Answer:
D
The answer cannot be found until it is known whether q is greater than, less than, or equal to 45°.
Explanation:
Since block moves with constant speed
So, frictional force
f = FCosq
Work done by friction
W = - fd
W = - fd Cos q
The answer may be greater or less than - fdSinq. It depends on the value of q which is less than, or equal to 45°.
There are two possible alignments of a dipole in an external electric field where the dipole is in equilibrium: when the dipole moment is parallel to the electric field and when the dipole moment is oriented opposite the electric field.
Part A
Are both alignments stable? (Consider what would happen in each case if you gave the dipole a slight twist.)
a) Yes
b) No
Part B
Based on your answer to the previous part and your experience in mechanics, in which orientation does the dipole have less potential energy?
a) The arrangement with the dipole moment parallel to the electric field has less potential energy.
b) The arrangement with the dipole moment opposite the electric field has less potential energy.
c) Both arrangements have the same potential energy.
Answer:
A. (b)
B. (a)
Explanation:
The electric dipole moment is the product of charge and the length of the dipole.
The torque on the dipole placed in the external electric field is given by
torque = p E sin A
where, p is the electric dipole moment, E is the electric field, A is the angle between the field and dipole moment.
When the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium.
When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.
So, the option (b) is correct.
Teh energy is given by
U = - p E cos A
When the angle A is zero , the potential energy is negative and it is minimum.
In this exercise we have to use the knowledge about dipole to be able to mark the correct alternative for each question, in this way we find that:
A) Letter b
B) Letter a
So knowing that the electric dipole moment is the product of charge and the length of the dipole and the torque on the dipole placed in the external electric field is given by:
[tex]torque = p E sin (A)[/tex]
where:
p: the electric dipole momentE: the electric fieldA: the angle between the field and dipole momentWhen the dipole moment is parallel to the electric field, the net torque is zero and it is said to be in stable equilibrium. When the dipole moment is anti parallel to the electric field, the net torque is zero but the dipole is in unstable equilibrium.
Now the energy is given by:
[tex]U = - p E cos (A)[/tex]
We can say that when the angle A is zero , the potential energy is negative and it is minimum.
See more about dipole at brainly.com/question/12757739
A ball is thrown from ground level with an initial speed of 24.5 m/s at an angle of 35.5 degrees above the horizontal. The ball hits a wall that is 25.8 meters horizontally from where it started. How high (meters) does the ball hit on the wall?
6.07 m
Explanation:
Given:
[tex]v_0=24.5\:\text{m/s}[/tex]
[tex]\theta_0 = 35.5°[/tex]
First, we need to find the amount of time it takes to travel a horizontal distance of 25.8 m. We know that
[tex]x = v_{0x}t \Rightarrow t = \dfrac{x}{v_0 \cos \theta_0}[/tex]
or
[tex]t = 1.29\:\text{s}[/tex]
To find the vertical height where the ball hit the wall, we use
[tex]y = v_{0y}t - \frac{1}{2}gt^2[/tex]
[tex]\:\:\:\:=(24.5\:\text{m/s})\sin 35.5(1.29\:\text{s}) \\ - \frac{1}{2}(9.8\:\text{m/s}^2)(1.29\:\text{s})^2[/tex]
[tex]\:\:\:\:=6.07\:\text{m}[/tex]
A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity.
a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:
Answer:
a) W_total = mg (2h + d) , b) E_total = - mg (h + d)
Explanation:
a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed
Let's look for work for the part that is in free fall
y = y₀ + v₀ t - ½ g t²
when he jumps out of a plane his vertical speed is zero
y =y₀ - ½ g t²
dy = 0 - ½ g 2t dt
the work in this first part is
W₁ = ∫ F dy
W₁ = mg ∫ g t dt
W₁ = m g² t² / 2
the time it takes to travel the distance y₀-y = h is
y₀-y = ½ g t²
t =[tex]\sqrt{2h/g}[/tex]
we substitute
W₁ = m g² 2h / g
W₁ = m g 2h
now we look for the work for the part with constant speed
since the velocity is constant let's use the uniform motion ratio
W₂ = F d
W₂ = mg d
the total work is
W_total = W₁ + W₂
W_total = 2mgh + m gd
W_total = mg (2h + d)
b) The change in gravitational potential energy
U = mg Δy
in the part with accelerated movement
U₁ = mg h
in the part with uniform movement
U₂ = mg d
the total potential energy is
E_total = U₁ + U₂
E_total = - mg (h + d)
what is the time taken by moving body with acceleration 0.1m/s2 if the initial or finak velocities are 20m/s and 30m/s respectively?
