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Nimfa-mama [501]

3 years ago

12

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.480 with the floor. If

the train is initially moving at a speed of 61.0 km/h, in how short a distance can the train be stopped at constant acceleration without causing the crates to slide over the floor?

1 answer:

elena-14-01-66 [18.8K]3 years ago

7 0

Answer:30.50 m

Explanation:

Given

coefficient of friction between railroad and crates $\mu =0.48$

Train initial velocity (u) $=61 kmph \approx 16.94 m/s$

To get the shortest distance brakes applied should be order of friction force between crates and railroad floor

$a=\mu g=0.48\times 9.8=4.704 m/s^2$

using $v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration or deceleration

s=distance

here v=0 , u=16.94 m/s

$0-(16.94)^2=2\times (-4.704)\times s$

$s=\frac{(16.94)^2}{2\times 4.704}$

s=30.502 m

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A 250. mL sample of gas at 1.00 atm and 20.0°C has the temperature increased to 40.0°C and the volume increased to 500. mL. What

ladessa [460]

Answer:

New pressure is 0.534 atm

Explanation:

Given:

Initial volume of the gas, V₁ = 250 mL

Initial pressure of the gas, P₁ = 1.00 atm

Initial temperature of the gas, T₁ = 20° C = 293 K

Final volume of the gas, V₂ = 500 mL

Final pressure of the gas = P₂

Final temperature of the gas, T₁ = 40° C = 313 K

now,

we know for a gas

PV = nRT

where,

n is the moles

R is the ideal gas constant

also, for a constant gas

we have

(P₁V₁/T₁) = (P₂V₂/T₂)

on substituting the values in the above equation, we get

(1.00 × 250)/293 = (P₂ × 500)/313

or

P₂ = 0.534 atm

Hence, the <u>new pressure is 0.534 atm</u>

5 0

3 years ago

A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil

Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}$

Put all the values,

$h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m$

So, the child is able to go at a height of 4.04 m.

7 0

3 years ago

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

alex41 [277]

Answer:

30643 J

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

t = Time taken = 1 ns

c = Speed of light = $3\times 10^8\ m/s$

$E_0$ = Maximum electric field strength = $1.52\times 10^{11}\ V/m$

A = Area = $1\ mm^2$

Magnitude of magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T$

Intensity is given by

$I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2$

Power, intensity and time have the relation

$E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J$

The energy it delivers is 30643 J

4 0

3 years ago

A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.

Xelga [282]

Answer:

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

Explanation:

As we know that magnetic field due to torroid is given as

$B = \frac{\mu_0 N i}{2\pi b}$

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

$\phi = B.A$

$\phi = \frac{\mu_0 N i}{2\pi b}(a^2)$

now total flux of N number of coils is given as

$\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}$

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

$L = \frac{\phi}{i}$

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

3 0

3 years ago

Compare the gravity between these pairs, each consisting of an Earth-like planet and its star. You are given the mass of the pla

Maru [420]

Answer:

The answer is below

Explanation:

Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:

$F=G\frac{m_1m_2}{r^2} \\\\Where\ F=force,G=gravitational\ constant, m_1\ and\ m_1=mass\ of\ objects,r\ =distance\ between \ the\ two\ objects.$

The masses and distances for this question is in common units, Therefore the result would be in ratios

a) 4 MEarth / 2 MSolar / 3 AU

The force (F) = (4 * 3) / 3² = 4/3

b) 1 MEarth / 1 MSolar / 1 AU

The force (F) = (1 * 1) / 1² = 1

c) 1 MEarth / 2 MSolar / 2 AU

The force (F) = (1 * 2) / 2² = 1/2

6 0

3 years ago