Answer:
a) 4.2 m/s
b) 13.6 m/s^2
Explanation:
She is jumping, and when her feet no longer touch the ground she is in free fall, only affected by the acceleration of gravity.
The equation for position under constant acceleration is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
We set up a reference system that has its origin at the point her center of mass is when she is standing and the positive Y axis points upwards, then:
Y0 = 0 m
a = -9.81 m/s^2
The equation for speed under constant acceleration is:
V(t) = Vy0 + a * t
We know that when she reaches her highes point her vertical speed will be zero because that is wehn her movement changes direction. We'll call this moment t1.
0 = Vy0 + a * t1
a * t1 = -Vy0
t1 = -Vy0/a
If we replace this value on the position equation we can find her initial speed:
Y(t1) = Y0 - Vy0 * Vy0/a + 1/2 * a * (-Vy0/a)^2
Y(t1) = - Vy0^2/a + 1/2 * Vy0^2/a
Y(t1) = -1/2 * Vy0^2 / a
Vy0^2 = -2 * a * Y(t1)
[tex]Vy0 = \sqrt{-2 * a * Y(t1)}[/tex]
[tex]Vy0 = \sqrt{-2 * (-9.81) * 0.9} = 4.2 m/s[/tex]
I assume her center of mass is at half her height, so when she is standing it would be at 93 cm of the grouind, and when she is crouching at 46.5 cm.
Therefore when she jumps her centr of mass moves 0.465 m before leaving the ground.
During that trajectory she moves with acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a *t^2
In this case her initial position is
Y0 = -0.465
Her initial speed is
Vy0 = 0
At t=t0 her position will be zero
The equation for speed under constan acceleration is
Vy(t) = Vy0 + a * t
Her speed at t0 will be 4.2 m/s
4.2 = a * t0
t0 = 4.2 / a
0 = -0.465 - 1/2 * 9.81 * (4.2 / a)^2
0.465 = 4.9 * 17.6 / a^2
a^2 = 86.2 / 0.465
[tex]a = \sqrt{185.4} = 13.6 m/s^2[/tex]
Amy initially 5.0 mi west of the United Center's Michael Jordan statue is running with a constant velocity of 5.0 mi/h due east. Alejandro is initially 4.0 mi east of the statue and is running with a constant velocity of 7.0 mi/h due west. How far are the runners from the statue when they meet?
Answer:
The statue is 1.67 miles west of both the runners
Explanation:
Wherever Amy and Alejandro meet they will have covered a total distance of 5+4 = 9 mi
They will also have run the same amount of time if they started at the same moment = t
Speed of Amy = 5 mi/h
Speed of Alejandro = 7 mi/h
Distance = Speed × Time
Distance travelled by Amy = 5t
Distance travelled by Alejandro = 7t
Total distance run by Amy and Alejandro is
5t+7t = 9
[tex]\\\Rightarrow 12t=9\\\Rightarrow t=\frac{12}{9}\\\Rightarrow t=\frac{4}{3}\ hours[/tex]
Distance travelled by Amy
[tex]5\times t=5\times \frac{4}{3}=\frac{20}{3}\\ =6.67\ miles[/tex]
The distance of Amy from the statue would be
6.67 - 5 = 1.67 miles
So, the statue is 1.67 miles west of both the runners
How do resistors in parallel affect the total resistance?
Answer:
They're going to increase the total resistance as [tex]R_{T} = \sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1}[/tex]
Explanation:
If the resistors are in parallel, the potential difference is the same for each resistor. But the total current is the sum of the currents that pass through each of the resistors. Then
[tex]I = I_1 + I_2 + ... + I_N[/tex]
where
[tex]I_i = \frac{V_i}{R_i}[/tex]
but
[tex]V_i = V_j = V[/tex] for [tex]i,j= 1, 2,..., N[/tex]
so
[tex]I = \frac{V}{R_1}+ \frac{V}{R_2} + ... + \frac{V}{R_N} = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)V = \frac{V}{R_T}[/tex]
where
[tex]R_T = \left(\frac{1}{R_1} +\frac{1}{R_2} + ... + \frac{1}{R_N}\right)^{-1} =\sum\limits_{i=1}^N \left(\frac{1}{R_i} \right)^{-1} [/tex]
In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is water, it floats fully submerged. In the second liquid it floats with height h/7 above the liquid surface. What is the relative density (the density relative to that of water) of the second liquid?
Answer:
The relative density of the second liquid is 7.
Explanation:
From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.
Let us assume that the volume of the object is 'V'
Thus for the liquid in which the block is completely submerged
The buoyant force should be equal to weight of liquid
Mathematically
[tex]F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)[/tex]
Thus for the liquid in which the block is 1/7 submerged
The buoyant force should be equal to weight of liquid
Mathematically
[tex]F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)[/tex]
Comparing equation 'i' and 'ii' we see that
[tex]\rho_{2}=7\times \rho _{1}[/tex]
Since the first liquid is water thus [tex]\rho _{1}=1gm/cm^3[/tex]
Thus the relative density of the second liquid is 7.
