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3 years ago

11

How can you decrease the amount of input force of a wheel and axle?

1 answer:

nasty-shy [4]3 years ago

7 0

Explanation:

hmm by the increasing the size of wheel and decreasing axle

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Iodine molecules separates into atoms after absorbing light of 450 nm. If one

OlgaM077 [116]

Answer:E = hc/? = 4.41 x 10-19 J

Energy absorbed by each atom :

E (atom) = 2.205 x 10-19 J

Now Bond Energy of each molecule (B) = 3.98 x J

So, for each atom 1.99 x 10-19 J

So now

KE of each atom = E(atom) - B (atom)

= 2.15 x 10-19 J

5 0

3 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

4 years ago

An object has 90,000 J of kinetic energy and is moving at 12 m/s. What is the objects mass?

trapecia [35]

Answer:

plug it into your calculator

Explanation:

Kinetic energy = KE = Joules = J

KE= 1/2mv^2

90,000 = 1/2 m (12^2)

180,000 = m (144)

m = 180,000 / 144

4 0

4 years ago

A truck of mass 500kg moving at 4m/s collides with another truck of mass 1500kg moving in the same direction at 2m/s. What is th

VMariaS [17]

Answer:

m1u1+m2u2=(m1+m2)v.

500×4+1500×2=(500+1500)V.

2000+3000=2000V.

5000=2000V.

V=2.5m/s

8 0

3 years ago

A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j

nika2105 [10]

To solve this problem we will use the kinematic equations of descriptive motion of a projectile for which both the height reached and the distance traveled are defined. From this type of movement the lion reaches a height (H) of 3m and travels a horizontal distance (R) of 10 m. Mathematically the equations that describe this movement are given as,

$H = \frac{v_0^2sin^2\theta}{2g}$

$R = \frac{v_0^2 sin 2\theta}{g}$

Dividing the two equation we have that

$\frac{H}{R}=\frac{\frac{v_0^2sin^2\theta}{2g}}{\frac{v_0^2 sin 2\theta}{g}}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{sin2\theta}$

$\frac{H}{R}= \frac{sin^2\theta}{2}*\frac{1}{2sin\theta cos\theta}$

$\frac{H}{R}= \frac{1}{4} \frac{sin\theta}{cos\theta}$

$\frac{H}{R}= \frac{1}{4} tan\theta$

Substituting values of H and R, we get

$\frac{3}{10} = \frac{1}{4} tan\theta$

$\theta = tan^{-1} \frac{12}{10}$

$\theta = 50.2\°$

Substituting the value of \theta in equation we get,

$H = \frac{v_0^2sin^2\theta}{2g}$

$v_0^2 = \frac{H 2g}{sin^2\theta}$

$v_0^2 = \frac{3*2*9.8}{sin^2(50.2)}$

$v_0^2 = 99.62$

$v_0 = \sqrt{99.62}$

$v_0 = 9.98m/s$

Therefore the speed of the mountain lion just as it leaves the ground is 9.98m/s at an angle of 50.2°

5 0

3 years ago