Classify the following matter

a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration

slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0

2 years ago

A circle graph shows that one-fourth of the students in a particular school ride the bus every day. What is this equivalent meas

Aleonysh [2.5K]

Answer:

it's 1.57... radians and 90°

3 0

2 years ago

8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer

Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

$M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \ by \ the \ effort}{disatnce \ moved \ by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m$

Therefore, the distance that the object is raised above its initial position is 5.625 m.

3 0

2 years ago

Can someone solve this problem and explain to me how you got it

evablogger [386]

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

4 0

3 years ago

Needing the answers ... i’ll make you brainliest ! Thank you✅

otez555 [7]

1) 4°C : It has the highest density as shown on the graph.

2) Water expands when it freezes, making it less dense than just water.

3) The ice would sink to the bottom, then the rest of the water would freeze as well, the entire lake/river/whatever will freeze eliminating the organisms that live there.

7 0

2 years ago