Answer:
7
Explanation:
Let the density of second liquid is d.
Density of water = 1 g/cm^3
In case of equilibrium, according to the principle of flotation, the weight of the body is balanced by the buoyant force acting on the body.
Let A be the area of cross section of block and D be the density of material of block and h be the height.
For first liquid (water):
Weight of block = m x g = A x h x D x g .... (1)
Buoyant force in water = A x h x 1 x g ..... (2)
Equating (1) and (2) we get
A x h x D x g = A x h x 1 x g
D = 1 g/cm^3
For second liquid:
Weight of block = m x g = A x h x D x g .... (1)
Buoyant force in second liquid = A x h/7 x d x g ..... (2)
Equating (1) and (2) we get
A x h x D x g = A x h/7 x d x g
D = d/7
d = 7 D = 7 x 1 = 7 g/cm^3
Thus, the relative density of second liquid is 7.
A closed system consisting of 4 lb of gas undergoes a process in which the relation between pressure and volume is pVn = constant. The process begins with p1 = 15 psi, v1 = 1.25 ft3/lb and ends with p2 = 53 psi and v2 = 0.5 ft3/lb. Determine: a) the volume in ft3 occupied by the gas at states 1 and 2, and b) the value of n.
Answer:
V1=5ft3
V2=2ft3
n=1.377
Explanation:
PART A:
the volume of each state is obtained by multiplying the mass by the specific volume in each state
V=volume
v=especific volume
m=mass
V=mv
state 1
V1=m.v1
V1=4lb*1.25ft3/lb=5ft3
state 2
V2=m.v2
V2=4lb*0.5ft3/lb= 2ft3
PART B:
since the PV ^ n is constant we can equal the equations of state 1 and state 2
P1V1^n=P2V2^n
P1/P2=(V2/V1)^n
ln(P1/P2)=n . ln (V2/V1)
n=ln(P1/P2)/ ln (V2/V1)
n=ln(15/53)/ ln (2/5)
n=1.377
A grasshopper makes four jumps. The displacement vectors are (1) 31.0 cm, due west; (2) 26.0 cm, 44.0 ° south of west; (3) 22.0 cm, 56.0 ° south of east; and (4) 23.0 cm, 75.0 ° north of east. Find (a) the magnitude and (b) direction of the resultant displacement. Express the direction as a positive angle with respect to due west.
Answer:
(a) 34.47 cm
(b) [tex]24.09^\circ[/tex] south of west
Explanation:
Let us draw a figure representing the individual displacement vectors in the four jumps as shown in the figure attached with this solution.
Now, let us try to write the four displacement vectors in in terms of unit vectors along the horizontal and the vertical axis.
[tex]\vec{d}_1= 31\ cm\ west = -31\ cm\ \hat{i}\\\vec{d}_2= 26\ cm\ south\ of\ west = -26\cos 44^\circ\ \hat{i} -26 \sin 44^\circ\ \hat{j}=(-18.72\ \hat{i}-18.06\ \hat{i})\ cm\\\vec{d}_3= 22\ cm\ south\ of\ east = 22\cos 56^\circ\ \hat{i} -22 \sin 56^\circ\ \hat{j}=(12.30\ \hat{i}-18.23\ \hat{i})\ cm\\\vec{d}_4= 23\ cm\ north\ of\ east = 23\cos 75^\circ\ \hat{i} +23\sin \sin 75^\circ\ \hat{j}=(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\[/tex]
Now, the vector sum of all these vector will give the resultant displacement vector.
[tex]\vec{D} = \vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} = -31\ cm\ \hat{i}+(-18.72\ \hat{i}-18.06\ \hat{i})\ cm+(12.30\ \hat{i}-18.23\ \hat{i})\ cm+(5.95\ \hat{i}+22.22\ \hat{i})\ cm\\\Rightarrow \vec{D} =(-31.47\ \hat{i}-14.07\ \hat{i})\ cm[/tex]
Part (a):
The magnitude of the resultant displacement vector is given by:
[tex]D=\sqrt{(-31.47)^2+(-14.07)^2}\ m = 34.47\ m[/tex]
Part (b):
Since the resultant displacement vector indicates that the final position of the vector lies in the third quadrant, the vector will make some positive angle in the direction south of west given by:
[tex]\theta = \tan^{-1}(\dfrac{14.07}{31.47})= 24.09^\circ[/tex]
To find the resultant displacement of the grasshopper, we can break down the vectors into their x and y components, and then sum up the components separately. After performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction is approximately 38.3° south of west.
Explanation:To find the resultant displacement of the grasshopper, we need to add the individual displacement vectors. We can do this by breaking down each vector into its x and y components.
For vector (1) with a magnitude of 31.0 cm due west, the x component is -31.0 cm and the y component is 0.
Similarly, for the other vectors, the x and y components are:
(2): x = -26.0*cos(44.0) cm, y = -26.0*sin(44.0) cm(3): x = 22.0*cos(56.0) cm, y = -22.0*sin(56.0) cm(4): x = 23.0*cos(75.0) cm, y = 23.0*sin(75.0) cmNow, we can sum up the x components and y components separately to find the resultant displacement.
