Answer:
[tex]N=0.37N[/tex]
Explanation:
Mass [tex]m=55g=>0.055kg[/tex]
Soapstone Density [tex]\rho_s=3000kg/m^2[/tex]
Whisky Density [tex]\rho_w=940kg/m^2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=U+N[/tex]
Therefore
[tex]N=m*g-(\frac{m}{\rho_s})*\rho_w*g[/tex]
[tex]N=0.055*9.81 - {(\frac{0.055}{3000})*940*9.81}[/tex]
[tex]N=0.37N[/tex]
the Period T of oscillation of a Single Pendulum depends on the length l, and acceleration g. Determine the exact form of the dependence.
Answer:
[tex]{ \tt{check \: in \: the \: pic}}[/tex]
Two charged particles exert an electric force of 27 N on each other. What will the magnitude of the force be if the distance between the particles is reduced to one-third of the original separation
Answer:
243 N
Explanation:
The formula for electromagnetic force is F= Kq1q2/r^2
where r is the distance between the charges, if the distance between the charges is reduced by 1/3 then F will increase by 9 [(1/3r)^2 becomes 1/9r which is 9F] so 27*9 is 243N
High-speed stroboscopic photographs show that the head of a -g golf club is traveling at m/s just before it strikes a -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at m/s. Find the speed of the golf ball just after impact.
The question is incomplete. The complete question is :
High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a 50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.
Solution :
We know that momentum = mass x velocity
The momentum of the golf club before impact = 0.200 x 60
= 12 kg m/s
The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.
Now the momentum of the club after the impact is = 0.2 x 40
= 8 kg m/s
Therefore the momentum of the ball is = 12 - 8
= 4 kg m/s
We know momentum of the ball, p = mass x velocity
4 = 0.050 x velocity
∴ Velocity = [tex]$\frac{4}{0.050}$[/tex]
= 80 m/s
Hence the speed of the golf ball after the impact is 80 m/s.
Two objects are at rest on a frictionless surface. Object 1 has a greater mass than object 2.
(a) When a constant force is applied to object 1, it accelerates through a distance d. The force is removed from object 1 and is applied to object 2. At the moment when object 2 has accelerated through the same distance d, which statements are true? (Select all that apply.)
K1 < K2 p1 = p2 p1 < p2 p1 > p2 K1 > K2 K1 = K2
(b) When a force is applied to object 1, it accelerates for a time interval ?t. The force is removed from object 1 and is applied to object 2. Which statements are true after object 2 has accelerated for the same time interval ?t? (Select all that apply.)
K1 > K2 K1 = K2 p1 = p2 p1 > p2 K1 < K2 p1 < p2
Answer:
Look at explanation
Explanation:
a) Kinetic energy= ΔW. W=Fd, and since in both scenarios the same force and same distance is travelled. K1=K2. I am assuming that the objects are at non zero height so by P=mgh, P1>P2
b. Again I am assuming that the objects are at non zero height so by P=mgh, P1>P2. A heavier mass, a constant force means a smaller acceleration. So a1<a2. We can then use x-x0=v0t+1/2at² and since v0=0, x-x0(d)=1/2at². Solve for t²=2d/a. Since t is the same for both but a1<a2, d1<d2. And since Kinetic Energy=ΔW, W=Fd and F is constant while d1<d2, K1<K2.
According to the question,
Potential energy be "P".Kinetic energy be "K".(a)
Word done towards both the block will be similar.
So,
→ [tex]P1 = P2[/tex]
→ [tex]K1= K2[/tex]
(b)
We know,
→ [tex]a = \frac{F}{M}[/tex]
or,
→ [tex]V = a\times t[/tex]
Now,
→ [tex]K = \frac{1}{2} MV^2[/tex]
[tex]= 0.5\times M\times V^2[/tex]
[tex]=0.5\times M\times (\frac{F^2}{M^2} )\times t^2[/tex]
[tex]= 0.5\times F^2\times \frac{t^2}{M}[/tex]
The force and t will be same. So K of the smaller mass will be greater than the larger mass.
hence,
→ [tex]K1<K2[/tex]
Thus the above responses are correct.
