Ask question

Ask question

All categories

3 years ago

8

A 0.60-kg object is suspended from the ceiling at the end of a 2.0-m string. When pulled to the side and released, it has a spee

d of 4.0 m/s at the lowest point of its path. What maximum angle does the string make with the vertical as the object swings up?

1 answer:

Brrunno [24]3 years ago

3 0

Answer:

ø = 53.7º

Explanation:

The measurement of kinetic energy in an object is calculated based on the object's mass and velocity.

KE=1/2mv²

where m is mass

v is velocity

given data

m=0.60 kg

v=4.0 m/s

So

KE = 1/2 . 0.6 . 4^2

KE= 4.8 J

All the KE is converted into GPE ("Gravitational potential energy (GPE) - energy stored in an object when moving the object to a height")

4.8 = 0.6 . 9.8 . ∆h

∆h = 0.816 m

This is the height that the object rises.

cos ø = (2 – 0.816 ) / 2 = 0.592

ø = 53.7º

You might be interested in

Which organisms break down dead matter and waste into nonliving elements?

Alekssandra [29.7K]

Decomposers is the correct answer. ( I got your back bro)

4 0

3 years ago

Read 2 more answers

Suppose humans have weights which are normally distributed with mean 170 lbs and SD 50 lbs. If 400 humans are selected at random

MrRa [10]

Answer:

$P(x< 175)= 0.9772$

Explanation:

given,

mean weight of human = μ = 170 lbs

standard deviation = SD = 50 lbs

N = 400 humans

By using central limit theorem,

P (x< 175)

$P(x< 175)= P(z$

$P(x< 175)= P(z$

$P (x< 175)= P(z$

$P (x< 175)= P(z$

using z-table

$P (x< 175)= 0.9772$

hence, the probability that total weight is less tan 175 lbs is equal to

$P(x< 175)= 0.9772$

7 0

3 years ago

How to calculate percent error

olga_2 [115]

Answer:

Steps to Calculate the Percent Error

Subtract the accepted value from the experimental value.

Take the absolute value of step 1.

Divide that answer by the accepted value.

Multiply that answer by 100 and add the % symbol to express the answer as a percentage.

Explanation:

3 0

3 years ago

A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an

elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)= $(2.4-x^2)\hat{i}N$

Initial position of the block=x=0

Initial velocity of block= $v_i=0$

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy= $K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0$

Work energy theorem:

$K_f-K_i=W$

Where $K_f=$ Final kinetic energy

$K_i$ =Initial kinetic energy

$W=Total work done$

Substitute the values then we get

$K_f-0=\int_{0}^{2}F(x)dx$

Because work done= $Force\times displacement$

$K_f=\int_{0}^{2}(2.4-x^2)dx$

$K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}$

$K_f=2.4(2)-\frac{8}{3}=2.13 J$

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy = $K=2.4x-\frac{x^3}{3}$

When the kinetic energy is maximum then $\frac{dK}{dx}=0$

$\frac{d(2.4x-\frac{x^3}{3})}{dx}=0$

$2.4-x^2=0$

$x^2=2.4$

$x=\pm\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2x$

Substitute x= $\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2\sqrt{2.4}$

Substitute x= $-\sqrt{2.4}$

$\frac{d^2K}{dx^2}=2\sqrt{2.4}>0$

Hence, the kinetic energy is maximum at x= $\sqrt{2.4}$

Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by

$K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx$

$k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}$

$K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J$

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0

4 years ago

¿Qué valor de voltaje se requiere para que pasen 35 mA a través de una resistencia de 15 kΩ? Si se aplica ese mismo voltaje a un

ehidna [41]

Answer:

perdón yo no hablo inglés

8 0

3 years ago