The magnitude of the resultant displacement can be found using the formula:
resultant magnitude = sqrt((sum of x components)^2 + (sum of y components)^2)
The direction of the resultant displacement can be found using the formula:
resultant direction = atan2((sum of y components), (sum of x components))
Plugging in the values and performing the calculations, we find that the magnitude of the resultant displacement is approximately 39.4 cm and the direction of the resultant displacement is approximately 38.3° south of west.
A red train traveling at 72km/h and a green train traveling at 144km/h are headed towards each other along a straight, level track. When they are 950m apart, each engineer sees the other's train and applies the breaks. The breaks slow each train at the rate of 1.0m/s^2. is there a collision? if so, give the speed of the red train and the speed of the green train at impact, respectively. If not, give the separation between the trains when they stop.
Answer:
given,
speed of red train = 72 km/h = 72× 0.278 = 20 m/s
speed of green train = 144 km/h = 144 × 0.278 = 40 m/s
deceleration of both the train = 1 m/s²
distance between the train when they start decelerating = 950 m
using equation of motion
v² = u² + 2 a s
distance taken by the red train to stop
v² = u² + 2 a s
0 = 20² - 2×1×s
s = 200 m
distance taken by the blue train to stop
v² = u² + 2 a s
0 = 40² - 2×1×s
s = 800 m
so, both train will be cover 200 + 800 = 1000 m
hence, there will be collision between both the trains.
distance traveled by the green train will be 750 m
distance traveled by the red train will be 200 m
so, velocity of the red train will be zero
velocity of the green train will be
v² = u² + 2 a s
v² = 40² - 2 × 1× 750
v = 10 m/s
hence, velocity of the green train will be 10 m/s.
A highway curve forms a section of a circle. A car goes around the curve. Its dashboard compass shows that the car is initially heading due east. After it travels 830. m, it is heading 15.0° south of east. Find the radius of curvature of its path. (Use the correct number of significant figures.)
Answer:
R = 3170.36m or R = 186.5m
Explanation:
For this problem, we have either trajectory (a), assuming that the car was going south-east, or trajectory (b), assuming the car was going north-east.
In both cases, we know that S = 830m = θ * R. Finding θ, will lead us to the value of R.
For option a:
θ = 15° = 0.2618 rad
[tex]R = \frac{S}{\theta} = 3170.36m[/tex]
For option b:
θ = 270° - 15° = 4.45 rad
[tex]R = \frac{S}{\theta} = 186.5m[/tex]
A spring stretches 0.2 cm per newton of applied force. An object is suspended from the spring and a deflection of 3 cm is observed. If g = 9.81 m/s?, what is the mass of the object, in kg?
Final answer:
The mass of the object is 1.53 kg.
Explanation:
To find the mass of the object, we need to use Hooke's Law which states that the force exerted by a spring is directly proportional to its extension. In this case, the spring stretches 0.2 cm per newton of applied force. The deflection of 3 cm corresponds to an applied force of 15 newtons (0.2 cm per newton * 3 cm).
Using the equation F = mg, where F is the force, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2), we can find the mass:
15 newtons = m * 9.81 m/s^2
m = 15 newtons / 9.81 m/s^2 = 1.53 kg
Therefore, the mass of the object is approximately 1.53 kg.
The mass of the object is approximately 1.53 kg.
Given that the spring stretches 0.2 cm per newton of applied force. Thus, we can say that
for F = - kx.
1 N = - k (0.2 cm)
or, k = spring constant of the given spring = [tex]\frac{F}{x}[/tex] = 5 N/cm
Now, for an object producing deflection of 3 cm, we can say that:
F = - k x = 5 N/cm × 3 cm
or, F = 15 N
This concludes that the weight of the object is 15 N.
Now, W = F = mg
hence, [tex]m = \frac{F}{g}[/tex]
or, m = [tex]\frac{15 \hspace{0.6mm} N}{9.8 \hspace{0.5mm} m/s^2}[/tex]
or, m ≈ 1.53 kg
A ball is thrown vertically upward with a speed of 15.0 m/s. (a) How high does it rise? m (b) How long does it take to reach its highest point? s (c) How long does the ball take to hit the ground after it reaches its highest point? s (d) What is its velocity when it returns to the level from which it started? (Assume the positive direction is upward. Indicate the direction with the sign of your answer.) m/s
Answer:
a) 0.76 m
b) 1.53 s
c) 0.39 s
d) 3.8 m/s
Explanation:
This is a problem in which we need to use the equations pertaining Uniformly Accelerated Motion, as the acceleration during all this process is constant: The gravitational pull on the ball, 9.8 m/s².
