Two equal forces act on two different objects, one of which has a mass ten times as large as the other. The larger object will have _________ acceleration that the less massive object.
Answer:
The larger object will have smaller acceleration that the less massive object.
Explanation:
Generally force is mathematically represented as
[tex]F = ma[/tex]
=> [tex]m = \frac{F}{a }[/tex]
at constant force we have
[tex]m \ \alpha \ \frac{1}{a}[/tex]
So if m is increasing a will be decreasing which means the object with the larger mass will have less acceleration
Based on the information in the table, which elements are most likely in the same periods of the periodic table?
Answer:
Just to help, periods on the periodic table are those running horizontally from left to right
Answer:
The answer is A.Boron and carbon are likely together in one period because they have very close atomic numbers, while gallium and germanium are likely together in another period because they have very close atomic numbers.
Explanation:
just took test
help me get the answer in Physical Science.
Answer:
lithium
Explanation:
I took physical science 2 years ago and passed with an A
A lamp of mass m hangs from a spring scale which is attached to the ceiling of an elevator. When the elevator is stopped at the fortieth floor, the scale reads mg. What does it read as the elevator slows down to stop at the ground floor?
a. more than mg
b. mg
c. less than mg
d. zero
e. can't tell
Answer:
The correct answer is (a)
Explanation:
A spring scale measures the weight of an object not the mass because according to hooke's law the extension of a spring is directly proportional to the load or force attached/applied to it. The force of gravity acting on the mass of any substance as it goes up actually reduces and increases as it comes down.
If F = ma, as a increases, F will also increase and vice versa
Where F = force
m = mass
a = acceleration (due to gravity in this case)
From the above explanation, it can be deduced that the scale will read more than mg as it gets to the ground because of an increase in the force of gravity (which also increases a) as it approaches the ground.
The emf of the battery is 1.5 V. In Nichrome there are 9 × 1028 mobile electrons per m3, and the mobility of mobile electrons is 7 × 10−5 (m/s)/(N/C). Each thick wire has length 29 cm = 0.29 m and cross-sectional area 9 × 10−8 m2. The thin wire has length 6 cm = 0.06 m and cross-sectional area 1.3 × 10−8 m2. (The total length of the three wires is 64 cm.) In the steady state, calculate the number of electrons entering the thin wire every second. Do not make any approximations, and do not use Ohm's law or series-resistance equations.
Answer:
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Explanation:
Given;
emf of the battery, V = 1.5 V
electron density, = 9 × 10²⁸ mobile electrons per m³
mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)
length of thin wire, L = 6 cm = 0.06 m
cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²
The magnitude of the electric field in the thin wire is given by;
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
the number of electrons entering the thin wire every second is given by;
[tex]e/s = mobility \ x \ Electric \ field\\\\number \ of \ electrons \ per \ second =\frac{7*10^{-5} (m/s)}{N/C} *25 (N/C)\\\\number \ of \ electrons \ per \ second = 1.75*10^{-3} \ m/s[/tex]
Therefore, the number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
Calculation of the number of electrons:Since
emf of the battery, V = 1.5 V
electron density, = 9 × 10²⁸ mobile electrons per m³
mobility of electron, u = 7 × 10⁻⁵ (m/s)/(N/C)
length of thin wire, L = 6 cm = 0.06 m
cross sectional area of the thin wire, A = 1.3 x 10⁻⁸ m²
So here the magnitude should be
E = V/L
E = (1.5) / (0.06)
E = 25 N/C
Now the number of electrons should be
= 7 × 10⁻⁵ *25
= 1.75 x 10⁻³ mobile
hence, The number of electrons entering the thin wire every second is 1.75 x 10⁻³ mobile electrons / second
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A small compass is held horizontally, the center of its needle has a distance of 0.270 m directly north
of a long wire that is perpendicular to the Earth's surface. When there is no current in the wire, the
compass needle points due north, which is the direction of the horizontal component of the Earth's
magnetic field at that location. This component is parallel to the Earth's surface. When the current in
the wire is 26.3 A, the needle points 22.9∘ east of north.
(a) Does the current in the wire flow toward or away from the Earth's surface? ( 2 marks)
(b) What is the magnitude of the horizontal component of the Earth's magnetic field at the location of
the compass? (3 marks)
Answer:
Explanation:
The needle is showing north south direction . when current starts flowing in the wire which is held vertical to the ground , it deflects towards east .
a )
Therefore a magnetic field towards east has been created . It is possible only if current flows towards the surface in the vertical wire .
b )
magnetic field created at the magnetic needle B = 10⁻⁷ x 2I / d where I is current and d is distance .