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A point on a rotating wheel (thin loop) having a constant angular velocityy of 300 rev/min, the wheel has a radius of 1.5m and a mass of 30kg. (I = mr^2)
a.) Determine the linear regression
b.) At this given angular velocity, what is the rotational kinetic energy?
Answer:
Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J
Explanation:
a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)
a= ω^2*r
ω= 300rev/min
convert into rev/s
300/60= 5rev/s
a= 18.75m/s^2
b) use Krot= 1/2 Iω^2
plug in gives
1/2(30*2.25)(25)= 843.75 J
In the early 1900's ____
began leading the automobile exploration in the US automotive industry.
-Karl Benz
-Henry Ford
-Gottlieb Daimler
-None of the above
Answer:
Henry Ford
Explanation:
he built the first ford
A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m along the axis and above the center of the loop
Answer:
[tex]B=2.91\ \mu T[/tex]
Explanation:
Given that,
The current in the loop, I = 2 A
The radius of the loop, r = 0.4 m
We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :
[tex]B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}[/tex]
Put all the values,
[tex]B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T[/tex]
So, the required magnetic field is equal to [tex]2.91\ \mu T[/tex].
If the moon started it's orbit around the Earth from a spot in line with a certain star, it will return to that same spot in about _______.
Answer:
1 month
Explanation:
Al and Ben are on roller skates and Al rolls into Ben. Al exerts a force of 30 N on Ben when they
collide. Explain what force Ben exerts on AI.
Answer:
Reaction force
Explanation:
Newton´s 3rd law says that every force exerted in nature has an equal and opposite force.
For example here, when Al exerts force on Ben, Ben exerts the same amount of force (30N) on Al.
Al exerts the action force and Ben exerts the reaction force.
An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made from the power source and a 47.2 H inductor, determine the inductive reactance and the rms current through the inductor.
Answer:
The inductance is 17784.96 ohm and rms current is 4.77 mA.
Explanation:
Voltage, V = 120 V
frequency, f = 60 Hz
Inductance, L = 47.2 H
The rms voltage is
[tex]V_{rms}=\frac{V_o}{\sqrt 2}\\\\V_{rms}=\frac{120}{\sqrt 2}\\\\V_{rms} = 84.87 V[/tex]
The reactance is given by
[tex]X_L = 2\pi f L\\\\X_L = 2\times 3.14\times 60\times 47.2 \\\\X_L = 17784.96 ohm[/tex]
The rms current is
[tex]I_{rms} =\frac{V_{rms}}{X_L}\\\\I_{rms}=\frac{84.87}{17784.96}\\\\I_{rms} = 4.77\times 10^{-3} A = 4.77 mA[/tex]
TIME REMAINING
45:43
What are possible units for impulse? Check all that apply.
kg • m
kg • meters per second
N • s
N • m
StartFraction Newtons times meters per second EndFraction
Answer:
n.m maby
Explanation:
i think or its kg m/s
Answer:
answer (B) & (C)
Explanation:
kg • /N • s
A small ball of uniform density equal to 1/2 the density of water is dropped into a pool from a height of 5m above the surface. Calculate the maximum depth the ball reaches before it is returned due to its bouyancy. (Omit the air and water drag forces).
Answer:
1.67 m
Explanation:
The potential energy change of the small ball ΔU equals the work done by the buoyant force, W
ΔU = -W
Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5
ΔU = mgΔh
ΔU = ρVgΔh
Now, W = ρ'VgΔh' where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h
So, ΔU = -W
ρVgΔh = -ρ'VgΔh'
ρVg(h - 5) = -ρ'Vgh
ρ(h - 5) = -ρ'h
Since the density of the small ball equals 1/2 the density of water,
ρ = ρ'/2
ρ(h - 5) = -ρ'h
(ρ'/2)(h - 5) = -ρ'h
ρ'(h - 5)/2 = -ρ'h
(h - 5)/2 = -h
h - 5 = -2h
h + 2h = 5
3h = 5
h = 5/3
h = 1.67 m
So, the maximum depth the ball reaches is 1.67 m.