To make things easier, we can divide this process in two parts: The first one (A) is from the moment the ball is thrown, until the moment it reaches it highest point and momentarily stops. The second one (B) from the moment it starts descending until it hits the ground.
a) During part A, we use the formula Vf²=Vi² +2*a*x . Where Vf is the final velocity (0 m/s, as the ball stopped in midair), Vi is the initial velocity (15 m/s), a is the acceleration (-9.8 m/s², it has a minus sign, as it goes against the direction of the movement) and x is the distance; thus, we're left with:
0 m/s=(15.0 m/s)²+2*(-9.8 m/s²)*x
We solve for x
x = 0.76 m
b) The formula is Vf=Vi +a*t, where t is the time. We're left with:
0 m/s=15.0 m/s + (-9.8 m/s²)*t
We solve for t
t= 1.53 s
c) Now we focus on part B, and use the formula x=Vi * t + [tex]\frac{at^{2}}{2}[/tex] , with the difference that the Vi is 0 m/s. We already know the value of x in exercise a). Note that a does not have a negative sign, as the direction of movement is opposite to the direction of part A
0.76 m=0 m/s * t +[tex]\frac{9.8\frac{m}{s^{2} } *t^{2} }{2}[/tex]
Solve for t
0.76=4.9t²
t=0.39 s
d) Once again we use the formula Vf=Vi +a*t, using the value of t previously calculated in exercise c).
Vf=0 m/s + 9.8 m/s² * 0,39 s
Vf=3.8 m/s
The ball rises to a height of 11.48 m, takes 1.53 s to reach its highest point, the same amount of time to fall back down, and has a velocity of -15.0 m/s when it returns to the starting level.
The question involves concepts from physics, specifically the kinematics of one-dimensional motion under constant acceleration due to gravity. Here is how we can solve each part of the given problem:
(a) How high does it rise?
To find the maximum height reached by the ball, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
v² = u² + 2as
At the highest point, the final velocity (v) is 0 m/s, the initial velocity (u) is 15.0 m/s (upward), the acceleration due to gravity (a) is -9.8 m/s² (downward), and s represents the height. Solving for s:
0 = (15.0)² + 2(-9.8)s
s = (15.0)² / (2 * 9.8)
s = 11.48 m
(b) How long does it take to reach its highest point?
To find the time (t) it takes for the ball to reach its highest point, we can use the equation:
v = u + at
Since the final velocity at the highest point is 0 m/s:
0 = 15.0 + (-9.8)t
t = 15.0 / 9.8
t = 1.53 s
(c) How long does the ball take to hit the ground after it reaches its highest point?
The time for the ball to fall back down is the same as the time taken to reach the highest point, so this is also 1.53 s.
(d) What is its velocity when it returns to the level from which it started?
The velocity on returning to the starting level will be the same magnitude as the initial velocity but in the opposite direction, so it will be -15.0 m/s, with the negative sign indicating the downward direction.
Electron kinetic energies are often measured in units of electron-volts (1 eV 1.6 x 10-19 J), which is the kinetic energy of an electron that is accelerated through a 1 volt potential. When an aluminum plate is irradiated with UV light of 253.5 nm wavelength the ejected electrons are observed to have an average kinetic energy of about 0.8 eV. Use these results to determine the electron binding energy (or "work function") o of aluminum (in eV units).
Answer:
4.1 eV
Explanation:
Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J
wavelength, λ = 253.5 nm = 253.5 x 10^-9 m
According to the Einstein energy equation
[tex]E = W_{o}+K[/tex]
Where, E be the energy incident, Wo is the work function and K is the kinetic energy.
h = 6.634 x 10^-34 Js
c = 3 x 10^8 m/s
[tex]E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J[/tex]
So, the work function, Wo = E - K
Wo = 7.85 x 10^-19 - 1.28 x 10^-19
Wo = 6.57 x 10^-19 J
Wo = 4.1 eV
Thus, the work function of the metal is 4.1 eV.
A 0.010 kg ball is shot from theplunger of a pinball machine.
Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?
Answer:
v = 0.85 m/s
Explanation:
Given that,
Mass of the ball, m = 0.01 kg
Centripetal force on the ball, F = 0.025 N
Radius of the circular path, r = 0.29 m
Let v is the speed of the ball. The centripetal force of the ball is given by :
[tex]F=\dfrac{mv^2}{r}[/tex]
[tex]v=\sqrt{\dfrac{Fr}{m}}[/tex]
[tex]v=\sqrt{\dfrac{0.025\times 0.29}{0.01}}[/tex]
v = 0.85 m/s
So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.
Use the work-energy theorem to determine the force required to stop a 1000 kg car moving at a speed of 20.0 m/s if there is a distance of 45.0 m in which to stop it.
Answer:
4.44 kN in the opposite direction of acceleration.
Explanation:
Given that, the initial speed of the car is, [tex]u=20m/s[/tex]
And the mass of the car is, [tex]m=1000 kg[/tex]
The total distance covered by the car before stop, [tex]s=45m[/tex]
And the final speed of the car is, [tex]u=0m/s[/tex]
Now initial kinetic energy is,
[tex]KE_{i}=\frac{1}{2}mu^{2}[/tex]
Substitute the value of u and m in the above equation, we get
[tex]KE_{i}=\frac{1}{2}(1000kg)\times (20)^{2}\\KE_{i}=20000J[/tex]
Now final kinetic energy is,
[tex]KE_{f}=\frac{1}{2}mv^{2}[/tex]
Substitute the value of v and m in the above equation, we get
[tex]KE_{f}=\frac{1}{2}(1000kg)\times (0)^{2}\\KE_{i}=0J[/tex]
Now applying work energy theorem.