B = 10⁻⁷ x 2 x 26.3 / .27
= 194.81 x 10⁻⁷ T
angle of deflection of solenoid = 22.9°
Tan 22.9 = B /H
.422 = 194.81 x 10⁻⁷ / H
H = 461.63 x 10⁻⁷ T
= .46 x 10⁻⁴ T .
A) The current in the wire flows towards the Earth's surface
B) The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
A) The compass needle held horizontally points in a North-south direction of the earth and also deflects eastwards when current is allowed to flow through it. The deflection of the needle indicates the presence/generation of a magnetic field on the earth surface. which is facilitated by the flow of the current in the wire towards the Earth's surface
B) Determine The magnitude of the horizontal component of the Earth's magnetic field
B ( magnetic field ) = 10⁻⁷ * 2I / d ---- ( 1 )
where : l = 26.3 A, d = 0.27 m
Back to equation ( 1 )
B = 10⁻⁷ * 2 * 26.3 / 0.27
= 194.81 * 10⁻⁷ T
Final step : Calculate the magnitude of horizontal component ( H )
Tan ∅ = B / H ---- ( 2 )
where : ∅ ( angle of deflection ) = 22.9°
∴ H = B / Tan ( 22.9° )
= ( 194.81 * 10⁻⁷ ) / 0.422
= 0.46 x 10⁻⁴ T
Hence we can conclude that The current in the wire flows towards the Earth's surface and The magnitude of the horizontal component of the Earth's magnetic field is : 0.46 x 10⁻⁴ T
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An FM radio station, 20 miles away, broadcast at a 93.4 MHz frequency(a) What is the wavelength of the radio wave associated with this signal ?(b) How long does it take for the signal to reach your radio from the station ?
Answer:
(a) Wavelength = 3.21 m (b) Time = [tex]1.07\times 10^{-4}\ s[/tex]
Explanation:
Given that,
The frequency of FM radio station, f = 93.4 MHz
(a) We need to find the wavelength of the radio wave associated with this signal. The relation between wavelength and frequency is given by :
[tex]c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{93.4\times 10^6}\\\\\lambda=3.21\ m[/tex]
(b) It is given that, an FM radio station, 20 miles away. Let t is time taken for signal to reach your radio from the station. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{20\times 1609.34}{3\times 10^8}\\\\t=1.07\times 10^{-4}\ s[/tex]
Hence, this is the required solution.
A coin rests on a record 0.15 m from its center. The record turns on a turntable that rotates at variable speed. The coefficient of static friction between the coin and the record is 0.30.
Required:
What is the maximum coin speed at which it does not slip?
Answer:
0.66m/sExplanation:
We are expected to solve for the velocity with no slip condition
we know that the expression that relate coefficient of friction and velocity is given as
μs = v^2/rg
Given
coefficient of friction μs = 0.3
radius r= 0.15
assume g=9.81m/s^2
substituting into the expression we have
0.3= v^2/0.15*9.81
v^2=0.3*0.15*9.81
v^2=0.44145
v=√0.44145
v=0.66
therefore the velocity is 0.66m/s
Momentum of the 2 kg mass moving with velocity 10 m/s is *
A. 2 kg*m/s
B. 20 kg*m/s
C. 200 kg*m/s
D. 20000 kg*m/s
If it takes you 5 minutes to dry your hair using a 1200-W hairdryer plugged into a 120-V power outlet, how many Coulombs of charge pass through your hair dryer
Answer:
The charge pass through your hair dryer is 3000 C.
Explanation:
Given that,
Power = 1200 W
Voltage = 120 V
Flow time = 5 min
We need to calculate the current
Using formula of power
[tex]P=VI[/tex]
[tex]I=\dfrac{P}{V}[/tex]
Put the value into the formula
[tex]I=\dfrac{1200}{120}[/tex]
[tex]I=10\ A[/tex]
We need to calculate the charge pass through your hair dryer
Using formula of current
[tex]I=\dfrac{Q}{t}[/tex]
[tex]Q=It[/tex]
Put the value into the formula
[tex]Q=10\times5\times60[/tex]
[tex]Q=3000\ C[/tex]
Hence, The charge pass through your hair dryer is 3000 C.
