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3 years ago

PLEASE HELPPPP ❗️❗️❗️

Physics

1 answer:

mojhsa [17]3 years ago

8 0

Answer:

Explanation:

There are 3 main forces at work here, gravity, normal and friction. The gravity pulls the car straight down and is what keeps the car on the ground. Normal force is straight up from the points where the car is touching, so since the wheels are the only parts of the car touching the street, this is where all the normal force is. Friction force opposes any and all motion, the car wants to slide down the hill and would slide down the hill if there was no friction, so the friction force is in the opposite direction of the cars intended motion.

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IF YOU ANSWER THESE 2 QUESTIONS! I WILL GIVE YOU BRAINEST!!! 15 POINTS!!!

lisabon 2012 [21]

Answer:

1. a

2. c [and maybe before]

Explanation:

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3 years ago

A viscous fluid is flowing through two horizontal pipes. The pressure difference P1 - P2 between the ends of each pipe is the sa

777dan777 [17]

Answer:half of shorter Pipe

Explanation:

Fluid is Flowing through two horizontal pipes with pressure difference

$P_1-P_2=\Delta P$

Both pipes have same radius

Length of one Pipe is twice of other

Let Longer Pipe be denote by 1 and smaller by 2

From Hagen Poiseuille equation

$\Delta P=\frac{128\mu L\cdot Q}{\pi D^4}$

Where $\mu =$ viscosity of medium

L=length of Pipe

Q=discharge

D=diameter

For longer Pipe

$\Delta P=\frac{128\mu 2L\cdot Q_1}{\pi \cdot D^4}$ ----1

For smaller Pipe

$\Delta P=\frac{128\mu L\cdot Q_2}{\pi \cdot D^4}$ ------2

From 1 & 2 we get

$2L\cdot Q_1=L\cdot Q_2$

$Q_2=2Q_1$

volume flow rate of longer pipe is half of smaller pipe

6 0

3 years ago

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial

V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF= $2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current, $I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

$t=0.5ns=0.5\times 10^{-9} s$

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$

5 0

3 years ago

Calculate the speed of a 500.0 kilogram car that possesses 400,000 joules of kinetic energy.

lozanna [386]

Answer: 40m/s

Explanation:

Given that,

speed of car (v) = ?

Mass of car (m) = 500.0 kilogram

Kinetic energy (KE) = 400,000 joules

Recall that kinetic energy is possessed by a moving object, and it depends on the mass and speed by which the object moves

i.e Kinetic energy = 1/2mv²

400,000J = 1/2 x 500.0kg x v²

400,000J = 250.0kg x v²

v² = 400,000J / 250.0kg

v² = 1600

Find the square root of v²

v = √1600

v = 40m/s

Thus, the speed of the car is 40m/s

6 0

3 years ago

A 42 kg surfer jumps off the back of a 22 kg surfboard that is moving forward with a velocity of 5.2 m/s. After the surfer leave

Fittoniya [83]

Answer:

The surfer leave the surferboard with a velocity of 4.72[m/s]

Explanation:

This problem is related to the Conservation of Momentum, and it can be calculated using the following equation.

$m_{1}*v_{1}+m_{2}*v_{2}=m_{1}*v_{1}^{'}+ m_{2}*v_{2}^{'} \\$

Where:

m1 = mass of the surfer = 42[kg]

v1 = velocity of the surfer before jumping = 5.2 [m/s]

m2 = mass of the surfboard = 22 [kg]

v2 = velocity of the surfboard before jumping = 5.2 [m/s]

Now after jumping

m1 = mass of the surfer = 42[kg]

v1' = velocity of the surfer after jumping = x

m2 = mass of the surfboard = 22 [kg]

v2' = velocity of the surfboard after jumping = 6.1 [m/s]

Now replacing in the equation.

$(42*5.2)+(22*5.2)= (42*X)+(22*6.1)\\42*X = 198.6\\x = 4.72[m/s]$

8 0

3 years ago