Which subatomic particle can be absent from an atom?

jolli1 [7]

A tiltmeter is an object to measure small movements.

8 0

3 years ago

The elastic energy stored in your tendons can contribute up to 35%% of your energy needs when running. Sports scientists find th

zzz [600]

Answer:

11.16 J

Explanation:

Elastic energy (E) stored in the tendon is given by 1/2ke^2

SPRINTERS

k = 31N/mm = 31N/mm × 1000mm/1m = 31,000N/m

e = 41mm = 41mm × 1m/1000mm = 0.041m

E = 1/2 × 31,000 × 0.041^2 = 26.06J

NON-ATHLETES

k = 31,000N/m

e = 31mm = 31mm × 1m/1000 = 0.031m

E = 1/2 × 31,000 × 0.031^2 = 14.90 J

Difference = 26.06 J - 14.90 J = 11.16 J

4 0

3 years ago

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

gayaneshka [121]

Force applied on the car due to engine is given as

$F_1 = 300 N$ towards right

Also there is a force on the car towards left due to air drag

$F_2 = 150 N$ towards left

now the net force on the car will be given as

$\vec F_{net} = \vec F_1 + \vec F_2$

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

$F_{net} = F_1 - F_2$

$F_{net} = 300 - 150$

$F_[net} = 150 N$

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

5 0

4 years ago

A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th

irina [24]

The angular momentum is defined as,

$L=I\omega$

Acording to this text we know for conservation of angular momentum that

$L_i=L_f$

Where $L_i$ is initial momentum

$L_f$ is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

$L_i=mvl$

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

$L_f=(I_b+I_s)\omega$

Substituting

$L_f=(ml^2+\frac{10ml^2}{3})\omega$

$L_f=\frac{13}{10}ml^2w$

$m(0.65)l=\frac{13}{10}ml^2 \omega$

$\omega=\frac{1}{2l}$

Applying conservative energy equation we have

$\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'$

$\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)$

Replacing the values and solving

$l=\frac{13}{0.54g}$

Substituting

l=\frac{13}{0.54(9.8)}

$l=2.45cm$

7 0

4 years ago

An object is launched upwards at an initial speed of 3 m/s. What is the maximum height reached by the object from where it was l

ollegr [7]

Answer:

the maximum height reached by the object from where it was launched is 0.4591 m

Explanation:

initial speed of the object, u= 3 m/s

The velocity at the maximum height will always be 0.

Therefore, final velocity, v= 0 m/s

Using the Newton's equation of motion,

v^2 - u^2 = 2*g*h(max)

0 - u^2= 2*g*h(max)

h(max) = -u^2 /2* g

where g is the gravitational acceleration.

g= - 9.8 m/s^2

substituting the values in equation,

h(max)= - (3*3) / 2*(-9.8)

h(max) = 0.4591 m

the maximum height reached by the object from where it was launched is 0.4591 m

learn more about maximum height reached here:

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4 0

2 years ago