For both resonance curves and Fourier spectra, amplitude is plotted vs frequency, but these two types of plots are not the same. Describe how they are different.
Answer:
he peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
Explanation:
In a resonance experiment, the amplitude of the system is plotted as a function of the frequency, finding maximums for the values where some natural frequency of the system coincides with the excitation frequency.
In a Fourier transform spectrum, the amplitude of the frequencies present is the signal, whereby each peak corresponds to a natural frequency of the system.
From this explanation we can see that in the first case the peaks are the natural frequencies that coincide with the excitation frequencies and in the second case they are the natural frequencies that make up the wave.
the peripheral nervous system is responsible for both sending and receiving signals to and from the brain
Answer:
its true trust me
Explanation:
Answer: true
Explanation: edge
a jogger travels at 4 m/s for 100 s what is the distance covered
400m
Explanation:
given,
v= 4m/s
t= 100s
d= ?
since, v = d / t
therefore, d = v * t (velocity multiplied by time)
=> d = 4 * 100
= 400m.
I am a cell. I am long and thin. I reach all the way from the brain
to the tip of a finger. I have a special coat of fat that helps me do
my job. My job is to send electrical signals from one part of the
body to another.
Answer:
Neurons
Explanation:
We humans have a nervous system that coordinates our behavior and transmits signals between different parts of our body.
Now, this nervous system contains a lot of nerve cells which we call Neurons. These Neurons have a cell like body and their job is to transmit signals from one part of our body to another.
Thus, the cell is called Neurons.
When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1061 N. The mass of the sky diver is 93.4 kg. Take upward to be the positive direction. What is his acceleration, including sign
Explanation:
According to newton's second law of motion.
[tex]\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N[/tex]
m is the mas of the sky diver = 93.4kg
a is the acceleration of the skydiver
From the formula above;
[tex]a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2[/tex]
Hence the acceleration of the sky diver is 1.563m/s²
A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various energies. If the electric field strength is 2.2 x 104 N/C, what should be the value of the magnetic field (in tesla) to select protons of velocity 6.4 x 105 m/s
Answer:
The value is [tex]B = 0.034 \ T [/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 2.2*10^{4} \ N/C[/tex]
The velocity is [tex]v = 6.4 *10^{5} \ m/s[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \frac{E}{v}[/tex]
=> [tex]B = \frac{2.2*10^{4}}{6.4 *10^{5}}[/tex]
=> [tex]B = 0.034 \ T [/tex]
. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is to the east.
Answer:
Explanation:
The total distance is how far you walk from the starting point.
Distance through west = 18.0m
Distance through north = 25.0m
Total distance covered = 18.0+25.0m
Total distance covered = 43.0m
This means that I am 43.0m from the starting point
Displacement is the distance covered in a specified direction. The displacement will be gotten using the Pythagoras theorem as shown:
[tex]d^2 = 25^2 + 18^2\\d^2 = 625+324\\d^2 = 949\\d = \sqrt{949}\\ d = 30.81m[/tex]
The direction of your displacement is 30.81m
Direction is gotten according to the formula;
[tex]\theta = tan ^{-1}{\frac{y}{x} }\\\theta = tan ^{-1}{\frac{25}{-18} }\\\theta = tan ^{-1}-1.3889}\\\theta = -60.27^0\\\theta = 180-60.27\\\theta = 119.7^0[/tex]
Note that the direction to the west is negative, that is why the x is -18.0m
The distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
Given-
Distance travel through the west is 18 m.
Distance travel through the north is 25 m.
Distance from starting point-
To know the total distance, add both the covered distance. Thus total distance x is,
[tex]x=18+25[/tex]
[tex]x=43[/tex]
Hence, the distance from the starting point is 43 m.
The displacement vector-
Displacement is calculated as the shortest distance between starting and final point. This shortest distance can be calculated using the Pythagoras theorem which states that in a right-angled triangle, the square of the hypotenuse [tex]d[/tex] is equal to the sum of the squares of the other two sides. Therefore,
[tex]d^2=18^2+25^2[/tex]
[tex]d^2=324+625[/tex]
[tex]d^2=949[/tex]
[tex]d=\sqrt{949}[/tex]
[tex]d=30.81[/tex]
The displacement vector is 30.81 m.
