PHYSICS COLLEGE

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answers

Answer 1

Answer:

The minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Given data:

The weight of ladder is, W = 100 N.

The length of ladder is, L = 8.0 m.

The coefficient of static friction between ladder and floor is, $\mu =0.40$ .

Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,

$N = mg$

And force along the horizontal direction is,

$F= \mu * N\n\nF = \mu * mg$

Now, taking the torque along the either end of ladder as,

$-mgcos \theta * (L)/(2)+Fsin \theta * L =0\n\nmgcos \theta * (L)/(2) = Fsin \theta * L$

Solving as,

$mgcos \theta * (L)/(2) = (\mu mg) * sin \theta * L\n\ncos \theta * (1)/(2) = (\mu) * sin \theta\n\ntan \theta = (1)/(2 * 0.40 )\n\n\theta = tan^(-1)(1/0.80)\n\n\theta = 51.34^(\circ)$

Thus, we can conclude that the minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

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Answer 2

Answer:

Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

$F_(y)=n-mg$

$N-mg=0$

$N=mg$

We need to calculate the normal force

Using Newtons law in horizontal direction

$F_(s)=f_(s)-P$

$f_(s)-P=0$

$f_(s)=P$

$P=\mu mg$

We need to calculate the minimum angle

Using torque about the point A then

$-mg\cos\theta* AB+P\sin\theta* AC=0$

Put the value into the formula

$mg\cos\theta*((L)/(2))=\mu mg\sin\theta* L$

$\cos\theta*(1)/(2)=\mu\sin\theta$

$(1)/(2)=\mu*\tan\theta$

$\theta=\tan^(-1)((1)/(2*0.40))$

$\theta=51.34^(\circ)$

Hence, The minimum angle is 51.34°

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List Five examples from daily life in which you see periodic motion caused by a pendulum(Marking Brainliest)

Answers

Answer:

by a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and a water wave.

Explanation:

You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

Answers

Answer:

$d'=√(v^2+w^2)$

Explanation:

Rate of Change

When an object moves at constant speed v, the distance traveled at time t is

$x=v.t$

We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

$y=v.t$

and the horizontal distance is

$x=w.t$

The distance between both friends is computed as the hypotenuse of the triangle

$d^2=x^2+y^2$

We need to find d', the rate of change of the distance between both friends.

Plugging in the above relations

$d^2=(v.t)^2+(w.t)^2$

$d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2$

Solving for d

$d=√((v^2+w^2)t^2)$

$d=√((v^2+w^2)).t$

Differentiating with respect to t

$\boxed{d'=√(v^2+w^2)}$

Final answer:

The problem is solved using Pythagoras' Theorem, representing the two travel paths forming a right triangle. The rate at which the distance increases between two points moving perpendicularly can be found by differentiating the resulting equation, which yields the expression sqrt[(v^2)+(u^2)].

Explanation:

The question is about the rate at which the distance between you and your friend is increasing at time t. It's a typical problem in kinematics. Because the roads are perpendicular to each other, we can solve the problem using Pythagoras' Theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's denote the distance you've traveled as D1 = v*t (because distance = speed * time) and the distance your friend has travelled D2 = u*t. The distance between you can be computed using Pythagoras' Theorem as D = sqrt(D1^2 + D2^2). Hence, D = sqrt[(v*t)^2 + (u*t)^2]. Differentiating D with respect to t using the chain rule will give us the rate at which the distance between you is increasing, which is sqrt[(v^2)+(u^2)].

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An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?

Answers

Final answer:

The final speed at the bottom of the incline can be calculated using the conservation of energy principle. There is no work done against friction as the object is moving on a frictionless surface. The speed does not change when the spring pushes it back towards the base of the incline due to lack of friction and it moves to a certain height given the angle of the incline and the initial speed.

