All categories

3 years ago

URGENT!! Question in attachment. thanks

Physics

2 answers:

Roman55 [17]3 years ago

6 0

Answer:9

Explanation:beacause

jarptica [38.1K]3 years ago

6 0

Answer:

a) describe, step by step, what happens

You might be interested in

Yurem is pulling a wagon across the playground with a force of 10 N. He asks Elianna to help. She agrees and pushes the back of

salantis [7]

Answer:

22 N applied force

Explanation: Since they are both pushing the wagon in the same direction the force adds up.

7 0

3 years ago

Read 2 more answers

The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to

xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x Formula (1)

Where;

F is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21 inches

Problem development

We replace data in formula 1 to calculate K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

8 0

3 years ago

Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r

ella [17]

Answer:

$\frac{dR(t)}{dt}=0.06\Omega$

Explanation:

Since $R(t)=\frac{V}{I(t)}$ , we calculate the resistance rate by deriving this formula with respect to time:

$\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})$

Deriving what is left (remember that $(\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)$ ):

$\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}$

So we have:

$\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}$

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

$\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega$

8 0

3 years ago

A coin is placed on a large disk which rotates uniformly at a rate of 1 rot/s. The coefficient of friction between the coin and

s2008m [1.1K]

Answer:

r = 0.02 m

Explanation:

from the question we have :

speed = 1 rps = 1x 60 = 60 rpm

coefficient of friction (μ) = 0.1

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

maximum distance without falling off (r) = ?

to get how far from the center of the disk the coin can be placed without having to slip off we equate the formula for the centrifugal force with the frictional force on the turntable force

mv^2 / r = m x g x μ

v^2 / r = g x μ .......equation 1

where

velocity (v) = angular speed (rads/seconds) x radius

angular speed (rads/seconds) = (\frac{2π}{60} ) x rpm

angular speed (rads/seconds) = (\frac{2 x π}{60} ) x 60 = 6.28 rads/ seconds

now

velocity = 6.28 x r = 6.28 r

now substituting the value of velocity into equation 1

v^2 / r = g x μ

(6.28r)^2 / r = 9.8 x 0.1

39.5 x r = 0.98

r = 0.02 m

6 0

3 years ago

A 50 kg mass is placed 2 meters from the fulcrum. To balance the lever, a second object is placed 4 meters

julia-pushkina [17]

Answer:

mass =25 kg

using clockwise moment = anticlockwise moment

8 0

3 years ago

URGENT!! Question in attachment. thanks ​

URGENT!! Question in attachment. thanks