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2 years ago

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You find it takes 220 n of horizontal force to move an unloaded pickup truck along a level road at a speed of 2.4 m/s. you then

load up the pickup and pump up its tires so that its total weight increases by 42% while the coefficient of rolling friction decreases by 19%.

1 answer:

Vinvika [58]2 years ago

3 0

Kinetic friction for the truck is given by:

F = μ*N = μ*m*g

μ coefficient of friction
m mass of the truck
g gravitational constant = 9.81

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You're driving your pickup truck around a curve that has a radius of 17 m.

creativ13 [48]

Answer:

$v = 10.8 m/s^{2}$

Explanation:

The centripetal force experienced by the steel toolbox is provided by the friction between the toolbox and the bed of the truck.

Therefore,

$\frac{mv^{2} }{r} = f$ --------------------------------(1)

Here,

v - speed of truck

m - mass of toolbox

r - radius of curve around which the truck is turning

f - frictional force

$f \leq$ μmg

In order to obtain Vmax, frictional force must also be max

Thus,

f = μmg--------------------------------------------------(2)

Substitute (2) in (1)

$\frac{mv^{2} }{r} =$ μmg

$v^{2} =$ μgr

g - 9.8 m/ $s^{2}$

The standard value of coefficient of static friction between steel and steel is

μ ≈ 0.7

Therefore,

$v = 10.8 m/s^{2}$

7 0

2 years ago

A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles i

Tresset [83]

Answer:

$\theta_1 = 0.400^o$

$\theta_2 =0.378^o$

Explanation:

From the question we are told that

The number of slits per cm is k = $161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m$

The order of the maxima is n = 1

The wavelength are $\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m$

The spacing between the slit is mathematically represented as

$d = \frac{ 0.01}{k}$

=> $d = \frac{ 0.01}{161}$

=> $d = 6.211 *10^{-5} \ m$

Generally the condition for constructive interference is

$n\lambda = d \ sin \theta$

At $\lambda_1$

$\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_1 = 0.400^o$

At $\lambda_2$

$\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ]$

$\theta_2 =0.378^o$

4 0

2 years ago

See the person on the right side of the front car. Six reference points could be used to show that the person is in / is NOT in

Rasek [7]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

1. According to person standing on ground ~

The person is in motion

2. According to The car ~

The person is not in motion

3. According to the Seat ~

The person is not in motion

4. According to another person on ride ~

The person is not in motion

5. According to the track ~

The person is in motion

6. According to the Sun ~

The person is in motion

I hope that's what you were looking for, goodluck for your assignment ~

7 0

2 years ago

A tall tube is evacuated and its stopcock is closed. The open end of the tube is immersed in a container of water (density 10^3

LuckyWell [14K]

Answer:

10.19 m

Explanation:

Water will rise to equalize the pressure inside and outside the tube.

The equation of pressure is given by

$p=\rho gh$

Where,

p = Pressure of air = 10⁵ N/m²

ρ = Density of water = 10³ kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Height the water will rise

$h=\frac{p}{gh}\\\Rightarrow h=\frac{10^5}{9.81\times 10^3}\\\Rightarrow h=10.19\ m$

∴ The water will rise by 10.19 m.

7 0

2 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

2 years ago