Explanation:

"Speed at the bottom of the incline:" This can be calculated using conservation of energy. The potential energy at the top (m*g*h) will convert into kinetic energy at the bottom (1/2*m*v^2). Here, m is the mass, g is acceleration due to gravity, h is the height, and v is the velocity. Using this, we can solve for v.
Work of friction on the incline: As per the question, the surface is frictionless. Therefore, the work done by friction is automatically 0 as there is no force of friction.
Speed of the object when it reaches the base of the incline again: As the surface is frictionless, the object reaches the incline with the same speed with which it left as there are no opposing forces to reduce its momentum.
Vertical distance it moves back up the incline: This can be calculated using the principles of conservation of energy and kinematic equations, taking into account the angle of the incline and the velocity of the object.

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The fundamental force that is responsible for beta decay is the: A) weak force. B) strong force. C) dark energy force. D) gravitational force E) electromagnetic force.

Answers

Answer:

A. Weak forces

Explanation:

The fundamental forces responsible for beta decay is the weak force. Weak forces are among the four fundamental forces of universe the electromagnetic, gravitational and strong forces. The weak forces are responsible for the decaying. The fundamental work of weak forces is covert neutron into proton and electron into neutrino. weak forces operate at very low distances as low as fermi meter.

Answer:

The answer is

dark energy force.

hope this helps u stay safe

Explanation:

Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

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When some stars use up their fuel, they undergo a catastrophic explosion called a supernova. This explosion blows much or all of a star's mass outward, in the form of a rapidly expanding spherical shell. As a simple model of the supernova process, assume that the star is a solid sphere of radius R that is initially rotating at 2.4 revolutions per day. After the star explodes, find the angular velocity, in revolutions per day, of the expanding supernova shell when its radius is 4.3R. Assume that all of the star's original mass is contained in the shell.

Answers

Answer:

0.0768 revolutions per day

Explanation:

R = Radius

$\omega$ = Angular velocity

As the mass is conserved the angular momentum is conserved

$I_1\omega_1=I_2\omega_2\n\Rightarrow (I_1)/(I_2)=(\omega_2)/(\omega_1)$

Moment of intertia for solid sphere

$I_1=(2)/(5)MR^2\n\Rightarrow I_1=0.4MR^2$

Moment of intertia for hollow sphere

$I_2=(2)/(3)M(4.3R)^2\n\Rightarrow I_2=12.327MR^2$

Dividing the moment of inertia

$(I_1)/(I_1)=(0.4MR^2)/(12.327MR^2)\n\Rightarrow (I_1)/(I_2)=0.032$

From the first equation

$\omega_2=\omega_1(I_1)/(I_2)\n\Rightarrow \omega_2=2.4* 0.032\n\Rightarrow \omega_2=0.0768\ rev\day$

The angular velocity, in revolutions per day, of the expanding supernova shell is 0.0768 revolutions per day

Final answer:

To find the angular velocity of the expanding supernova shell, we can use the principle of conservation of angular momentum. The initial angular momentum of the star can be equated to the final angular momentum of the shell. By substituting the given information and solving the equation, we can find the angular velocity of the shell.

Explanation:

When a star undergoes a supernova explosion, a large amount of its mass is blown outward in the form of a rapidly expanding shell. To find the angular velocity of the expanding shell, we can use the principle of conservation of angular momentum. Assuming that all of the star's original mass is contained in the shell, we can equate the initial angular momentum of the star to the final angular momentum of the shell.

The angular velocity of the star before the explosion can be calculated using the equation:

angular velocity before = 2 * pi * initial frequency

where the initial frequency is given as 2.4 revolutions per day.

After the explosion, the radius of the expanding shell is given as 4.3 times the radius of the star. Using the principle of conservation of angular momentum, we can set the initial angular momentum of the star equal to the final angular momentum of the shell:

initial angular momentum of the star = final angular momentum of the shell

Since the final angular momentum of the shell is given by:

final angular momentum of the shell = moment of inertia of the shell * angular velocity of the shell

where the moment of inertia of the shell is given by:

moment of inertia of the shell = 2/5 * mass of the shell * (radius of the shell)^2

and the angular velocity of the shell is what we are trying to find, we can rewrite the equation as:

initial angular momentum of the star = 2/5 * mass of the shell * (radius of the shell)^2 * angular velocity of the shell

By substituting the expression for the initial angular momentum of the star and solving for the angular velocity of the shell, we can find the answer.

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