During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ± 0.001

Question

During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away
from another trolley, Y, of mass (2.561 ± 0.001) kg with a speed of (3.2 ± 0.01) ms
-1
. The
second trolley is moving away with a distance of (2.5 ± 0.01) ms-1
.
What is the absolute uncertainty of the ratio of momentum of the two trolleys
X
Y
?
Please I really need the answer of this question. Can somebody here do it for me?

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Thiên Di 5 months 2021-09-05T01:01:14+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-09-05T01:03:10+00:00

    Answer: The ratio X/Y is (1.172 ± 97.667)

    Explanation: Absolute uncertainty is the value that when combined with a reported number, gives the range of the number.

    Momentum is a quantity of motion an object has. It is calculated as a relation between mass of an object and its velocity: p = m.v

    For the trolley X, momentum is:

    p_{x} = (2.34 ± 0.01)*(3.2 ± 0.01)

    and for trolley Y, momentum is:

    p_{y} = (2.561 ± 0.001)*(2.5 ± 0.01)

    To solve the multiplications:

    For x:

    p_{x} = 2.34*3.2 = 7.5

    relative uncertainty = \frac{0.01}{2.34} + \frac{0.01}{3.2} = 0.0074

    absolute uncertainty = 7.5*0.0074 = 0.055

    p_{x} = 7.5 ± 0.055

    For y:

    p_{y} = 2.561*2.5 = 6.4

    relative uncertainty = \frac{0.001}{2.561} + \frac{0.01}{2.5} = 0.0044

    absolute uncertainty = 6.4*0.0044 = 0.028

    p_{y} = 6.4 ± 0.028

    The ratio of momentum:

    \frac{p_x}{p_y} = (7.5 ± 0.055) ÷ (6.4 ± 0.028)

    Dividing absolute uncertainty, the rules are the same for multiplication.

    \frac{p_x}{p_y} = \frac{7.5}{6.4} = 1.172

    relative uncertainty = \frac{0.0055}{7.5} + \frac{0.028}{6.4} = 0.012

    absolute uncertainty = 1.172*0.012 = 97.667

    \frac{p_x}{p_y} = 1.172 ± 97.667

    The ratio of momentum of the 2 trolleys is 1.172 ± 97.667.

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