Question

Please, someone help me with my assignment. It's really urgent.

Answer 1

$Given :-$

Average of first 40 results is 40, and that of next 60 results is 60

$To \: Find :-$

The average of all 100 results

$Solution :-$

We know that

$\bf\red{Average = \dfrac{Sum \; of \; observation}{No. \; of \; observation}$

For the first 40 results

Let the sum be x

$\sf 40 = \dfrac{x}{40}$

$\sf 40 \times 40 = x$

$\sf 1600 = x$

For the last 60 result

Sum = y

$\sf 60 = \dfrac{y}{60}$

$\sf 60 \times 60 = y$

$\sf 3600 = y$

$\bf \pink{Total \; Avg. = \dfrac{x+y}{100}$

$\sf Total \; Average = \dfrac{1600 + 3600}{100}$

$\sf Total\; average = \dfrac{5200}{100}$

$\sf Total \; Average = 52$

Answer 2

☆Concept:

》as we release the toy car from the spring,its spring potential energy will get converted to kinetic energy which will give velocity to it to move forward.

》as the friction is neglected, the bottom point will possess same velocity throughout.

☆Solution:

1) see the attachment!

▪︎here ,as the Normal is a contact force it will always be the reaction given by the circular loop to the toy car, so it will always be directed inwards.

▪︎whereas,the centrifugal(mv²/r) will be directed outside (if we see it as particle frame of reference)

2) let's velocity at bottom- $\ v_{b}$

velocity at top - $\ v_{t}$

☆total kinetic energy at bottom:

》 the spring energy = the initial kinetic it possess.

• $\dfrac{1}{2}kx² = \dfrac{1}{2}mv_{b}²$

given: - k = 2500N/m, x=d=2cm=$\ 2×10^{-2}$m

• $\dfrac{1}{2}kx² = K.E_{at\;bottom}$

• $\dfrac{1}{2}(2500)(2×10^{-2})² =K.E_{at\;bottom}$

•$\dfrac{1}{2}(2500)(4×10^{-4}) =K.E_{at\;bottom}$

》KE at bottom = 0.5joules.

☆total energy at topmost point:

》at top its possess kinetic energy + some potential

》here we can say that initial loss of some kinetic will be converted to final potential + kinetic(by energy theorem)

》initial energy = final energy

• $\dfrac{1}{2}mv_{b}² = \dfrac{1}{2}mv_{t}² + mg(2R)$

• $\dfrac{1}{2}mv_{b}² = K.E_{at\;top} + mg(2R)$

• $\ K.E_{at\;top} = \dfrac{1}{2}mv_{b}²- mg(2R)$

• $\ K.E_{at\;top} = \dfrac{1}{2}kd²- mg(2R)$

▪︎▪︎above is the eq. which is required in k,d,m,g form.

•$\ K.E_{at\;top} = 0.5- mg(2R)$

m=0.1kg,g=10,R=10cm=0.1m

• $\ K.E_{at\;top} = 0.5- 0.1(10)(2×0.1)$

• $\ K.E_{at\;top} = 0.5- 0.2$

• K.E at top = 0.3 joules.

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