Work done= change in kinetic energy
Therefore,
[tex]F.S=KE_{f}-KE_{i}\\F\times 45=(0-200000)J\\F=\frac{-200000J}{45}\\ F=-4444.44N\\F=-4.44kN[/tex]
Here, the force is negative because the force and acceleration in the opposite direction.
To make a secure fit, rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.871 cm in diameter is to be placed in a hole 1.869 cm in diameter in a metal at 24 ∘C.To what temperature must the rivet be cooled if it is to fit in the hole?
Answer:
-65 degC
Explanation:
ΔD=[tex]D_{0}[/tex]αΔT
1.869-1.871=1.871(1.2E-5)ΔT
ΔT≈-89 degC
24-89=-65 degC
You drive a car 690 ft to the east, then 380 ft to the north. a) What is the magnitude of your displacement?
b) What is the direction of your displacement?
Answer:displacement =787.71 m
Explanation:
Given
Driver driver the car 690 ft to the east
then turn 380 ft to the north
(a)magnitude of acceleration is [tex]=\sqrt{380^2+690^2}=\sqrt{620500}[/tex]
displacement=787.71 m
(b)direction of displacement
[tex]tan\theta =\frac{380}{690}=0.5507[/tex]
[tex]\theta =28.84^{\circ}[/tex] with east direction
The magnitude of the displacement is approximately 775 ft and its direction is roughly 29 degrees north of east.
Explanation:Your total displacement after driving a car 690 ft to the east and then 380 ft to the north is calculated using the Pythagorean theorem, which states that the magnitude of the hypotenuse (displacement) of a right triangle (formed by the eastwards and northwards journeys of the car) can be found by sqrt((eastwards travel)^2 + (northwards travel)^2). Applying the theorem, we obtain sqrt((690 ft)^2 + (380 ft)^2) = 775 ft approximately.
The direction of the displacement can be found using the tangent of the angle, which is the ratio of the opposite (northwards) to the adjacent side (eastwards). Applying the inverse tangent function, we get tan^-1(380/690), which gives us approximately 29 degrees. Therefore, the direction of the displacement is 29 degrees north of east.
Learn more about Displacement here:https://brainly.com/question/33459975
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Your throw a ball straight upward at an initial speed of 5 m/s. How many times does the ball pass a point 2 m above the point you launched it from? Draw an x-t and a v-t diagram for the motion of this ball.
Answer:
The ball never passes 2m high, Hmax=1.27m
Explanation:
we assume the ball doesn't bounce when it hits the ground.
We calculate the maximum height, Vf = 0.
[tex]v_{o}^{2}=2gH_{max}\\H_{max}=v_{o}^{2}/(2g)=5^{2}/(2*9.81)=1.27m[/tex]
So, the ball never passes 2m high.
Kinematics equations:
[tex]x(t)=v_{o}t-1/2*g*t^{2}\\v(t)=v_{o}-gt[/tex]
Find annexed the graphics of x(t) and v(t)
Suppose you have two identical capacitors. You connect the first capacitor to a battery that has a voltage of 21.2 volts, and you connect the second capacitor to a battery that has a voltage of 12.8 volts. What is the ratio of the energies stored in the capacitors?
Answer:
r=2.743
Explanation:
The energy stored on a capacitor is of type potencial, therfore depends on the capacity to "store" energy. Inthe case of the capacitor, it stores charge (Q), and the equations you use to calculate it are:
[tex]E_p=\frac{Q^2}{2C}=\frac{QV}{2}=\frac{CV^2}{2}[/tex]
In this case we know V and C, therefore we use the last expression:
[tex]E_{p1}=\frac{CV_1^2}{2}[/tex]
[tex]E_{p2}=\frac{CV_2^2}{2}[/tex]
[tex]\frac{E_{p1}}{E_{p2}}=r=\frac{\frac{CV_1^2}{2}}{\frac{CV_2^2}{2}} \\r=(\frac{V_1}{V_2})^2\\r=(\frac{21.2}{12.8})^2[/tex]
r=2.743
For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the object at the time specified. For all of the motions, the positive y axis is upward.
Part A
An elevator is moving downward when someone presses the emergency stop button. The elevator comes to rest a short time later. Give the signs for the velocity and the acceleration of the elevator after the button has been pressed but before the elevator has stopped.
Enter the correct sign for the elevator's velocity and the correct sign for the elevator's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Part B
A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball immediately after the ball leaves the child's hand?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Part C
A child throws a baseball directly upward. What are the signs of the velocity and acceleration of the ball at the very top of the ball's motion (i.e., the point of maximum height)?
Enter the correct sign for the baseball's velocity and the correct sign for the baseball's acceleration, separated by a comma. For example, if you think that the velocity is positive and the acceleration is negative, then you would enter +,- . If you think that both are zero, then you would enter 0,0 .