Answer:
t= 100s
Explanation:
use v=v0+at
plug in givens and solve for t
30=20+0.1*t
t= 100s
Assume that I = E/(R + r), prove that 1/1 = R/E + r/E
[tex]\implies {\blue {\boxed {\boxed {\purple {\sf { \frac{1}{I} = \frac{R}{E} + \frac{r}{E} }}}}}}[/tex]
[tex]\large\mathfrak{{\pmb{\underline{\orange{Step-by-step\:explanation}}{\orange{:}}}}}[/tex]
[tex]I = \frac{ E}{ R + r} \\[/tex]
[tex] ➺\:\frac{I}{1} = \frac{E}{R + r} \\[/tex]
Since [tex]\frac{a}{b} = \frac{c}{d} [/tex] can be written as [tex]ad = bc[/tex], we have
[tex]➺ \: I \: (R + r) = E \times 1[/tex]
[tex]➺ \: \frac{1}{I} = \frac{R + r}{E} \\ [/tex]
[tex]➺ \: \frac{1}{I} = \frac{R}{E} + \frac{r}{E} \\ [/tex]
[tex]\boxed{ Hence\:proved. }[/tex]
[tex]\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}[/tex]
Need ur help,,, :-[ :-{
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Answer:
Graph B express the magnetic relationship of magnetic flux and electronic flow
The gravitational field strength due to its planet is 5N/kg What does it mean?
Answer:
The weight of an object is the force on it caused by the gravity due to the planet. The weight of an object and the gravitational field strength are directly proportional. For a given mass, the greater the gravitational field strength of the planet, the greater its weight.
Weight can be calculated using the equation:
weight = mass × gravitational field strength
This is when:
weight (W) is measured in newtons (N)
mass (m) is measured in kilograms (kg)
gravitational field strength (g) is measured in newtons per kilogram (N/kg)
When an apple falls towards the earth,the earth moves up to meet the apple. Is this true?If yes, why is the earth's motion not noticeable?
Answer:
because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.
write down the following units in the ascending of their value A) mm nm cm um B) 1m 1cm 1km 1mm. convert the following units into SI without changing their values? A)3500g B)2.5km C)2h
Answer:
A) nm, um, mm, cm
B) 1mm, 1cm, 1m, 1km
A) 3500g, B) 2500m, C) 7200 seconds
Notice that all the initial spring potential energy was transformed into gravitational potential energy. If you compressed the spring to a distance of 0.200 mm , how far up the slope will an identical ice cube travel before reversing directions
Answer:
The correct answer will bs "2.41 m".
Explanation:
According to the question,
M = 50 g
or,
= 0.050 kg
[tex]\Theta = 25^{\circ}[/tex]
k = 25.9 N/m
Δx = 0.200 m
Let the traveled distance be "x".
By using trigonometry, the height will be:
⇒ [tex]h = l Sin \Theta[/tex]
hence,
⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]
[tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]
By putting the values, we get
[tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]
[tex]l=2.41 \ m[/tex]
four equal magnitude point charges 3 microcoulomb is placed at the corners of a square that is 40cm inside find the force on any one of the charges
Answer:
Approximately [tex]0.97\; \rm N[/tex]. This force would point away from the center of the square (to the left at [tex]45^\circ[/tex] above the horizontal direction.)
Explanation:
Coulomb's constant: [tex]k \approx 8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].
By Coulomb's Law, the electrostatic force between two point charges [tex]q_1[/tex] and [tex]q_2[/tex] that are separated by [tex]r[/tex] (vacuum) would be:
[tex]\displaystyle F = \frac{k \cdot q_1 \cdot q_2}{r^2}[/tex].
Consider the charge on the top-left corner of this square.
Apply Coulomb's Law to find the electrostatic force between this charge and the charge on the lower-left corner.
Convert quantities to standard units:
[tex]q_1 = q_2 = 3 \times 10^{-6}\; \rm C[/tex].
[tex]r = 0.40\; \rm m[/tex].
[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(0.40\; \rm m)^{2}} \\ &\approx 0.506\; \rm N\end{aligned}[/tex].
As the two charges are of the same sign, the electrostatic force on each charge would point away from the other charge. Hence, for the charge on the top-left corner of the square, the electrostatic force from the charge below it would point upwards.
Similarly, the charge to the right of this charge would exert an electrostatic force with the same magnitude (approximately [tex]0.506\; \rm N[/tex]) that points leftwards.
