Two tiny spheres with charges +20×10^-8 C and -5.0×10^-8 C with 10cm apart. Point P in the middle of it.

How much work is accplished when a force of 250 n pushes a box accross the floor for a distance of 50 meters

7nadin3 [17]

You just multiply these two numbers, it's 1250J

4 0

3 years ago

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0

2 years ago

Lúc 7g bạn an đi từ nhà đến trường với tóc độ trung bình là 20km/h . Bạn đến trường lúc 7g20. Tính khoảng cách từ nhà tới trường

Len [333]

Answer:

Distance = 6.667 kilometres

Explanation:

Given the following data;

Speed = 20 km/h

Departure time = 7:00

Arrival time = 7:20

Time taken = 20 minutes

To calculate the distance travelled from home to school;

First of all, we would have to convert the value of time in minutes to hours.

Conversion:

60 minutes = 1 hour

20 minutes = X hours

Cross-multiplying, we have;

X = 20/60 = 1/3 hours

Mathematically, the distance travelled by an object is calculated by using the formula;

Distance = speed * time

Distance = 20 * 1/3

Distance = 20/3 =

Distance = 6.667 kilometres

7 0

3 years ago

An object starts at the 100 m mark and ends up at the -100 m mark, 50

horsena [70]

It traveled 200 m in 50 seconds. 200/50 can be simplified to 4 m/s!
The velocity is -4 m/s (negative because it travelled from 100 to -100 or backwards)

8 0

2 years ago

A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)