The Direction of displacement-The direction of displacement [tex]\theta[/tex] with these two sides can be calculated with the formula,
[tex]\theta=tan^{-1}\dfrac{25}{-18}[/tex]
Here due to the west direction(opposite side), the sign is taken negatively.
[tex]\theta=tan^{-1}(-1.389)[/tex]
[tex]\theta=-60.27^o[/tex]
For the other quarter,
[tex]\theta=180-60.27=119.7^o[/tex]
Hence, the distance from the starting point is 43 m, the displacement vector is 30.81 m and the direction of the displacement is 119.7 degrees.
For more about the displacement, follow the link below-
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A microwave oven operates at 2.50 GHzGHz . What is the wavelength of the radiation produced by this appliance? Express the wavelength numerically in nanometers.
Answer:
The wavelength is [tex]\lambda = 1.2 * 10^8 nm[/tex]
Explanation:
From the question we are told that
The frequency of operation of the microwave is [tex]f = 2.50 GHz = 2.50 *10^{9} \ Hz[/tex]
Generally the wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]\lambda = \frac{3.0 *10^{8}}{ 2.50 *10^{9}}[/tex]
=> [tex]\lambda = 0.12 \ m [/tex]
converting to nanometer
[tex]\lambda = 1.2 * 10^8 nm[/tex]
If a car is traveling at an average speed of 70 kilometers per hour how long does it take the car to travel 14 kilometers
Answer:
Explanation:
O.20 hour A
If a car is traveling at an average speed of 70 kilometers per hour, 0.2 hours it takes the car to travel 14 kilometers.
What is average speed?By multiplying the distance that an item travels in one unit by the amount of time it takes to go that distance, one may determine the speed of the object. The speed of the item on this voyage, denoted by the letter "s," is equal to s = D/T if "D" is indeed the distance traveled in certain time "T."
Understanding average speed will help you better comprehend the pace of a travel. On a travel, the pace could occasionally change. Knowing the average speed then becomes crucial to getting an idea of how quickly the route will be finished.
Distance covered = average speed × Time travelled
14=70× Time travelled
Time travelled = 0.2 hours
Therefore, 0.2 hours it takes the car to travel 14 kilometers.
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If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 20.0 km? Take the mass of the sun to be 1.99×10^30, the gravitational constant to be G = 6.67×10^−11Nm^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.p
Answer:
8.93*10^13 N.
Explanation:
Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:[tex]F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }[/tex]
where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.Replacing by the values, we get:[tex]F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N[/tex]
Fg = 8.93*10^13 N.Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of south. The resultant vector A+ B is given by
Answer:
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
Explanation:
Let [tex]\vec A = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})[/tex] and [tex]\vec B = 4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex], both measured in meters. The resultant vector [tex]\vec R[/tex] is calculated by sum of components. That is:
[tex]\vec R = \vec A+\vec B[/tex] (Eq. 1)
[tex]\vec R = 6\cdot (\cos 30^{\circ}\,\hat{i}+\sin 30^{\circ}\,\hat{j})+4\cdot (-\sin 30^{\circ}\,\hat{i}-\cos 30^{\circ}\,\hat{j})[/tex]
[tex]\vec R = (6\cdot \cos 30^{\circ}-4\cdot \sin 30^{\circ})\,\hat{i}+(6\cdot \sin 30^{\circ}-4\cdot \cos 30^{\circ})\,\hat{j}[/tex]
[tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex]
The resultant vector [tex]\vec R = \vec A+\vec B[/tex] is given by [tex]\vec R = 3.196\,\hat{i}-0.464\,\hat{j}\,\,\,[m][/tex].
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]
A plane travelling at 100 m/s accelerates at 5 m/s² for a distance of 125 m. What is the final velocity of the plane?
Analyzing the question:
We are given:
initial velocity (u) = 100 m/s
final velocity (v) = v m/s
distance (s) = 125 m
acceleration (a) = 5 m/s²
Solving for Final Velocity (v):
from the third equation of motion:
v² - u² = 2as
v² - (100)² = 2(5)(125)
v² - 10000 = 1250
v² = 1250 + 10000
v² = 11250
v = 106.06 m/s
PLEASE HELP
A sharpshooter fires a 0.22 caliber rifle horizontally at 100 m/s at a target 75m away. How far does the
bullet drop by the time it reaches the target?
