Can a force directed north balance a force directed east

On a straight road (taken to be in the +x direction) you drive for an hour at 50 km per hour, then quickly speed up to 90 km per

dalvyx [7]

Answer:

a) 230 Km b) 76.7 km/h c) Please see below

Explanation:

a) If we can neglect the time while the driver accelerated, the movement can be divided in two parts, each of them at a constant speed:

$x = x1 + x2 \\\\x1 = 50 km/h* 1hr = 50 km, \\x2 = 90 Km/h*2hr = 180 km\\$

⇒ $x = 50 km + 180 km = 230 km$

b) The average x component of velocity, can be calculated applying the definition of average velocity, as follows:

$vavg,x = \frac{xf-xo}{t-to}$

If we choose t₀ = 0 and x₀ = 0, replacing xf and t by the values we have already found, we can find vavg,x as follows:

$vavg,x =\frac{230 km}{3 hr} =76.7 km/h$

c) The found value of avg,x is not the same as the arithmetic average of the initial and final values of vx (70 Km/h) due to the time traveled at both velocities was not the same.

If the driver had droven half of the time (1.5 h) at 50 km/h and the other half at 90 km/h, total displacement would have been as follows:

$x = 50 km/h*1.5 h + 90 km/h*1.5 hr = 210 km$

Applying the definition of average velocity once more:

$vavg,x =\frac{210 km}{3 hr} =70 km/h$

which is the same as the arithmetic average of the initial and final values of vₓ.

7 0

3 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

The potential (relative to infinity) at the midpoint of a square is 3.0 V when a point charge of +Q is located at one of the cor

Sveta_85 [38]

Answer:

d) 12 V

Explanation:

Due to the symmetry of the problem, the potential (relative to infinity) at the midpoint of the square, is the same for all charges, provided they be of the same magnitude and sign, and be located at one of the corners of the square.

We can apply the superposition principle (as the potential is linear with the charge) and calculating the total potential due to the 4 charges, just adding the potential due to any of them:

V = V(Q₁) + V(Q₂) +V(Q₃) + V(Q₄) = 4* 3.0 V = 12. 0 V

4 0

3 years ago

A train's velocity is slowly decreasing as it approaches a station. About 1 km before it reaches the station, several stowaways

Mars2501 [29]

Answer:

Increases

Explanation:

From Newton's second law:

F = ma

If F stays the same and m decreases, then a increases.

Therefore, the magnitude of the train's acceleration increases (the acceleration becomes more negative).

6 0

3 years ago

As a volunteer, you have to comply with the patient privacy act

shutvik [7]

Correct you can't share people's information without their permission.

7 0

3 years ago