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3 years ago

As you approach your vehicle and perform checks it is not necessary to always

1 answer:

4 0

1.) Have your keys in hand before approaching or entering your car.

2. Be alert to other pedestrians and drivers.

3.) Search for signs of movement between, beneath and around objects to both sides of your vehicle

4.) check the spare tire for proper inflation

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A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

iogann1982 [59]

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

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

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$

4 0

3 years ago

What is the net force on the block shown above? 50 N Up 50 N Down 0 N 5 N Down

WINSTONCH [101]

Answer:

the net force will be 50 N Up

Explanation:

because left and right forces balance each other and by calculating up and down forces , force exerting upward is 50 more than down .

hence, net force is 50N Up

3 0

2 years ago

Read 2 more answers

A light flashes at position x=0m. One microsecond later, a light flashes at position x=1000m. In a second reference frame, movin

padilas [110]

Answer:

To the right relative to the original frame.

Explanation:

In first reference frame S,

Spatial interval of the event, $\rm \Delta x=1000\ m-0\ m=1000\ m.$

Temporal interval of the event, $\rm \Delta t = 1\ \mu s=10^{-6}\ s.$

In the second reference frame S', the two flashes are simultaneous, which means that the temporal interval of the event in this frame is $\rm \Delta t'=0\ s.$

The speed of the frame S' with respect to frame S = v.

According to the Lorentz transformation,

$\rm \Delta t'=\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right ).\\\\Since,\ \Delta t'=0,\\\therefore \dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\left( \Delta t-\dfrac{v\Delta x}{c^2}\right )=0\\\Rightarrow \Delta t-\dfrac{v\Delta x}{c^2}=0\\\dfrac{v\Delta x}{c^2}=\Delta t\\v=\dfrac{c^2\Delta t}{\Delta x }.\\\\Also, \ \Delta t,\ \Delta x>0\ \Rightarrow v>0.$

And positive v means the velocity of the second frame S' is along the positive x-axis direction, i.e., to the right direction relative to the original frame S.

7 0

2 years ago

The decrease of PE for a freely falling object equals its gain in KE, in accord with the conservation of energy. By simple algeb

Delvig [45]

Answer:

$v = \sqrt{20h}$

Explanation:

The potential energy (PE) we are looking here is gravitational potential energy (GPE).

GPE= mgh,

where m is the mass of an object,

g is the gravitational field strength

h is the height of the object

KE= ½mv²,

where m is the mass and v is the velocity

loss in GPE= gain in KE

mgh= ½mv²

gh= ½v² (divide by m throughout)

Assuming that the object is on earth, then g= 10N/Kg

½v²= 10h (substitute g=10)

v²= 20h (×2 on both sides)

v= √20h (square root both sides)

4 0

2 years ago

a piece of metal and a piece of brick are both sitting in the sun. If both are the same mass, what property of matter will make

Ghella [55]

Answer:

The answer is conduction.

5 0

2 years ago