To find how far it travels, we just have to multiply
[tex]1.66\times2.5=4.15[/tex]It travels 4.15 meters.
During this interval, there are emitted 2.5 waves.
A 12 ohm hair dryer is plugged into a 240 V power supply. What is the current?
Given,
The resistance of the hairdryer, R=12 Ω
The voltage of the power supply, V=240 V
From Ohm's law,
[tex]V=IR[/tex]Where I is the current.
On substituting the known values,
[tex]\begin{gathered} 240=I\times12 \\ \Rightarrow I=\frac{240}{12} \\ =20\text{ A} \end{gathered}[/tex]Thus the current is 20 A.
An AC generator produces an output voltage E = 200 sin96.084 t volts. What is the frequency of the voltage ?
We can write the output voltage E in the following form:
E = E0*sin(ωt), where ω is angular period.
ω = 96.084
We also know that ω = 2πf, where f is frequency.
96.084 = 2πf; f = 96.084/2π
f = 15.2922 Hz
A 2.0 m long frictionless pendulum of mass 1.6 kg is released from point A at an angle & of
15 degrees. What is the speed of the pendulum at Point C.
The speed of pendulum at point C would be 0.47 m/sec when a 2 meter long frictionless pendulum of mass 1.6 kg is released from point A.
What is speed and how it is calculated out to be 0.47 meter/second?Speed is a quantity used to measure the distance travelled per unit time of the given time period.Here is given the length of rope be 2 meter and the mass of pendulum 1.6 kg released from point A.Using L cos theta = L ( 1 - cos theta) , we can write v^2/2g and putting all the numerical values we will get the speed be 0.47 m/s.Putting the angle that is theta be 15 degrees and getting the v the speed be 0.47 m/s.Hence the speed of the pendulum at point C would be 0.47 m/s.To know more about speed visit:
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Compute, in centimeters and in meters, the height of a basketball player who is 6 ft 1 in. tall into cm and m
We will have that:
6ft 1in is:
[tex]6ft(\frac{12in}{1ft})+1in=73in[/tex]Then:
[tex]73in(\frac{2.54cm}{1in})=185.42cm[/tex]So, the solution is both meters and cm are respectively.
1.8542 m and 185.42 cm.
A man can crack a nut by applying a force of 100 N a lever of length 0.5 m. What should be the length of the lever if he wants to use a force of 75 N to crack the nut?
Answer:
So, to apply the force of 75N length of nut cracker should be 66.6 cm
Answer:
Explanation:
Given:
F₁ = 100 N
L₁ = 0.5 m
F₂ = 75 N
__________
L₂ - ?
According to the rule of moments:
M₁ = M₂
F₁·L₁ = F₂·L₂
The length of the lever:
L₂ = F₁·L₁ / F₂
L₂ = 100·0.5 / 75 ≈ 0.67 m
Suppose that the Towngas supply pressure is 8.5 kPa (gauge pressure) andthe total volume is 2.4 m? enters a building from outside where thetemperature is 10 °C and passes into a building where the temperature is 35°C, if the pressure was reduced to 2 kPa. What would be the new volume ofthe gas?
Given,
The initial pressure of the gas, P₁=8.5 kPa
The initial volume of the gas, V₁=2.4 m³
The initial temperature of the gas, T₁=10 °C=282.15 K
The temperature of the building, i.e., the temperature of the gas after entering the building, T₂=35 °C=308.15 K
The pressure of the gas after entering the building, P₂=2 kPa
Let us assume the new volume of the gas is V₂
From the combined gas law,
[tex]\begin{gathered} \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} \\ \Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2} \end{gathered}[/tex]On substituting the known values,
[tex]undefined[/tex]An archerfish squirts water with a speed 2 m/s at an angle 50 degrees above the horizontal, and aims for a beetle on a leaf 3cm above the water surface. (A) At what horizontal distance from the beetle should the archerfish fire if it is to hit its target in the least time? (B) How much time does the beetle have to react?