Part A: -,+
The elevator is moving downward, this is what determines the direction of the velocity, as it will follow the direction of the movement. As we are told that the positive direction is upward, then the velocity has negative direction. Also, after the button is pressed, the elevator starts to stop, in other words, its velocity starts to decreased. This means that the acceleration has an opposite direction to the velocity, therefore, its sign is +.
Part B: +, -
The ball is moving upward, and as said before, this is what determines the direction of the velocity, as it will follow the direction of the movement. Then, velocity has a + sign.
Also, after the ball is thrown, there is no other force other than gravity, which will oppose to the movement of the ball, trying to make it come back to the ground. This means that the acceleration has an opposite direction to the velocity, in other words, it's directed downward, therefore, its sign is -.
Part C: 0, -
The acceleration of the ball since it was thrown until it fell to the ground will always be the gravity, which will always go downward (-).
After being thrown, the ball's velocity will start to decrease because of gravity. When its velocity has turned to 0, the ball will have reached maximum height . At this point it will start to fall again, accelerated by gravity. But at the very top, the velocity of the ball is 0.
The charge per unit length on a long, straight filament is -92.0 μC/m. Find the electric field 10.0 cm above the filament.
Answer:
E = 1.655 x 10⁷ N/C towards the filament
Explanation:
Electric field due to a line charge is given by the expression
E = [tex][tex]\frac{\lambda}{2\pi\times\epsilon_0\times r}[/tex][/tex]
where λ is linear charge density of line charge , r is distance of given point from line charge and ε₀ is a constant called permittivity and whose value is
8.85 x 10⁻¹².
Putting the given values in the equation given above
E = [tex]\frac{92\times10^{-6}}{2\times3.14\times8.85\times10^{-12}\times10^{-1}}[/tex]
E = 1.655 x 10⁷ N/C
Two tiny conducting sphere are identical and carry charges of -20 μC and +50 μC. They are separated by a distance of 2.50 cm. What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive?
Answer:
Force between two spheres will be 14400 N
And as the both charges of opposite nature so force will be attractive
Explanation:
We have given two conducting spheres of charges [tex]q_1=-20\mu C=-20\times 10^{-6}C\ and\ q_2=50\mu C=50\times 10^{-6}C[/tex]
Distance between the spheres = 2.5 cm =0.025 m
According to coulombs law we know that force between two charges is given by [tex]F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}=\frac{9\times 10^9\times 20\times 10^{-6}\times 50\times 10^{-6}}{0.025^2}=14400N[/tex]
As the both charges of opposite nature so force will be attractive
An archer standing on a 15 degree slope shoots an arrow at an angle of 26 degrees above the horizontal. How far below its original point of release does the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground?
Answer:
The arrow will hit 112.07 m from the point of release.
Explanation:
The equation for the position of an object in a parabolic movement is as follows:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
Where:
x0 = initial horizontal position
v0 = initial velocity
α = launching angle
y0 = initial vertical position
t = time
g = acceleration due to gravity
We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).
Then:
y = y0 + v0 · t · sin α + 1/2 · g · t²
Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t² (g is downward)
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s ( the negative value is discarded)
Now, with this time we can calculate the horizontal distance:
x = x0 + v0 · t · cos α (x0 = 0, the same as y0)
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
The object comes to 112.07 m below its original point of release the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground.
The equation for the position of an object in a parabolic movement is as follows:
r = (x₀ + v₀ t cos α, y₀ + v₀ t sin α + 1/2 g t²)
y = y₀ + v₀ t sin α + 1/2 g t²
Since the origin of the frame of reference is located at the point where the arrow is released, y₀ = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)
-1.88 m = 33 m/s sin 41° t - 1/2 9.8 m/s² t²
0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m
Solving the quadratic equation:
t = 4.5 s
Now, with this time we can calculate the horizontal distance:
x = x₀ + v₀ t cos α
x = 33 m/s · 4.5 s · cos 41° = 112.07 m
The object comes to 112.07 m below its original point of release the arrow hit if it is shot with a speed of 33 m/s from a height of 1.88 m above the ground.
To know more about the horizontal distance:
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What wavelength photon is required to excite a hydrogen atom from the n = 1 state to the n = 3 state?
Answer:
The wavelength required is 102.9 nm.
Explanation:
The energy levels for the hydrogen atom are
[tex]E(n) = \frac{-13.6 \ eV}{n^2}[/tex]
So, for a transition from the first level to the third level we got
[tex]\Delta E = E(3) - E (1)[/tex]
[tex]\Delta E = \frac{-13.6 \ eV}{ 3 ^2} - \frac{-13.6 \ eV}{1^2}[/tex]
[tex]\Delta E = \frac{-13.6 \ eV}{ 9} - \frac{-13.6 \ eV}{1}[/tex]
[tex]\Delta E = \frac{8}{9} 13.6 \ eV[/tex]
[tex]\Delta E = 12.09 \ eV[/tex]
[tex]\Delta E = 12.09 \ eV * \frac{1.6 \ 10^-19 \ Joules}{1 \ eV}[/tex]
[tex]\Delta E = 1.93 \ 10^-18 \ Joules[/tex]
So we need a photon with this energy.