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The bullet drops "2.76 m" by the time it reaches the target.
First, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here to find the total time to reach the target:
[tex]s = vt\\\\t = \frac{s}{v}[/tex]
where,
s = distance = 75 m
v = velocity = 100 m/s
t = time = ?
Therefore,
[tex]t = \frac{75\ m}{100\ m/s}[/tex]
t = 0.75 s
Now, we will analyze the vertical motion of the bullet. We will use the second equation of motion in the vertical direction to find the height dropped by the bullet.
[tex]h = v_it+\frac{1}{2}gt^2[/tex]
where,
h = height dropped = ?
vi = initial vertical speed = 0 m/s
t = time interval = 0.75 s
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]h = (0\ m/s)(0.75\ s)+\frac{1}{2}(9.81\ m/s^2)(0.75\ s)^2[/tex]
h = 2.76 m
Learn more about equations of motion here:
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The attached picture shows the equations of motion in the horizontal and vertical directions.
What are two ways that an object can have kinetic energy?
Answer:
The object has to have mass and speed
Explanation:
You can increase both speed and mass to increase the kinetic energy, hope this answers your question.
Happy Halloween!
You release a ball from rest at the top of a ramp. 6 s later it is moving at 4.0
m/s. What is the acceleration? (in meters per second squared) *
Your answer
[tex]a = \frac{vf - vi}{t} [/tex]
here initial velocity vi=0 as ball release from rest
the final velocity is vf=4.0
time is t=6
so putting all these values in above equation
[tex]a = \frac{ 4.0- 0}{6} [/tex]
[tex]a = 0.6667m \s {}^{2} [/tex]
While making some observations at the top of the 66 m tall Astronomy tower, Ron
accidently knocks a 0.5 kg stone over the edge. How long will a student at the bottom
have to get out of the way before being hit?
Analysing the question:
Since the stone was dropped, there was no initial velocity applied on it and hence it's initial velocity of the stone is 0 m/s
We are given:
height of the tower (h) = 66 m
mass of the stone (m) = 0.5 kg
initial velocity of the stone (u) = 0 m/s
time taken by the stone to reach the ground (t) = t seconds
acceleration due to gravity = 10 m/s²
** Neglecting air resistance**
Finding the time taken by the stone to reach the ground:
from the second equation of motion
h = ut + 1/2at²
replacing the variables
66 = (0)(t) + 1/2 (10)(t)²
66 = 5t²
t² = 13.2
t = 3.6 seconds
I initially wanted to subtract the height of the student from the height of the tower since the time i calculated is the time taken by the stone to reach the ground and that means that the stone has already hit the student before 3.6 seconds
but since we were NOT given the height of a student, the person who posed this question wants the time taken by the stone to reach the ground and that is what we solved
A force of 41 N acts on an object which has a mass of 2.4 kg. What acceleration (in m/s2) is produced by the force
Answer:
The acceleration is [tex] a = 17.083 \ m/s^2 [/tex]
Explanation:
From the question we are told that
The force is [tex]F = 41 \ N[/tex]
The mass of the object is [tex]m = 2.4 \ kg[/tex]
Generally the force is mathematically represented as
[tex]F = m* a[/tex]
=> [tex] 41 = 2.4* a[/tex]
=> [tex] a = 17.083 \ m/s^2 [/tex]
During a thunderstorm the electric field at a certain point in the earth's atmosphere is 1.07 105 N/C, directed upward. Find the acceleration of a small piece of ice of mass 1.08 10-4 g, carrying a charge of 1.05 10-11 C.
Answer:
The acceleration of a small piece of ice is 10.40 m/s².
Explanation:
The electric force is given by:
[tex]F = Eq[/tex]
Where:
E is the electric field = 1.07x10⁵ N/C
q is the charge = 1.05x10⁻¹¹ C
The electric force is equal to Newton's second law:
[tex] Eq = ma [/tex]
Where:
m is the mass = 1.08x10⁻⁴ g = 1.08x10⁻⁷ kg
a is the acceleration
Hence, the acceleration is:
[tex] a = \frac{Eq}{m} = \frac{1.07 \cdot 10^{5} N/C*1.05 \cdot 10^{-11} C}{1.08 \cdot 10^{-7} kg} = 10.40 m/s^{2} [/tex]
Therefore, the acceleration of a small piece of ice is 10.40 m/s².
I hope it helps you!
If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.