A ) The horizontal distance from the beetle, the archerfish should fire if it is to hit its target in the least time = 1.19 m
B ) The time beetle have to react = 0.93 s
T = ( [tex]u_{y}[/tex] + √ [tex]u_{y}[/tex]² - 2 g H ) / g
T = Total time taken
g = Acceleration due to gravity
H = Height above the ground
[tex]u_{y}[/tex] = Y-component of initial velocity
u = 2 m / s
θ = 50°
H = 3 cm
[tex]u_{y}[/tex] = u sin θ
[tex]u_{y}[/tex] = 2 * sin 50°
[tex]u_{y}[/tex] = 1.54 m / s
T = ( 1.54 + √ 1.54² - ( 2 * 9.8 * 3 ) ) / 9.8
T = ( 1.54 + √ 56.42 ) / 9.8
T = 9.1 / 9.8
T = 0.93 s
R = [tex]u_{x}[/tex] T
[tex]u_{x}[/tex] = u cos θ
[tex]u_{x}[/tex] = 2 * cos 50°
[tex]u_{x}[/tex] = 1.28 m / s
R = 1.28 * 0.93
R = 1.19 m
Therefore,
A ) The horizontal distance from the beetle, the archerfish should fire if it is to hit its target in the least time = 1.19 m
B ) The time beetle have to react = 0.93 s
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Identify as many different ways as you can for giving energy to a basketball? (Select all that apply)
To answer this question we need to remember each kind of energy:
• Potential energy is the energy held by an object because of its position relative to other objects.
,• Kinetic energy is the energy held by an object due to is motion.
,• Internal energy is the energy due to the movement of the molecules of the object.
With this in mind we conclude that the following are ways of giving energy to a basketball:
• You can give a basketball kinetic energy by pushing on it with your hand, as in throwing or dribbling.
• You can give a basketball kinetic energy by spinning it on your finger.
• You can give a basketball potential energy by lifting it upward with your hand, as when shooting a free throw.
,• You can give a basketball internal energy by heating it.
It takes 5 seconds for a 2 kg box to be pushed 10 meters from rest. What was the forceof the push?
Given data:
* The mass of the box is 2 kg.
* The time taken by the box to travel the given distance is 5 seconds.
* The distance traveled by the box is 10 meters.
* The initial velocity of the box is 0 m/s.
Solution:
By the kinematics equation, the distance traveled by the box in terms of its acceleration is,
[tex]S=ut+\frac{1}{2}at^2[/tex]where u is the initial velocity, t is the time taken, a is the acceleration, and S is the distance traveled,
Substituting the known values,
[tex]\begin{gathered} 10=0+\frac{1}{2}\times a\times(5)^2 \\ 10=\frac{25}{2}\times a \\ a=10\times\frac{2}{25} \\ a=0.8ms^{-2} \end{gathered}[/tex]By the Newton's second law, the force exerted on the box in terms of the acceleration is,
[tex]F=ma[/tex]where m is the mass of the box, a is the acceleration and F is the force,
Substituting the known values,
[tex]\begin{gathered} F=2\times0.8 \\ F=1.6\text{ N} \end{gathered}[/tex]Thus, the force of the push is 1.6 N.
A sailboat starts from rest and accelerates at a rate of 0.13 m/s? over a distance of 344 m.(a) Find the magnitude of the boat's final velocitym/s (b)find the time it takes the boat to travel this distance
We know that
• It starts from rest. (The initial velocity is zero).
,• The acceleration rate is 0.13 m/s^2.
,• The distance covered is 344 m.
To find the magnitude of the boat's final velocity, we have to use the following formula.
[tex]v^2_f=v^2_0+2ad[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} v^2_f=0^2+2(0.13(\frac{m}{s^2}))(344m) \\ v^2_f=89.44(\frac{m^2}{s^2}) \\ v_f=\sqrt[]{89.44(\frac{m^2}{s^2})} \\ v_f\approx9.46(\frac{m}{s}) \end{gathered}[/tex](a) Therefore, the final velocity is 9.46 meters per second.Now, to find the time it takes the boat to travel this distance, we use the following formula.
[tex]d=v_0\cdot t+\frac{1}{2}at^2[/tex]Using the given magnitudes, we have the following.