Astronauts in space move a toolbox from its initial position ????????→=<15,14,−8>m to its final position ????????→=<17,14,−1>m. The two astronauts each push on the box with a constant force. Astronaut 1 exerts a force ????1→=<18,7,−12>???? and astronaut 2 exerts a force ????2→=<16,−10,16>????.
Required:
What is the total work performed on the toolbox?
If both forces are measured in Newtons, then the net force is
F = (18, 7, -12) N + (16, -10, 16) N = (34, -3, 4) N
The toolbox undergoes a displacement (i.e. change in position) in the direction of the vector
d = (17, 14, -1) m - (15, 14, -8) m = (2, 0, -9) m
The total work done by the astronauts on the toolbox is then
F • d = (34, -3, 4) N • (2, 0, -9) m = (68 + 0 - 36) N•m = 32 J
The work done by the two astronauts is equal to 96 J.
What is work done?work done?Work done is defined as the product of force applied and the distance moved by the force.
Work done = Force × DistanceThe forces applied = 18+16 N, 7+ -10 N, and -12 + 16N
Forces = 34 N, -3 N, and 4N
Distances = (17 - 15, 14 - 14, -1 - - 8) m
Distances = 2, 0, 7
Work done = 34 × 2 + -3 × 0 + 4 × 7
Work done = 96 J
Therefore, the work done by the two astronauts is equal to 96 J.
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A water-balloon launcher with mass 2 kg fires a 0.75 kg balloon with a
velocity of 14 m/s to the west. What is the recoil velocity of the launcher?
What is the answer
Answer:
5.25 m/s to the east
Explanation:
Applying,
MV = mv.............. Equation 1
Where M = mass of the launcher, V = recoil velocity of the launcher, m = mass of the balloon, v = velocity of the balloon
make V the subject of the equation
V = mv/M............ Equation 2
From the question,
M = 2 kg, m = 0.75 kg, v = 14 m/s
Substitute these values into equation 2
V = (0.75×14)/2
V = 5.25 m/s to the east
* 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 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is the spring constant of the bungee cord, assuming it has negligible mass compared to that of the jumper in N/m
Answer:
2.80N/m
Explanation:
Given data
mass m= 56kg
perios T= 11.2s
The expression for the period is given as
T=2π√m/k
Substitute
11.2= 2*3.142*√56/k
square both sides
11.2^2= 2*3.142*56/k
125.44= 351.904/k
k=351.904/125.44
k= 2.80N/m
Hence the spring constant is 2.80N/m
While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases
Answer:
a
Explanation:
The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.
A 100-m long transmission cable is suspended between two towers. If the mass density is 18.2 g/cm and the tension in the cable is 6543 N, what is the speed (m/s2) of transverse waves on the cable
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
The AM radio station WDRJ broadcasts news and sports at a frequency of 704 kHz (kilohertz). What is the wavelength of the radio waves this station broadcasts? _____ meters
Give your answer to the nearest hundredth of a meter (two places after the decimal). Just enter the number; do NOT use scientific notation.
Answer:
AM broadcasts occur on North American airwaves in the medium wave frequency range of 525 to 1705 kHz (known as the “standard broadcast band”). The band was expanded in the 1990s by adding nine channels from 1605 to 1705 kHz.
A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. If the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill?
A) 3.57 m.
B) 4.28 m.
C) 3.14 m.
D) 2.68 m.
Answer:
A(3.56m)
Explanation:
We have a conservation of energy problem here as well. Potential energy is being converted into linear kinetic energy and rotational kinetic energy.
We are given ω= 4.27rad/s, so v = ωr, which is 6.832 m/s. Place your coordinate system at top of the hill so E initial is 0.
