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kaheart [24]
3 years ago
5

Can anyone explain and get the answer pls?

Physics
1 answer:
Veronika [31]3 years ago
6 0

FL₁ =Fl₂

80.x₁ = 30.7

x₁=2.625 m

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Give 10 examples of units you might use or see in any given day
Ierofanga [76]
Cups
teaspoon
tablespoon
liters
milliliters
gallons
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tons
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3 0
4 years ago
In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm, and are 5.60
noname [10]

a) Minimum speed of the proton: 6.05\cdot 10^6 m/s

b) Angle of the velocity: \theta=-5.1^{\circ}

Explanation:

a)

The proton experiences a vertical force due to the electric field, given by:

F=qE

where

q=1.6\cdot 10^{-19}C is the proton charge

E=610,000 N/C is the magnitude of the electric field

The vertical acceleration of the proton is therefore

a=\frac{qE}{m}

where

m=1.67\cdot 10^{-27}kg is its mass

Therefore, the vertical position of the proton at time t is

y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2

where we assumed that the initial vertical velocity is zero (because the proton is fired horizontally) and the initial vertical position, halfway between the two plates, is the origin.

The horizontal motion of the proton instead is uniform, so the horizontal position is given by

x(t)=v_0 t

where v_0 is the initial speed. This equation can be rewritten as

t=\frac{x(t)}{v_0}

And substituting into the eq. for y,

y(t) = \frac{1}{2}\frac{qE}{m} \frac{x^2}{v_0^2}

Solving for the initial speed,

v_0 = \sqrt{\frac{qEx^2}{2my}}

The proton just misses one of the plate when

x = 5.60 cm = 0.056 m (length of the plates)

y = 0.25 cm = 0.0025 m (half the distance between the plates)

Therefore, we find the initial speed:

v=\sqrt{\frac{(1.6\cdot 10^{-19})(610,000)(0.056)^2}{2(1.67\cdot 10^{-27})(0.0025)}}=6.05\cdot 10^6 m/s

b)

In order to find the angle, we just need to analyze the horizontal and vertical component of the final velocity of the proton.

The horizontal velocity is constant so it is:

v_x = v_0 = 6.05\cdot 10^6 m/s

The vertical velocity is given by:

v_y^2 - u_y^2 = 2ay

where:

u_y=0 (initial vertical velocity is zero)

a=\frac{qE}{m} (acceleration)

y = 0.0025 m (vertical displacement)

Solving for v_y,

v_y = \sqrt{2ay}=\sqrt{2\frac{qEy}{m}}=5.4\cdot 10^5 m/s

Therefore, the final angle of the velocity with respect to the horizontal is:

\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.4\cdot 10^5}{6.05\cdot 10^6})=5.1^{\circ}

And since the electric field is downward (the proton just misses the lower field), it means that the angle is below the horizontal:

\theta=-5.1^{\circ}

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0
3 years ago
it takes 840s to walk completely around a circular track, moving at a speed of 1.20m/s? what is the radius of the track?
Tcecarenko [31]

Answer:

160.43 meters

Explanation:

T=(2*pi*r)/v

840=(2*pi*r)/1.2

1008=2*pi*r

504=pi*r

160.43=radius

6 0
3 years ago
What happens to the speed of an object when dropped at free fall
dedylja [7]

If it's not falling through air, water, smoke, or anything else,
and gravity is the only force on it, then its speed increases
at a constant rate ... 9.8 meters per second for every second
it falls.  (That's the number on Earth.  It's different in other places.)
3 0
4 years ago
One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet
Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

               \frac{G M m}{r^{2}}=m \omega^{2} r

The orbit’s period is given by,

               T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}

Where,

T_{e} = Earth’s period

T_{p} = planet’s period

M_{s} = sun’s mass

r_{e} = earth’s radius

Now,

             T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}

As, planet mass is equal to 0.7 times the sun mass, so

            T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}

Taking the ratios of both equation, we get,

             \frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}

            \frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}

            \left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}

           \frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}

Given T_{p}=9.5 \text { days } and T_{e}=365 \text { days }

          \frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}

         r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}

7 0
3 years ago
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