The energy of a photon its given by
[tex]E = h \nu = h \frac{c}{\lambda}[/tex]
So, the wavelength will be
[tex]\lambda = \frac{h c}{E}[/tex]
[tex]\lambda = \frac{6.62 \ 10^{-34} \ \frac{m^2 kg}{s} \ * 3.00 \ 10^8 \ \frac{m}{s}}{1.93 \ 10^-18 \ Joules}[/tex]
[tex]\lambda = 10.29 \ 10^{-8} m[/tex]
[tex]\lambda = 1.029 \ 10^{-7} m[/tex]
[tex]\lambda = 102.9 \ nm[/tex]
Three objects are dropped from the top of a building. The first is thrown straight down with a velocity v, the second is thrown straight up with a velocity 2v, and the third is simply dropped. Which one has the highest speed when it hits the ground?
Answer:
Second ball
Explanation:
When a ball is thrown up with a certain velocity when the object reaches the same point from where it was thrown the velocity of the object becomes equal to the velocity with which the ball was thrown.
First ball
[tex]v_g_1^2-u^2=2as\\\Rightarrow v_g_1=\sqrt{2as+u^2}\\\Rightarrow v_g_1=\sqrt{2as+v^2}[/tex]
Second ball
[tex]v_g_2^2-u^2=2as\\\Rightarrow v_g_2=\sqrt{2as+u^2}\\\Rightarrow v_g_2=\sqrt{2as+4v^2}[/tex]
Third ball
[tex]v_g_3^2-u^2=2as\\\Rightarrow v_g_3=\sqrt{2as+0^2}\\\Rightarrow v_g_3=\sqrt{2as}[/tex]
From the equations above it can be seen that the second ball will have the highest velocity when it hits the ground.
So, [tex]v_g_3<v_g_1<v_g_2[/tex]
A cube whose sides are of length = 1.8 m is placed in a uniform electric field of magnitude E = 5.8 ✕ 10^3 N/C so that the field is perpendicular to two opposite faces of the cube. What is the net flux through the cube (in N · m^2/C)?
Answer:
Zero
Explanation:
Electric flux is defined as the number of electric field lines which passes through any area in the direction of area vector.
The formula for the electric flux is given by
[tex]\phi =\overrightarrow{E}.\overrightarrow{dS}[/tex]
Here, E is the strength of electric field and dS be the area vector.
It is a scalar quantity.
According to the Gauss's theorem, the electric flux passing through any surface is equal to the [tex]\frac{1}{\epsilon _{0}}[/tex] times the total charge enclosed through the surface.
Here, the charge enclose is zero, so the total flux is also zero.
A startled armadillo leaps upward, rising 0.540 m in the first 0.216 s. (a) What is its initial speed as it leaves the ground? (b)What is its speed at the height of 0.540 m? (c) How much higher does it go? Use g=9.81 m/s^2.
Answer:
a) 3.6 m/s
b) 1.53 m/s
c) 0.12 m
Explanation:
t = Time taken = 0.216 s
u = Initial velocity
v = Final velocity
s = Displacement = 0.54 m
a = Acceleration due to gravity = 9.81 m/s² (negative upward)
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 0.54=u\times 0.216+\frac{1}{2}\times -9.81\times 0.216^2\\\Rightarrow u=\frac{0.54+\frac{1}{2}\times 9.81\times 0.216^2}{0.216}\\\Rightarrow u=3.6\ m/s[/tex]
Initial speed as it leaves the ground is 3.6 m/s
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 0.54+3.6^2}\\\Rightarrow v=1.53\ m/s[/tex]
Speed at the height of 0.540 m is 1.53 m/s
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-3.6^2}{2\times -9.81}\\\Rightarrow s=0.66\ m[/tex]
The total height the armadillo leaps is 0.66 m
So, the additional height is 0.66-0.54 = 0.12 m
Ethyl alcohol has a boiling point of 78.0°C, a freezing point of -114°C, a heat of vaporization of 879 kJ/kg, a heat of fusion of 109 kJ/kg, and a specific heat of 2.43 kJ/kg.K. How much energy must be removed from 0.651 kg of ethyl alcohol that is initially a gas at 78.0°C so that it becomes a solid at -114°C?
Answer:
946.92 kJ
Explanation:
This process has 3 parts:
1. The first part, where the temperature of Ethyl alcohol remains constant and it changes from gas to liquid.
2. The second part, where the temperature drops from 78°C to -114°C
3. The third parts, where the temperature remains constant and it changes from liquid to solid.
The energy lost in a phase change is:
Q = m*cl
The energy lost because of the drop in temperature is:
[tex]Q = m c(T_2-T_1)[/tex]
cl is the heat of vaporization or heat of fusion, depending on the type of phase change. c is the specific heat.