For the charge to the lower-right of the top-left charge, [tex]r = \sqrt{2} \times 0.40\; \rm m[/tex]. Therefore:
[tex]\begin{aligned}F &= \frac{k \cdot q_1 \cdot q_2}{r^2} \\ &\approx \frac{8.98755 \times 10^{-9}\; \rm N \cdot m^{2} \cdot C^{-2} \times (3 \times 10^{-6}\; \rm C)^{2}}{(\sqrt{2} \times 0.40\; \rm m)^{2}} \\ &\approx 0.253 \; \rm N\end{aligned}[/tex].
This force would point to the top-left of the top-left charge, which is [tex]45^\circ[/tex] above the horizontal direction. Decompose this force into two components that are normal to one another:
Horizontal component: approximately [tex]\sin(45^\circ) \times 0.253\; \rm N \approx 0.179\; \rm N[/tex].Vertical component: approximately [tex]\cos(45^\circ) \times 0.253\; \rm N \approx 0.179\; \rm N[/tex]Consider the net force on the top-left charge in two components:
Horizontal component: approximately [tex]0.506\; \rm N[/tex] from the charge on the top-right corner and approximately [tex]0.179\; \rm N[/tex] from the charge on the lower-right corner. Both components point to the left-hand side. [tex]F_x \approx 0.506\; \rm N + 0.179\; \rm N = 0.685\;\rm N[/tex] (to the left).Vertical component: approximately [tex]0.506\; \rm N[/tex] from the charge on the lower-left corner and approximately [tex]0.179\; \rm N[/tex] from the charge on the lower-right corner. Both components point upwards. [tex]F_y \approx 0.506\; \rm N + 0.179\; \rm N = 0.685\;\rm N[/tex] (upward).Combine these two components to find the magnitude of the net force on this charge:
[tex]\begin{aligned}F &= \sqrt{{F_x}^{2} + {F_y}^{2}} \\ &\approx \sqrt{0.685^2 + 0.685^2 }\; \rm N \\ &\approx 0.97\; \rm N\end{aligned}[/tex].
This force would point to the top-left of this charge (also at [tex]45^\circ[/tex] above the horizontal direction, away from the center of the square) because its horizontal and vertical components have the same magnitude.
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
Why is the melting of ice a physical change?
A. It changes the chemical composition of water.
B. It does not change the chemical composition of water.
C. It creates new chemical bonds.
D. It forms new products.
E. It is an irreversible change that forms new products.
It does not change the chemical composition of water.
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
Two cylindrical resistors are made from copper. The first one is of length L and of radius r . The 2nd resistor is of length 6L and of radius 2r. The ratio of these two resistances R1/R2 is:
Answer:
[tex]R1/R2=\frac{2}{3}[/tex]
Explanation:
From the question we are told that:
1st's Length [tex]l=L[/tex]
1st's radius [tex]r=r[/tex]
2nd's Length [tex]l=6L[/tex]
2nd's radius [tex]r=2r[/tex]
Generally the equation for Resistance R is mathematically given by
[tex]R=\frac{\rho L}{\pi r^2}[/tex]
Therefore
[tex]R_1=\frac{\rho L}{\pi r^2}[/tex]
And
[tex]R_2=\frac{\rho 6L}{\pi (2r)^2}[/tex]
Therefore
[tex]R1/R2=\frac{\frac{\rho L}{\pi r^2}}{\frac{\rho 6L}{\pi (2r)^2}}[/tex]
[tex]R1/R2=\frac{2}{3}[/tex]
I need help on this physics problem.
Answer:
the speed of the nerve impulse in miles per hour is 201.59 mi/hr
Explanation:
Given;
the speed of the nerve impulse, v = 90.1 m/s
To convert this speed in meters per second to miles per hour, we use the following method;
1,609 meter = 1 mile
3,600 s = 1 hour
[tex]v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr[/tex]
Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of
A source emits sound at a fixed constant frequency f. If you run towards the source, the frequency you hear is
Answer:
increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.
Explanation:
A cable is lifting a construction worker and a crate, as the drawing shows. The weights of the worker and crate are 965 and 1510 N, respectively. The acceleration of the cable is 0.620 m s 2 , upward. What is the tension in the cable (a) below the worker and ( b) above the worker
Answer:
Explanation:
a)
Below the worker , the tension in cable is pulling the crate . Let the tension be T₁ .
weight of crate is acting downwards .
Total weight 1510 N.
Net force acting on both = T₁ - 1510
Applying second law of Newton ,
T₁ - 1510 = 1510 / 9.8 x 0.62 [ 1510 / 9.8 = mass of crate ]
T₁ - 1510 = 95.5
T₁ = 1605.5 N.
b )
Above the worker , the tension in cable is pulling both the worker and the crate . Let the tension be T₂ .
weight of both worker and crate is acting downwards .