[tex]344m=\frac{1}{2}(0.13(\frac{m}{s^2}))t^2[/tex]Let's solve for t.
[tex]\begin{gathered} t=\sqrt[]{\frac{2\cdot344m}{0.13(\frac{m}{s^2})}} \\ t=\sqrt[]{\frac{688m}{0.13}}\sec \\ t\approx72.75\sec \end{gathered}[/tex](b) Therefore, it takes 72.75 seconds.A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 219 m, and the car completes the turn in 32.0 s.
(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors î and ĵ.
(b) Determine the car's average speed.
(c) Determine its average acceleration during the 32.0-s interval.
a.) the acceleration when the car is at B located at an angle of 35.0° is 2.689i -0.42818j
b.) the car's average speed is v= 6.84375 m/s
c.) average acceleration during the 32.0-s interval is (−0.181 i+0.181 j)m/s²
What is acceleration?Acceleration is described as the rate of change of the velocity of an object with respect to time.
(a) The car’s speed around the curve is found from
v= 219/32.0
v= 6.84375 m/s
This is the answer to part (b) of this problem. We calculate the radius of the curve from
(1/4) X 2πr = 219 m
which gives r = 139.4 m
The car’s acceleration at point B is then
ar = (V²/r ) towards the center
= ( 6.84375)² / 139.4 at 35.0° north of west
= (2.9761 m/s²)(cos 35.0)(-i) + (sin 35.0j)
= -2.689i -0.42818j
(b) From part (a), v= 6.84375 m/s
(c) We find the average acceleration from
A avg = (Vf - Vi)/ change in t
A avg = ( 6.84375 j - 6.84375 i ) / 32.0 s
A avg = (−0.181 i+0.181 j)m/s²
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Full working out…….2.A vibrating mass-spring system has a frequency of 0.56 Hz. How much energy ofthis vibration is carried away in a one-quantum change?
ANSWER
3.7128 x 10⁻³⁴ J
EXPLANATION
The energy carried in a one-quantum change is the product of Planck's constant, h, and the frequency of vibration, f,
[tex]E=hf[/tex]Planck's constant is 6.63 x 10⁻³⁴ J*s and, in this case, the frequency of vibration is 0.56 Hz. So, the energy carried away is,
[tex]E=0.56Hz\cdot6.63\cdot10^{-34}J\cdot s=3.7128\cdot10^{-34}J[/tex]Hence, the energy carried away in a one-quantum change is 3.7128 x 10⁻³⁴ J.
How much time would it take a car to go from a speed of 10m/s to 30m/s if it acceleratesat a rate of 4.0m/s/s?
Given,
The initial velocity of the car, u=10 m/s
The final velocity of the car, v=30 m/s
The acceleration of the car, a= 4.0 m/s²
From one of the equations of the motion, we have
[tex]v=u+at[/tex]Where t is the time duration.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]t=\frac{30-10}{4.0}=5\text{ s}[/tex]Therefore the time that the car takes to accelerate is 5 s
2. A 14000 kg air jet accelerates from rest to 70 m/s before it takes off. What is the changein momentum of the jet?
Answer:
980,000 kg m/s
Explanation:
The change in momentum can be calculated as
[tex]\begin{gathered} \Delta p=m\Delta v \\ \Delta p=m(v_f-v_i) \end{gathered}[/tex]Where m is the mass, vf is the final velocity and vi is the initial velocity. Replacing m = 14000 kg, vf = 70 m/s and vi = 0 m/s, we get
[tex]\begin{gathered} \Delta p=14000\text{ kg \lparen}70\text{ m/s - 0 m/s\rparen} \\ \Delta p=14000\text{ kg \lparen70 m/s\rparen} \\ \Delta p=980000\text{ kg m/s} \end{gathered}[/tex]Therefore, the change in momentum of the jet is 980,000 kg m/s
Question 10 of 10If one of two interacting charges is doubled, the force between the chargeswillO A. decrease by 4 timesO B. increaseC. decreaseD. stay the sameSUBMIT
The force between the two charges is
[tex]Force\text{ }\propto\text{ charge}[/tex]Thus, if the charge is doubled, then its force will also be doubled.
