When is the particle in figure (a) speeding up? (enter your answer using interval notation.)

A long uniform wooden board (a playground see-saw) has a pivot point at its center. An older child of mass M=32 kg is sitting a

PIT_PIT [208]

Answer: $\frac{5L}{6}$

Explanation:

Given

Wooden board is pivoted at center and

Older child of mass $M=32\ kg$ is sitting at a distance of L from center

if two child of mass $\frac{M}{2}$ is sitting at a distance $\frac{L}{6}$ and $x$ (say) from pivot then net torque about pivot is zero

i.e.

$\Rightarrow \tau_{net}=MgL-\frac{M}{2}g\frac{L}{6}-\frac{M}{2}gx$

as $\tau_{net}=0$

Therefore

$MgL=\frac{M}{2}g\frac{L}{6}+\frac{M}{2}gx$

$L-\frac{L}{6}=x$

$x=\frac{5L}{6}$

Therefore another child is sitting at a distance of $\frac{5L}{6}$

3 0

3 years ago

Which property of substances help to make a thermometer?<br> I NEED HELP.

3241004551 [841]

Answer:

Mercury

Explanation:

Mercury streches when it is hot so that is why it goes up when the temperature is higher

6 0

3 years ago

The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic

Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight = $\frac{2uSin(A)}{g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = $\frac{2(4)Sin(60)}{10}$ = 0.7

ii) distance travel = Range = R = $\frac{u^{2}Sin(2A) }{g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = $\frac{4^{2}Sin(2*60) }{10}$ = 1.39

iii) Maximum Height = H = $\frac{u^{2}(Sin(A))^{2} }{2g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: $\frac{4^{2}(Sin60)^{2} }{2*10}$ = 0.6

4 0

3 years ago

a spiral spring of natural length 10.0cm has a scale pan hanging freely in its tower ends . When an object of mass of 20g is pla

weqwewe [10]

Answer:

Explanation:

There seems to be a typo in the problem statement. It says the spring stretches to a shorter length after more mass is added. Please check the problem statement. I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm. The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables. If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

0.0180 m + 0.0009 = 0.02050 m + 0.00041

0.00049 = 0.0025 m

m = 0.196

The pan has a mass of 0.196 kg, or 196 g.

4 0

3 years ago

During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.5 rad/s. Find the angular displ

zavuch27 [327]

Answer: angular displacement in rad = 3038.45 rad

angular displacement in rev = 483.589 rev

Explanation: mathematically

Angular velocity = angular displacement / time taken.

Angular velocity = 33.5 rad/s, time taken = 90.7s

33.5 = angular displacement /90.7

Angular displacement = 33.5 * 90.7 = 3038.45 rad

But 1 rev =2π

Hence 3038.45 rad to rev is

3038.45/2π = 483.599 rev

7 0

4 years ago

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