So, the energy lost in each part is:
1. [tex]Q_1 = 0.651kg*879 kJ/kg = 572.23 kJ[/tex]
2. [tex]Q_2 = 0.651kg*2.43 kJ/kgK(78.0^oC - (-114^oC)) = 303.73 kJ[/tex]
3. [tex]Q_3 = 0.651kg*109kJ/kg = 70.96 kJ[/tex]
Then, the total energy removed should be:
Q = Q1 + Q2 + Q3 = 572.23 kJ + 303.73kJ + 70.96kJ = 946.92 kJ
A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energy of the group of charges ?
Answer:
- 2.7 x 10^-6 J
Explanation:
q1 = 1 nC at x = 0 cm
q2 = - 1 nC at x = 1 cm
q3 = 4 nC at x = 2 cm
The formula for the potential energy between the two charges is given by
[tex]U=\frac{Kq_{1}q_{2}}{r}[/tex]
where r be the distance between the two charges
By use of superposition principle, the total energy of the system is given by
[tex]U = U_{1,2}+U_{2,3}+U_{3,1}[/tex]
[tex]U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}[/tex]
[tex]U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}[/tex]
U = - 2.7 x 10^-6 J
2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due north. The police officer begins to pursue the speeder - first there is a 0.800 s reaction time when the officer has no change in speed, then the officer accelerates at 5.00 m/s2. Including the reaction time, how long does it take for the police car to reach the same position as the speeding car
Answer:
11.1 s
Explanation:
Speed of the police car as given = v = 18 m/s
Speed of the car = V = 42 m/s
Reaction time = t = 0.8 s
Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m
Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m
Acceleration of the police car = a = 5 m/s/s
The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.
Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.
⇒ D = 14.4 + 18 t + 0.5 (5) t²
⇒ D = 14.4 + 18 t+2.5 t² → (1)
For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).
⇒ 14.4 + 18 t+2.5 t²= 33.6 + 42 t
⇒ 2.5 t² -24 t - 19.2 = 0
⇒ t = 10.3 s
Total time = t +0.8 s
⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s
Your hair grows at the rate of 0.0330 mm/hr. What is the rate of your hair growth in ft/year?
Answer: 0.95 ft/year
Explanation: In order to explain this question we have to convert the units so
if we have a rate equal to 0.033 mm/hr then 0.033 mm *24*365 hr/year
then 1 m=3.28 feet
8760*0.033 *10^-3m* 3.28 feet/m=1.08*10^-4 feet*8760=0.95 feet/year
A 10-cm-long thin glass rod uniformly charged to 11.0 nC and a 10-cm-long thin plastic rod uniformly charged to - 11.0 nC are placed side by side, 4.10 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
The electric field strengths at various points around a charged object can be calculated using Coulomb's Law. They depend on the charge of the object and the distance from it. Fields produced by multiple charges need to be added vectorially.
Explanation:You're asking about the electric field strengths at varying distances from a charged glass rod, located next to a plastic rod with a charge of equal magnitude, but of opposite sign. The electric field strength E at a point in the vicinity of a charged object can be derived using Coulomb's Law. This law states that the force F between two charges q1 and q2 is proportional to the product of the charges and inversely proportional to the square of the distance r between them.
Given this, the electric field E created by a charge q at a distance r from the charge is given by E = k|q| / r², where k is Coulomb's constant, approximately 9 × 10⁹ N•m²/C². The direction of the electric field is determined by the sign of the charge. When it's positive, the field lines are going outward, and when it's negative, they are directed inward.
To calculate the net electric field at a point on the line connecting the midpoints of the rods, we consider the electric fields produced by both rods. Since they're opposite in sign, the fields will have opposite directions, and they should be added vectorially. The magnitudes and directions of the fields will also depend on the specific distances (1.0 cm, 2.0 cm, and 3.0 cm) from the glass rod.
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A particular spiral galaxy can be approximated by a thin disk-like volume 62 Thousand Light Years in radius and 7 Hundred Light Years thick. If this Galaxy contains 1,078 Billion stars, estimate the average distance between the stars in this galaxy. Hint: calculate the average volume per star in cubic Light Years, and then estimate the approximate linear dimension across such a volume. (Indicate your answer to one decimal place.)
Answer:
Approximate linear dimension is 2 light years.
Explanation:
Radius of the spiral galaxy r = 62000 LY
Thickness of the galaxy h = 700 LY
Volume of the galaxy = πr²h
= (3.14)(62000)²(700)
= (3.14)(62)²(7)(10)⁸
= 84568×10⁸
= [tex]8.45\times 10^{12}[/tex] (LY)³
Since galaxy contains number of stars = 1078 billion stars ≈ [tex]1.078\times 10^{12}[/tex]
Now volume covered by each star of the galaxy = [tex]\frac{\text{Total volume of the galaxy}}{\text{Number of stars}}[/tex]
= [tex]\frac{8.45\times 10^{12} }{1.078\times 10^{12}}[/tex]
= 7.839 Light Years
Now the linear dimension across the volume
= [tex](\text{Average volume per star})^{\frac{1}{3}}[/tex]
= [tex](7.839)^{\frac{1}{3}}[/tex]
= 1.99 LY
≈ 2 Light Years
Therefore, approximate linear dimension is 2 light years.