Hence the correct option is increase.
A quantity of steam (650 g) at 116°C is condensed, and the resulting water is frozen into ice at 0°C. How much heat was removed?answer in:____ kcal
Total heat removed = 473.04 kCal
Explanation:Heat removed to convert the 116°C to 100°C steam
[tex]\begin{gathered} H=mc(\theta_2-\theta_1) \\ \\ H=650(1.996)(116-100) \\ \\ H=20758.4J \end{gathered}[/tex]Heat removed from 100°C of steam to 100°C of water (Latent heat of condensation)
[tex]\begin{gathered} H_c=650\times2257 \\ \\ H_c=1467050J \end{gathered}[/tex]Heat removed from 100°C water to 0°C water
[tex]\begin{gathered} H_w=650\times4.2\times100 \\ \\ H_w=273000J \end{gathered}[/tex]Heat removed from 0°C water to 0°C ice
[tex]\begin{gathered} H_i=mL_f \\ \\ H_i=650(336) \\ \\ H_i=218400J \end{gathered}[/tex]Total heat removed = 20758.4J + 1467050 + 273000 + 218400
Total heat removed = 1979208.4 J
Convert to kilocalorie
Total heat removed = 1979208.4/4184
Total heat removed = 473.04 kCal
A penguin runs 29,000 m/s how far will it travel in 10 seconds
Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.
[tex]s=\frac{d}{t}[/tex]s = speed
d = distance
t = time
[tex]\begin{gathered} d=st \\ d=29000\text{ m/s}\cdot10s \\ d=290000\text{ m} \end{gathered}[/tex]The distance would be 290,000 m
A ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22.0%,
If a ray of light travels from air into a liquid, as shown in figure below. The ray is incident upon the liquid at an angle of 30.0°. The angle of refraction is 22°, and the refractive index of the liquid would be 1.334.
What is refraction?It is the phenomenon of bending of light when it travels from one medium to another medium. The bending towards or away from the normal depends upon the medium of travel as well as the refractive index of the material.
By using Snell's law,
Refractive index of the liquid = sin(i) /sin(r)
=sin(30) /sin(22)
= 1.334
Thus, the refractive index of the liquid would be 1.334.
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What would the separation between two identical objects, one carrying
2 C
of positive charge and the other
2 C
of negative charge, have to be if the electrical force on each was precisely
2 N?
Please Help
The distance between the two charges is 134,164.1 m.
What is the distance between the two identical charges?
The distance between the two identical charges is determined by applying Coulomb's law as shown below.
F = kq²/r²
where;
K is Coulomb's constantq is the magnitude of the chargesr is the distance between the chargesF is the electric force between the two chargesr = √(kq²/F)
r = √(9 x 10⁹ x 2²) / 2)
r = 134,164.1 m
Thus, the distance between the two charges is determined by applying Coulomb's law.
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Ellipses have only one focus. Is this true or false?
In any ellipse there are two foci. This two points are fixed and are fundamental for the construction opf the ellipse. Therefore the statement is false.
1. Consider the example problem, but with the lower pressure reduced to 100 Pa. How much work would be done in a single cycle?1. 29,970 J2. 0 J3. -30,030 J4. 27,000 J
29970 J
Explanation:Given:
[tex]\begin{gathered} P_A=P_B=10^5Pa \\ P_c=P_D=100\text{ Pa\lparen The new lower pressure\rparen} \\ V_A=V_D=0.1m^3 \\ V_B=V_C=0.4m^3 \end{gathered}[/tex]Work done across AB
[tex]\begin{gathered} W_{AB}=P_{AB}(V_B-V_A) \\ \\ W_{AB}=10^5(0.4-0.1) \\ \\ W_{AB}=0.3\times10^5 \\ \\ W_{AB}=30000J \end{gathered}[/tex]Work done across BC = 0 J (Since there is no change in volume)
Work done across DA = 0 J (Since there is no change in volume)
Work done across CD
[tex]\begin{gathered} W_{CD}=P_{CD}(V_D-V_C) \\ \\ W_{CD}=100(0.1-0.4) \\ \\ W_{CD}=100(-0.3) \\ \\ W_{CD}=-30J \end{gathered}[/tex]Work done in one cycle:
[tex]\begin{gathered} W_{cycle}=W_{AB}+W_{BC}+W_{DA}+W_{CD} \\ \\ W_{cycle}=3000J+0+0-30 \\ \\ W_{cycle}=29970J \end{gathered}[/tex]29970J of work will be done in a single cycle
An object on a horizontal, frictionless surface is attached to a spring, displaced, and then released. If it isdisplaced 0.12m from its equilibrium position and released after 0.8s its displacement is found to be 0.12m onthe opposites side, and passed the equilibrium position once during the interval. Find:
Given:
The maximum displacement from the equilibrium position is A = 0.12 m.
