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3 years ago

Write the types of lever with 5.5 example

Physics

1 answer:

tekilochka [14]3 years ago

7 0

First class lever -- Teeter-totter, Oars on a boat, Catapult, Shoehorn, Scissors, Pair of pliers

Second class lever -- Wheelbarrow, Crowbar, Nut cracker, Doors or gates, bottle openers

Third class lever -- Tweezers, Stapler, Mousetrap, Broom, Hockey stick

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An electrical heater is used to supply 245.0 J of energy to a 15.0 g sample of aluminum, originally at 310.0 K. Determine the fi

Softa [21]

Answer:

$T_f=328.1 K$

Explanation:

To determine the final temperature of the sample, we use the specific heat formular as follows:

$Q=mCp(T_f-T_0)\\T_f=\frac{Q}{mCp} +T_{0}\\mol Al= \frac{15g}{26.98gmol^{-1}}=0.555 mol\\T_f=\frac{245.0J}{0.555mol\times24.35Jmol^{-1}K^{-1}} +310.0K\\T_f=328.1 K$

Finally, the temperature of the aluminium sample has raised 18 K.

8 0

4 years ago

Consider that a clay model of a tiger has a mass of 0.195 kg and glides on ice at a speed of 0.75 m/s. It hits another clay mode

juin [17]

Answer:

The final velocity after the collision is 0.27 m/s.

Explanation:

Given that,

Mass of tiger, m = 0.195 kg

Initial speed of tiger model, v = 0.75 m/s

Mass of another clay model, m' = 0.335 kg

Initially, second model is at rest, v' = 0

We need to find the final velocity after the collision. It is a case of inelastic collision. Using the conservation of linear momentum as :

$mv+m'v'=(m+m')V\\\\V=\dfrac{mv+m'v'}{(m+m')}\\\\V=\dfrac{0.195\times 0.75}{(0.195 +0.335 )}\\\\V=0.27\ m/s$

So, the final velocity after the collision is 0.27 m/s.

6 0

4 years ago

The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge

Eva8 [605]

Answer:

a) $v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) $v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) $v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) $v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) $v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

$KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)$ (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

$\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}$

We can write the mass of a proton in MeV/c².

$m_{p}=938.28 MeV/c^{2}$

Now we can calculate the speed in each stage.

a) Cockcroft-Walton (750 keV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{0.75 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.04$

$v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s$

b) Linac (400 MeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{400 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.71$

$v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s$

c) Booster (8 GeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{8000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.994$

$v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s$

d) Main ring or injector (150 Gev)

$\beta = \sqrt{1-\frac{1}{\left (\frac{150000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.999$

$v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s$

e) Tevatron (1 TeV)

$\beta = \sqrt{1-\frac{1}{\left (\frac{1000000 MeV}{938.28 MeV}+1 \right )^{2}}}$

$\beta = 0.9999$

$v = c \cdot 0.9999 = 2.999\cdot 10^{8} m/s$

Have a nice day!

4 0

4 years ago

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling

Natalka [10]

Answer:

H = 5 m

Explanation:

As the person leaves the slide horizontally so the time taken by the person to hit the water is given as

$t = 0.672 s$

so we can find the vertical velocity by which person will hit the water using kinematics

$v_y = u_y + at$

$v_y = 0 + (9.81)(0.672)$

$v_y = 6.6 m/s$

now the speed of the person at the end of the slide is given as

$v = \frac{L}{t}$

$v = \frac{5}{0.672}$

$v_x = 7.44 m/s$

now by energy conservation we can find the initial height

$mgH = \frac{1}{2}m(v_x^2 + v_y^2)$

$H = \frac{1}{2g}(v_x^2 + v_y^2)$

$H = \frac{1}{2(9.81)}(7.44^2 + 6.6^2)$

$H = 5 m$

6 0

3 years ago

Read 2 more answers

Consider a uniformly wound solenoid having N=210 turns, length l=0.18 m, and cross-sectional area A = 4.00 cm2. Assume l is much

strojnjashka [21]

Answer:

Explanation:

Number of turns

N = 210turns

Length of solenoid

l = 0.18m

Cross sectional area

A = 4cm² = 4 × 10^-4m²

A. Inductance L?

Inductance can be determined using

L = N²μA/l

Where

μ is a constant of permeability of the core

μ = 4π × 10^-7 Tm/A

A is cross sectional area

l is length of coil

L is inductance

Therefore

L = N²μA / l

L=210² × 4π × 10^-7 × 4 × 10^-4 / 0.18

L = 1.23 × 10^-4 H

L = 0.123 mH

B. Self induce EMF ε?

EMF is given as

ε = -Ldi/dt

Since rate of decrease of current is 120 A/s

Then, di/dt = —120A/s, since the current is decreasing

Then,

ε = -Ldi/dt

ε = - 1.23 × 10^-4 × -120

ε = 0.01478 V

ε ≈ 0.015 V

4 0

3 years ago

Write the types of lever with 5.5 example​

Write the types of lever with 5.5 example