Ef= Ug+Klin+Krot= -mgh+1/2mv^2+1/2Iω^2
Since it is a solid uniform disk I= 1/2MR^2, so Krot will be 1/4Mv^2(r^2ω^2= v^2).
Ef= -mgh+3/4mv^2
Since Ef=Ei=0
Mgh=3/4mv^2
gh=3/4v^2
h=0.75v^2/g
plug in givens to get h= 3.57m
The following contribute to accidents when a teen driver has other teens as passengers
Answer:
When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
What are the factors contribute to accidents when a teen driver has other teens as passengers?When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.
Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.
Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.
Overconfidence, lack of focus, and phone while driving are the factors contribute to accidents when a teen driver controls other teens as passengers,
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A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.
Answer:
Explanation:
Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .
A )
Intensity of sound at 1 m distance = 60 /4 π d²
d = 1 m
Intensity of sound at 1 m distance = 60 /(4 π 1²)
= 4.78 W m⁻² s⁻¹
B )
Intensity of sound at 1.5 m distance = 60 /4 π d²
d = 1.5 m
Intensity of sound at 1 m distance = 60 /(4 π 1.5²)
= 2.12 W m⁻² s⁻¹
C )
Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²
d = 1.5 m
= 4 x 60 /(4 π 1.5²)
= 8.48 W m⁻² s⁻¹
D )
Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .
.06 /4 π d² = 10⁻¹²
d² = .06 /4 π 10⁻¹²
d = 6.9 x 10⁴ m .
The total resistance of a parallel circuit is 25 ohms. If the total current is 100mA, how much current is through a 220 ohm resistor that makes up part of the parallel circuit?
Answer:
The current across the resistance is 0.011 A.
Explanation:
Total resistance, R = 25 ohms
Total current, I = 100 mA = 0.1 A
Let the voltage is V.
By the Ohm's law
V = I R
V = 0.1 x 25 = 2.5 V
Now the resistance is R' = 220 ohm
As they are in parallel so the voltage is same. Let the current is I'.
V = I' x R'
2.5 = I' x 220
I' = 0.011 A
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis
Answer: -5×10-3
Explanation:
E=kq/r
A child with a weight of 230 N swings on a playground swing attached to 2.20-m-long chains. What is the gravitational potential energy of the child-Earth system relative to the child's lowest position at the following times?
(a) when the chains are horizontal (in J)
(b) when the chains make an angle of 33.0° with respect to the vertical (in J)
(c) when the child is at his lowest position (in J)
Answer:
a) U = 506 J, b) U = 37.11 J, c) U = 0
Explanation:
The gravitational power energy is given by the expression
U = m g (y -y₀)
In general, a reference system is set that allows the expression to be simplified, in this case let's assume the reference system at the child's lowest point, therefore y₀ = 0
Let's use trigonometry to find the child's height
h = y = L - L cos θ
we substitute
U = m g L (1 - cos θ)
a) when the chain is horizontal θ = 90 and cos 90 = 0
U = mg L
weight and mass are related
W = mg
m = W / g
U = 230 2.20
U = 506 J
b) θ = 33.0º
cos 33 = 0.83867
U = 230 (1 - 0.83867)
U = 37.11 J
c) in this case θ = 0 cos 0 = 1
U = 0
the unit for
ΔL/L
is
Answer:
the unit for ΔL/L is "unitless".
Explanation:
Given;
ΔL/L
by physics convection, the above parameters can be defined as;
delta L (ΔL) is change in length, with SI unit as meters (m),
L is the original length of the material, with SI unit as meters (m)
The ratio of the change in length to the original length has no unit since both units cancel out during the division.
[tex]\frac{\Delta L }{L } = \frac{(m)}{(m)} = \ unitless[/tex]
This ratio (ΔL/L), is also called tensile strain and it has no unit.
Therefore, the unit for ΔL/L is "unitless".
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.
