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Zarrin [17]
3 years ago
11

A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir

ection.]
1. What force would a proton experience in the same field?
Physics
1 answer:
Pachacha [2.7K]3 years ago
3 0

Answer:

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

\vec{F} = q \vec{E}

11\hat{i} = -5 \times 10^{-6} \vec{E}

\vec{E} =-2.2 \times 10^{6}\hat{i}

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

\vec{F} = q \vec{E}

\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Force experienced by the photon in the same field is equal to \vec{F}= -3.52\times 10^{-13}\hat{i}\ N

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A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
Help on finding kinetic energy??
Jobisdone [24]
Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story. 
8 0
3 years ago
Read 2 more answers
(this is a question from my leap practice)
ki77a [65]
The answer is C. in sort of a way. You can't technically see black matter. However, it is holding the galaxies together.
4 0
4 years ago
A descent vehicle landing on the moon has a vertical velocity toward the surface of the moon of 36.5 m/s. At the same time, it h
sammy [17]

Answer:

66.26 m/s

Explanation:

Horizontal velocity, Vx = 55.3 m/s

Vertical velocity, Vy = 36.5 m/s

The value of the resultant velocity is given by the vector sum of the two velocities which are acting at 90°.

V=\sqrt{V_{x}^{2}+V_{y^{2}}}

V=\sqrt{55.3^{2}+36.5^{2}}}

V = 66.26 m/s

Thus, the velocity of the vehicle is 66.26 m/s along its descent path.

5 0
3 years ago
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