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3 years ago

Which of the following is true of work?

Physics

1 answer:

FinnZ [79.3K]3 years ago

3 0

Answer:

Explanation:

C: Force and Weight are related. This is not the answer.

B: Not true Much as we might feel this way after working, it is not the answer either. Not in Physics at least.

D: is treated the same way as B.

A: is the actual answer in physics. How you feel has nothing to do with it. Work is some Force moving a distance. Both are needed.

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Which of the following are not clues that a chemical reaction has taken place?

Sladkaya [172]

Im thinking a and d but if its just one answer im thinking mostly a

6 0

3 years ago

Read 2 more answers

A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

VMariaS [17]

Answer:

Hence, work done= 287.54 J

Explanation:

Given data:

angle of ramp with the ground θ =20°

force applied = 76 N

work done on the crate to slide down 4 m down the ramp

W= F×d cosθ ( only the cos component of the force will slide the crate down)

W= 76×4×cos20= 287.54 J

4 0

4 years ago

On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze

sp2606 [1]

Answer:

0.78333 m/s in the opposite direction

1.566 m/s in the same direction

Explanation:

$m_1$ = Mass of penny = 0.0025 kg

$m_2$ = Mass of nickel = 0.005 kg

$u_1$ = Initial Velocity of penny = 2.35 m/s

$u_2$ = Initial Velocity of nickel = 0 m/s

$v_1$ = Final Velocity of penny

$v_2$ = Final Velocity of nickel

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.0025-0.005}{0.0025+0.005}\times 2.35+\frac{2\times 0.5}{0.4005+0.5}\times 0\\\Rightarrow v_1=-0.78333\ m/s$

The final velocity of the penny is 0.78333 m/s in the opposite direction

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.0025}{0.0025+0.005}\times 2.35+\frac{0.005-0.0025}{0.005+0.0025}\times 0\\\Rightarrow v_2=1.566\ m/s$

The final velocity of the nickel is 1.566 m/s in the same direction

6 0

3 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

3 years ago

Check all that apply. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic

dedylja [7]

Answer:

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

Explanation:

The magnitude of the magnetic force exerted on a current-carrying wire due to a magnetic field is given by

$F=ILB sin \theta$ (1)

where I is the current, L the length of the wire, B the strength of the magnetic field, $\theta$ the angle between the direction of the field and the direction of the current.

Also, B, I and F in the formula are all perpendicular to each other. (2)

According to eq.(1), we see that the statement:

<em>"The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.</em>"

is correct, because when the current is perpendicular to the magnetic field, $\theta=90^{\circ}, sin \theta = 1$ and the force is maximum.

Moreover, according to (2), we also see that the statements

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field. "</em>

<em>"The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current. "</em>

because F (the force) is perpendicular to both the magnetic field and the current.

5 0

3 years ago