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laiz [17]
3 years ago
9

I need an answer ASAP this is Mass Times acceleration

Physics
1 answer:
GenaCL600 [577]3 years ago
8 0

Answer:

force = mass \times acceleration \\ 1200 = 400 \times a \\ a =  \frac{1200}{400}  \\ a = 3 \:  {ms}^{ - 2}

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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
Velocity vs time graph
allochka39001 [22]

Answer:

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5 0
3 years ago
Help me plsss! no links
Stolb23 [73]

Answer:

a is the answer to the question

3 0
3 years ago
A 0.40-kg mass, attached to the end of a 0.75-m string, is whirled around in a circular horizontal path. If the maximum tension
hichkok12 [17]

To solve this problem we will apply the concepts given from the circular movement of the bodies for which we have that the centripetal Force is defined as a product between the mass and the velocity squared at the rate of rotation, mathematically this is

F_c = \frac{mv^2}{r}

Where,

m = Mass

v = Velocity

r = Radius

Our values are given as

m = 0.4kg\\r = 0.75m\\F_c = 450N

Rearranging to find the velocity we have that,

F_c = \frac{mv^2}{r}

v = \sqrt{\frac{F_c * r}{m}}

v = \sqrt{\frac{450 * 0.75}{0.4}}

v = 29.0474m/s

Therefore the  maximum speed can the mass have if the string is not to break is 29m/s

3 0
3 years ago
Which type of earthquake waves are responsible for most of the damage caused by an earthquake? Why?
den301095 [7]

Answer:

Surface waves travel along the surface. There are two types of body waves: P-waves travel fastest and through solids, liquids, and gases; S-waves only travel through solids. Surface waves are the slowest, but they do the most damage in an earthquake.

5 0
3 years ago
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