity is one-third that on Earth. What would be the mass on Mars of a person who has a mass of 90 kilograms (kg) on Ea

rth?
Physics
1 answer:
snow_tiger [21]3 years ago
4 0

Answer: The person will still have a mass of 90kg on Mars

Explanation: The Truth is, the mass of a body remains constant from place to place. It is the weight which is equal to {mass of body * acceleration due to gravity{g}} that varies from place to place since it is dependent on {g}.

In this case the person will have a Weight of 90*9.8 = 882N on Earth.

{ "g" on Earth is 9.8m/s²}

And a Weight of 90*3.3 = 297N on Mars.

{ From the question "g" on Mars is {9.8m/s²}/3 which is 3.3m/s²}

From this analysis you notice that the WEIGHT of the person Varies but the MASS remained Constant at 90kg.

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Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.
Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

displacement = +1.0 m-0 m=+1 m

8 0
3 years ago
The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the
Studentka2010 [4]

Answer:

V(t)= 240V* H(t-5)

Explanation:

The heaviside function is defined as:

H(t) =1 \quad t\geq 0\\H(t) =0 \quad t

so we see that the Heaviside function "switches on" whent=0, and remains switched on when t>0

If we want our heaviside function to switch on when t=5, we need the argument to the heaviside function to be 0 when t=5

Thus we define a function f:

f(t) = H(t-5)

The -5 term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

V(t)= 240V* H(t-5)

I have made a sketch for you, and added it as attachment.  

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