Total weight = 965 + 1510 = 2475 N.
Net force acting on both = T₂ - 2475
Applying second law of Newton ,
T₂ - 2475 = 2475 / 9.8 x 0.62 [ 2475 / 9.8 = mass of both worker and crate ]
T₂ - 2475 = 156.6
T₂ = 2631.6 N.
Water is falling on the blades of a turbine at a rate of 100 kg/s from a certain spring. If the height of spring be 100m, then the power transferred to the turbine will be: a) 100 KW b) 10 KW c) 1 KW d) 100 W
Answer:
Natae Si Jordan Kaya Sya Napaihe
Explanation:
haha
15.1.2 Exam: Semester Exam
ŽA
This graph shows the energy of a reaction over time. Which statement is
true?
Potential energy
N
Reaction progress
O A. G represents the activation energy
OB. H represents the energy of the products,
O C. G represents the energy of the products.
< PREVIOUS
Answer: D. F represents the activation energy
Explanation:
The activation energy is the energy required to get the reactants to begin reacting with one another such that products are created. This energy ranges from the minimum to the maximum energy required.
F is therefore the activation energy because it shows the range between the minimum energy it took for the reaction to start and the maximum energy that was required to continue the reaction.
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Answer:
I DON'T UNDERSTAND
Explanation:
GUESS A MISUNDERSTANDING PLZ PUT A UNDERSTANDABLE QUESTION.
A proton is held at rest in a uniform electric field. When it is released, the proton will gain:_________
a) electrical potential energy.
b) kinetic energy.
c) both kinetic energy and electric potential energy.
d) either kinetic energy or electric potential energy.
If the force on an object is negative, what is known about the change in velocity?
O The velocity will decrease.
O The velocity will increase.
O The velocity will not change.
O The object will change direction.
What has a wind speed of 240 kph or greater?
Answer:
SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
The current in a conductor is 2.5A .explain the meaning of this statement
Answer:
In 1 second the amount of charge flowing through the conductor is 2.5 Q.
Explanation:
A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800-N firefighter has climbed 4.00 m along the ladder from the bottom. (b) If the ladder is just on the verge of slipping when the firefighter is 9.00 m from the bottom, what is the coefficient of static friction between ladder and ground
Answer:
a) fr = 266.92 N, fy = 1300 N, b) μ = 0.36
Explanation:
a) This is a balancing act.
Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive
-w x - W x₂ + R y = 0 (1)
usemso trigonometry to find distances
cos 60.08 = x / 7.5
x = 7.5 cos 60.08
x = 3.74 m
fireman
cos 60.08 = x₂ / 4
x2 = 4 cos 60
x2 = 2 m
wall support
sin 60.08 = y / 15
y = 15 are 60.08
y = 13 m
we substitute in equation 1
R y = w x + W x2
R = (w x + W x2) / y
R = (500 3.74 +800 2) / 13
R = 266.92 N
now let's write the expressions for the translational equilibrium
X axis
R -fr = 0
R = fr
fr = 266.92 N
Y Axis
Fy - w-W = 0
fy = 500 + 800
fy = 1300 N
b) ask the friction coefficient
the firefighter's distance is
cos 60.08 = x₃ / 9.00
x₃ = 9 cos 60
x₃ = 5.28 m
from equation 1
R = (w x + W x₃) / y
R = 500 3.74 + 800 5.28) / 13
R = 468.769 N
we saw that
fr = R = 468.769
The expression for the friction force is
fr = μ N
in this case the normal is the ratio to pesos
N = Fy
N = 1300 N
μ N = fr
μ = fr / N
μ = 468,769 / 1300
μ = 0.36
Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
Answer:
periodic
Explanation:
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward in 1.71 s. He breaks the tackle and runs straight forward another 24.0 m in 5.20 s. Calculate his average velocity (in m/s) for the entire motion. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answer:
Average Velocity = 3.63 m/s
Explanation:
First, we will calculate the total displacement of the quarterback, taking forward direction as positive:
Total Displacement = 15 m - 3 m + 24 m = 36 m
Now, we will calculate the total time taken for this displacement:
Total Time = 3 s + 1.71 s + 5.2 s = 9.91 s
Therefore, the average velocity will be:
[tex]Average\ Velocity = \frac{Total\ Displacement}{Total\ Time}\\\\Average\ Velocity = \frac{36\ m}{9.91\ s}[/tex]
Average Velocity = 3.63 m/s
