What type of tide occurs when the sun, moon, and earth form a right triangle in space?

A 1.60-kg object is held 1.05 m above a relaxed, massless vertical spring with a force constant of 330 N/m. The object is droppe

pentagon [3]

Answer:

(A) l = 0.39 m

(B) l =0.38 m

(C) l = 0.14 m

Explanation:

Answer:

Explanation:

Answer:

Explanation:

from the question we are given the following values:

mass (m) = 1.6 kg

height (h) = 1.05 m

compression of spring (l) = ?

spring constant (k) = 330 N/m

acceleration due to gravity (g) = 9.8 m/s^{2}

(A) initial potential energy of the object = final potential energy of the spring

potential energy of the object = mg(1.05 + l)

potential energy of the spring = 0.5 x k x l^{2} (k= spring constant)

therefore we now have

mg(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) = 0.5 x 300 x l^{2}

15.68 (1.05 + l) = 150 x l^{2}

16.5 + 15.68l = 150l^{2}

l = 0.39 m

(B) with constant air resistance the equation applied in part A above becomes

initial P.E of the object - air resistance = final P.E of the spring

mg(1.05 + l) - 0.750(1.05 + l) = 0.5 x k x l^{2}

1.6 x 9.8 x (1.05 + l) - 0.750(1.05 + l) = 0.5 x 300 x l^{2}

(16.5 + 15.68l) - (0.788 + 0.75l) = 150l^{2}

16.5 + 15.68l - 0.788 - 0.75l = 150l^{2}

15.71 + 14.93l = 150^{2}

l =0.38 m

(C) where g = 1.63 m/s^{2} and neglecting air resistance

the equation mg(1.05 + l) = 0.5 x k x l^{2} now becomes

1.6 x 1.63 x (1.05 + l) = 0.5 x 300 x l^{2}

2.608 (1.05 +l) = 0.5 x 300 x l^{2}

2.74 + 2.608l = 150 x l^{2}

l = 0.14 m

6 0

2 years ago

What is one way to model gravity and magnetism

Oxana [17]

Answer:

Gravity and magnetism are not the same thing. In fact, they are completely separate forces. Gravity is a force that acts between any two objects with mass. ... Magnetism can either pull the two objects together or push them apart, depending on which way the magnets point.

Explanation:

6 0

3 years ago

Read 2 more answers

At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta

Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

u = initial speed of the automobile = 0 m/s

a = constant acceleration of the automobile = $2.5\ m/s^2$
v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

$\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\$

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0

3 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0

3 years ago

If you have 80g of a radioactive substance whose half life is 2 days, how long will it take for the substance to decay to the po

Ber [7]

Answer:

6 days.

Explanation:

From radioactivity, The expression for half life is given as,

R/R' = 2⁽ᵃ/ᵇ)................... Equation 1

Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.

Given: R = 80 g, R' = 10 g, b = 2 days.

Substitute into equation 1

80/10 = 2⁽ᵃ/²⁾

8 = 2⁽ᵃ/²⁾

2³ = 2⁽ᵃ/²)

Equating the base and solving for a

3 = a/2

a = 2×3

a = 6 days.

5 0

3 years ago