Half of the time period is
[tex]T_{\frac{1}{2}}=\text{ 0.8 s}[/tex]To find the amplitude, time period, and frequency.
Explanation:
Amplitude is the maximum displacement from the equilibrium position.
Thus, the amplitude is A = 0.12 m.
One time period is the time taken from maximum displacement on one side(say A) to maximum displacement on the opposite side and back to the maximum displacement on the same side(A).
Thus, the time period is
[tex]\begin{gathered} T=2T_{\frac{1}{2}} \\ =2\times0.8 \\ =1.6\text{ s} \end{gathered}[/tex]The frequency will be
[tex]\begin{gathered} f=\frac{1}{T} \\ =\frac{1}{1.6} \\ =0.625\text{ Hz} \end{gathered}[/tex]Point charges 88μC,-55μC and 70 μC are placed in a straight line. The central one is 0.75m from each of the others. Calculate the net force on each due to the other two.
The net force on the charges is 139.04 N.
What is the net force between the charges?
The net force between the charges is calculated by applying Coulomb's law as follows;
F = kq₁q₂/r²
where;
k is Coulomb's constantq₁ is the first chargeq₂ is the second charger is the distance between the chargesF(12) = (9 x 10⁹ x 88 x 10⁻⁶ x 55 x 10⁻⁶) / (0.75²)
F(12) = 77.44 N
F(23) = (9 x 10⁹ x 55 x 10⁻⁶ x 70 x 10⁻⁶) / (0.75²)
F(23) = 61.6 N
The net force on the charges is calculated as follows;
F(net) = 77.44 N + 61.6 N
F(net) = 139.04 N
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There are three point charges placed in a straight line, so the net force on each due to the other two is 139.04 N.
What is a charge?Charged matter experiences a force when it comes into contact with an electromagnetic field because electric charge is a property of matter. An electric field can have either a positive or negative charge. Charges that are similar to one another and dissimilar to one another are drawn to one another.
Given information in the question,
Charge, q₁ = 88 μC
Charge, q₂ = -55 μC
Charge, q₃ = 70 μC
Distance, r = 0.75 meters.
F = kq₁q₂/r²
Put the values in the above formula,
F₁₂ = (9 x 10⁹ x 88 x 10⁻⁶ x 55 x 10⁻⁶) / (0.75²)
F₁₂ = 77.44 N
F₂₃ = (9 x 10⁹ x 55 x 10⁻⁶ x 70 x 10⁻⁶) / (0.75²)
F₂₃ = 61.6 N
Now, calculate the net force :
F(net) = 77.44 N + 61.6 N
F(net) = 139.04 N
Hence, the net force due to the other two is 139.04 N.
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When you eat cereal and thenlift weights, how is the energytransformed?A. The chemical energy in the cereal istransformed into mechanical energy.B. The thermal energy in the cereal istransformed into mechanical energy.C. The mechanical energy in the cereal istransformed into chemical energy.
in any food not just in cereal there are chemical energy is stored in the form of
carbon, protein, fats, etc.
when we eat the food our body transforms this energy into
energy by decomposing the food into its elementary
particles and store it in the form of chemical energy and when we need energy
to do work it (body) coverts the energy into the mechanical energy by various
process.
so the correct answer is option (A)
A 65.0-kg basketball player jumps vertically and leaves the floor with a velocity of 1.80 m/s upward. (a) What impulse does the player experience? (b) What force does the floor exert on the player before the jump? (c) What is the total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump?
