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3 years ago

14

the parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800.m/s. what is the velocity of

the vehicle?

1 answer:

Softa [21]3 years ago

4 0

Answer:

6.5 m/s

Explanation:

Momentum = mass × velocity

p = mv

7800 kg m/s = (1200 kg) v

v = 6.5 m/s

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If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?

Lunna [17]

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,

$W=T.S$

⇒ $W=270*160*cos\pi \\W=-43200 J$

Hence work done by the rope is - 43200 J.

Learn more about force problems on

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8 0

2 years ago

Read 2 more answers

Flasher units are being discussed. Technician A says that only a DOT-approved flasher unit should be used for turn signals. Tech

zmey [24]

Answer: C

Both Technicians A and B

Explanation:

Only a DOT-approved flasher unit should be used for turn signals. And a parallel (variable-load) flasher will function for turn signal usage, although it will not warn the driver if a bulb burns out.

3 0

3 years ago

Un niño amarra una soga a una piedra, y la hace girar como en la gráfica. la piedra realiza un M.C.U, girando con una rapidez de

inessss [21]

Answer:

$\Delta \theta = 56\,rad$

$\Delta \theta \approx 3208.564^{\circ}$

Explanation:

El ángulo barrido en el intervalo de tiempo dado es (The covered angle in the given time interval is):

$\Delta \theta = \omega \cdot \Delta t$

$\Delta \theta = \left(14\,\frac{rad}{s} \rjght)\cdot (4\,s)$

$\Delta \theta = 56\,rad$

$\Delta \theta \approx 3208.564^{\circ}$

4 0

3 years ago

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

3 years ago

a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature

igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber $V=1 m^3$

Initial Temperature $T_1=300 K$

Final Temperature $T_2=400 K$

Heat Supplied $Q=10 J$

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

$\Delta U=Q-W$

as the volume is fixed therefore work

$W=\int PdV=0$

thus $\Delta U=mc_v\Delta T=Q$

$c_v$ for mono-atomic gas is $12.471 J/K-mol$

$n\times 12.471\times (400-300)=10$

$n=0.008018 mol$

and 1 mole contains $6.022\times 10^{23} molecules$

thus No of molecules $=0.008018\times 6.022\times 10^{23}$

No of molecules $=4.82\times 10^{21} molecules$

3 0

3 years ago