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3 years ago

11

Jack has a weight of 300 N and sits 2.0 m from the pivot of see - saw. Jill has a weight of 450 N and sits 1.5 m from pivot. Who

will move down?

1 answer:

coldgirl [10]3 years ago

3 0

Answer:

Jill will move down first

Explanation:

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A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

3 years ago

A substance is round in one container and has a volume of one liter. In a much larger container, the substance is rectangular bu

UkoKoshka [18]

Answer: liquid

explanation: 1 liter is a measurement of liquids, not solids, or gases.
Liquids also have a set volume, but can flow to take the shape of the bottom of their container.

8 0

3 years ago

Your lab instructor has asked you to measure a spring constant using a dynamic methodâ€”letting it oscillateâ€”rather than a sta

yuradex [85]

Answer:

k = 6,547 N / m

Explanation:

This laboratory experiment is a simple harmonic motion experiment, where the angular velocity of the oscillation is

w = √ (k / m)

angular velocity and rel period are related

w = 2π / T

substitution

T = 2π √(m / K)

in Experimental measurements give us the following data

m (g) A (cm) t (s) T (s)

100 6.5 7.8 0.78

150 5.5 9.8 0.98

200 6.0 10.9 1.09

250 3.5 12.4 1.24

we look for the period that is the time it takes to give a series of oscillations, the results are in the last column

T = t / 10

To find the spring constant we linearize the equation

T² = (4π²/K) m

therefore we see that if we make a graph of T² against the mass, we obtain a line, whose slope is

m ’= 4π² / k

where m’ is the slope

k = 4π² / m'

the equation of the line of the attached graph is

T² = 0.00603 m + 0.0183

therefore the slope

m ’= 0.00603 s²/g

we calculate

k = 4 π² / 0.00603

k = 6547 g / s²

we reduce the mass to the SI system

k = 6547 g / s² (1kg / 1000 g)

k = 6,547 kg / s² =

k = 6,547 N / m

let's reduce the uniqueness

[N / m] = [(kg m / s²) m] = [kg / s²]

7 0

3 years ago

A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled

hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁) where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ = (v² - u²)/2a

h₂ = h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ = 1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ = 1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ = 1.3 m + 0.54 m

h₂ = 1.84 m

7 0

2 years ago

81. A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a

Irina18 [472]

Answer:

Explanation:

Given

mass of squirrel $m=560 gm$

Surface area of squirrel $A=930 cm^2$

and the area which face $A_f=\frac{A}{2}=465 cm^2$

height of tree $h=5 m$

Coefficient of drag $C=1$

drag Force $F_d=\frac{1}{2}C\cdot \rho \cdot A\cdot v^2$

Terminal velocity is given

$F_d=mg$

$\frac{1}{2}C\cdot \rho \cdot A\cdot v^2=mg$

$v=\sqrt{\frac{2\times m\times g}{\rho \times C\times A_f}}$

$v=\sqrt{\frac{2\times 0.560\times 9.8}{1.2\times 1\times 465\times 10^{-4}}}$

$v=13.9 m/s$

(b)Mass of person $m=56 kg$

$v^2-u^2=2gh$

$u=0$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 5}$

$v=9.89 m/s$

7 0

3 years ago

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