(a) The impulse experienced by the player is 117 Ns.
(b) The force the floor exert on the player before the jump is 637 N.
(c) The total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump is 260 N.
What impulse does the player experience?
The impulse experienced by the player is the change in the momentum of the player.
J = ΔP
J = m(v - u)
where;
m is the mass of the playerv is the final velocity of the player = 0u is the initial velocity of the player = - 1.8 m/s (negative because of upward direction)J = 65 x (0 + 1.8)
J = 117 Ns
The force the floor exert on the player before the jump is calculated as;
F = mg
where;
g is acceleration is due to gravityF = 65 x 9.8
F = 637 N
The total average force exerted by the floor on the player if the player is in contact with the floor for 0.450 s during the jump is calculated as;
F = ma
F = m(v/t)
F = (mv)/t
F = (65 x 1.8) / 0.45
F = 260 N
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The chart shows data for four moving objects.
Object
W
X
Y
Z
Which object has the greatest acceleration?
W
Initial Velocity
(m/s)
Ο Ζ
11
10
12
20
Final Velocity
(m/s)
29
34
40
28
Change in
Time (s)
6
12
7
8
Object Y has the greatest acceleration = 4m/s2
What is an acceleration ?
acceleration: the rate at which the speed and direction of a moving object vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.
Acceleration = (final velocity - initial velocity ) /time
For object W
Acceleration = ( 29-11)/6 = 3m/s2
For object X
Acceleration = (34-10)/12 = 2m/s2
For object Y
Acceleration = ( 40-12)/7 = 4m/s2
For object Z
Acceleration = (28-20)/8 = 1m/s2
Object Y has the greatest acceleration = 4m/s2
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Two Styrofoam blocks are brought near each other and are observed to repel each other. Each block has the same amount of charge on it. Which statement is true about this situation? A)the charge on each block must be positive B)the charge on each block must be negative C)the blocks are creating magnetic fields from non moving electric charges D)each block has the same kind of charge but we don't' know what kind it is
ANSWER:
D)each block has the same kind of charge but we don't' know what kind it is
STEP-BY-STEP EXPLANATION:
Two charges of the same type always repel each other.
In this case, it is not possible to know what type of load they are, only that they are the same.
Therefore, the correct answer is:
D)each block has the same kind of charge but we don't' know what kind it is
Two different masses have equal, non-zero kinetic energies. The momentum of the smaller mass isSelect one:A) smaller than the momentum of the larger mass.B) There is not enough information to answer the question.C) equal to the momentum of the larger mass.D) zero.E) Larger than momentum of the larger mass.
Let the mass of smaller object be m and mass of larger object be M.
As the kinetic energies are equal,
[tex]\frac{p_m^2}{2m}=\frac{p^2_{_M}}{2M}[/tex]On solving further,
[tex]\frac{p_m}{p_M}=\text{ }\sqrt[]{\frac{m}{M}}[/tex]Hence, option A is correct.
carts, bricks, and bands
5. What acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2
B. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
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cassy can get more force on the bricks she breaks with a blow of her bare hand when _______.
Answer:Her hand is made to bounce from the bricks
Explanation:
Cassy can get more force on the bricks that she breaks with a blow of her bare hand when her hand is made to bounce from the bricks,
What is Force?A force in physics is an effect that has the power to alter an object's motion. A mass-containing object's speed can vary, or accelerate, as a result of a force. Intuitively, a pull or a push can also be used to describe force. Being a vector quantity, a force also has magnitude and direction. The SI unit of newton is used to measure it (N). The letter F stands for force.
According to Newton's second law's original formulation, an object's net force is equal to the speed at which momentum is changing over time.
Objects' velocities can be altered by the concepts of push, drag, and torque. Thrust causes an object's velocities to increase, while torque causes an object's velocities to decrease. Each part of an extended body typically exerts pressures on its neighboring sections, and the internal mechanical force is determined by how these stresses are distributed.
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