1. A block of mass m = 10.0 kg is released with a speed v from a frictionless incline at height 7.00 m. The
block reaches the horizontal ground and then slides up another frictionless incline as shown in Fig. 1.1. If the
horizontal surface is also frictionless and the maximum height that the block can slide up to is 26.0 m, (a) what
is the speed v of the block equal to when it is released and (b) what is the speed of the block when it reaches
the horizontal ground? If a portion of length 1 2.00 m on the horizontal surface is frictional with coefficient
of kinetic friction uk = 0.500 (Fig. 1.2) and the block is released at the same height 7.00 m with the same
speed v determined in (a), (c) what is the maximum height that the block can reach, (d) what is the speed of the
block at half of the maximum height, and (e) how many times will the block cross the frictional region before
it stops completely?
1 = 2.00 m (frictional region)
Let A be the position of the block at the top of the first incline; B its position at the bottom of the first incline; C its position at the bottom of the second incline; and D its position at the top of the second incline. I'll denote the energy of the block at a given point by E (point).
At point A, the block has total energy
E (A) = (10.0 kg) (9.80 m/s²) (7.00 m) + 1/2 (10.0 kg) v₀²
E (A) = 686 J + 1/2 (10.0 kg) v₀²
At point B, the block's potential energy is converted into kinetic energy, so that its total energy is
E (B) = 1/2 (10.0 kg) v₁²
The block then slides over the horizontal surface with constant speed v₁ until it reaches point C and slides up a maximum height of 26.0 m to point D. Its total energy at D is purely potential energy,
E (D) = (10.0 kg) (9.80 m/s²) (26.0 m) = 2548 J
Throughout this whole process, energy is conserved, so
E (A) = E (B) = E (C) = E (D)
(a) Solve for v₀ :
686 J + 1/2 (10.0 kg) v₀² = 2548 J
==> v₀ ≈ 19.3 m/s
(b) Solve for v₁ :
1/2 (10.0 kg) v₁² = 2548 J
==> v₁ ≈ 22.6 m/s
Now if the horizontal surface is not frictionless, kinetic friction will contribute some negative work to slow down the block between points C and D. Check the net forces acting on the block over this region:
• net horizontal force:
∑ F = -f = ma
• net vertical force:
∑ F = n - mg = 0
where f is the magnitude of kinetic friction, a is the block's acceleration, n is the mag. of the normal force, and mg is the block's weight. Solve for a :
n = mg = (10.0 kg) (9.80 m/s²) = 98.0 N
f = µn = 0.500 (98.0 N) = 49.0 N
==> - (49.0 N) = (10.0 kg) a
==> a = - 4.90 m/s²
The block decelerates uniformly over a distance 2.00 m and slows down to a speed v₂ such that
v₂² - v₁² = 2 (-4.90 m/s²) (2.00 m)
==> v₂² = 490 m²/s²
and thus the block has total/kinetic energy
E (C) = 1/2 (10.0 kg) v₂² = 2450 J
(c) The block then slides a height h up the frictionless incline to D, where its kinetic energy is again converted to potential energy. With no friction, E (C) = E (D), so
2450 J = (10.0 kg) (9.80 m/s²) h
==> h = 25.0 m
(d) At half the maximum height, the block has speed v₃ such that
2450 J = (10.0 kg) (9.80 m/s²) (h/2) + 1/2 (10.0 kg) v₃²
==> v₃ ≈ 15.7 m/s
The block loses speed and thus energy as it moves between B and C, but its energy is conserved elsewhere. If we ignore the inclines and pretend that the block is sliding over a long horizontal surface, then its velocity v at time t is given by
v = v₁ + at = 22.6 m/s - (4.90 m/s²) t
The block comes to a rest when v = 0 :
0 = 22.6 m/s - (4.90 m/s²) t
==> t ≈ 4.61 s
It covers a distance x after time t of
x = v₁t + 1/2 at ²
so when it comes to a complete stop, it will have moved a distance of
x = (22.6 m/s) (4.61 s) + 1/2 (-4.90 m/s²) (4.61 s)² = 52.0 m
(e) The block crosses the rough region
(52.0 m) / (2.00 m) = 26 times
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec