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.
Two parallel plates have equal but opposite charges on their surface. The plates are separated by a finite distance. A fast moving proton enters the space between the two plates through a tiny hole in the left plate A. The electric potential energy of the proton increases as it moves toward plate B. (a) How is the speed of the proton affected as it moves from plate A to plate B
Answer:
Explanation:
The plates A and B are charged by opposite charges but which plate is positively charged and which is negatively charged is not clear.
Now a proton which is positively charged is moving from plate A to B . If it is attracted by plate B then its kinetic energy will be increased and potential energy will be decreased due to conservation of energy . In that case B will be negatively charged .
But in the given case it is stated that
potential energy of the proton increases . That means its kinetic energy decreases . In other words its speed decreases . It points to the fact that plate B is also positively charged.
So proton will be repelled and its speed will be decreased.
An IOLab device is moving with constant speed in a straight line. According to Newton's Laws of Motion, which of these statements can be true?
I. No force is acting.
II. Only one nonzero force is acting.
III. Two nonzero forces are acting.
IV. Three nonzero forces are acting.
select one:
(A) only I
(B) only I and II
(C) only II and III
(D) only I, III and IV
(E) only II, III and IV
Answer:
(A) only I
Explanation:
According to Newton's first law of motion, a particle in motion will continue moving in a straight line at constant speed or will stay at rest, if at rest, as long as there's no external force acting on it.
ALL of the statements except #2 COULD be true. (Choice-D)
A bead with a mass of 0.050 g and a charge of 20 nC is free to slide on a vertical rod. At the base of the rod is a fixed 20 nC charge. In equilibrium, at what height above the fixed charge does the bead rest?
Final answer:
The bead rests at a height of 9.03 x 10^7 meters above the fixed charge.
Explanation:
To find the height above the fixed charge where the bead rests in equilibrium, we need to consider the electric forces acting on the bead. The electric force is given by the equation:
F = k * (q1 * q2) / r^2
Where F is the force between the two charges, q1 and q2 are the charges, r is the distance between them, and k is the electrostatic constant. In this case, the two charges are the fixed charge at the base of the rod and the charge on the bead. Setting the gravitational force equal to the electric force, we can solve for the height.
First, we need to convert the given charge of 20 nC to coulombs by dividing it by 10^9:
q2 = 20 nC / 10^9 = 20 * 10^-9 C
Next, we calculate the gravitational force and the electric force:
F_gravity = m * g
F_electric = k * q1 * q2 / r^2
Since the bead is in equilibrium, the two forces must be equal:
m * g = k * q1 * q2 / r^2
Now, we can solve for the height:
h = sqrt(k * q1 * q2 / (m * g))
Plugging in the given values:
h = sqrt((9 * 10^9 N * m^2 / C^2) * (20 * 10^-9 C) / (0.050 x 10^-3 kg * 9.8 m/s^2))
Simplifying:
h = sqrt(4 * 10^11 / (0.049 x 10^-3))
h = sqrt(8.16 x 10^14)
h = 9.03 x 10^7 m
Therefore, the bead rests at a height of 9.03 x 10^7 meters above the fixed charge.
A football is kicked from ground level at an angle of 53 degrees. It reaches a maximum height of 7.8 meters before returning to the ground. How long will the football spend in the air, in seconds?
Answer:
1.61 second
Explanation:
Angle of projection, θ = 53°
maximum height, H = 7.8 m
Let T be the time taken by the ball to travel into air. It is called time of flight.
Let u be the velocity of projection.
The formula for maximum height is given by
[tex]H = \frac{u^{2}Sin^{2}\theta }{2g}[/tex]
By substituting the values, we get
[tex]7.8= \frac{u^{2}Sin^{2}53 }{2\times 9.8}[/tex]
u = 9.88 m/s
Use the formula for time of flight
[tex]T = \frac{2uSin\theta }{g}[/tex]
[tex]T = \frac{2\times 9.88\times Sin53 }{9.8}[/tex]
T = 1.61 second
Suppose that you are on an unknown planet in a distant galaxy, and you are trying to determine the acceleration of gravity of this planet. The length of a physical pendulum be 0.81 m and the measured period was 1.138 s. Determine the gravitational acceleration of this planet?
Final answer:
The acceleration due to gravity on this planet is approximately [tex]g = 9.82 m/s^2.[/tex]
Explanation:
To determine the acceleration due to gravity on an unknown planet using a pendulum, you can utilize the formula for the period of a simple pendulum: T = 2π[tex]\sqrt{(L/g}[/tex], where T is the period, L is the length, and g is the acceleration due to gravity.
With the given length of the pendulum being 0.81 m and the period being 1.138 s, you can rearrange the formula to solve for [tex]g = (4*22/7)^2[/tex][tex](L/T^2)[/tex]. Plugging in the known values, [tex]g = (4*22/7)^2[/tex][tex](0.81 m / 1.138^2 s^2)[/tex].
Computing the value, we find that the acceleration due to gravity on this planet is approximately [tex]g = 9.82 m/s^2.[/tex]
A square steel bar of side length w = 0.18 m has a thermal conductivity of k = 14.6 J/(s⋅m⋅°C) and is L = 1.7 m long. Once end is placed near a blowtorch so that the temperature is T1 = 88° C while the other end rests on a block of ice so that the temperature is a constant T2.Input an expression for the heat transferred to the cold end of the bar as a function of time, using A-w2 as the cross-s area of the bar.
The heat transferred to the cold end of the square steel bar as a function of time can be calculated using a derivation of the heat conduction formula, factoring in the steel bar's thermal conductivity, cross-sectional area, temperature differential, time, and length.
Explanation:This question relates to the transfer of heat through a square steel bar using conduction. The heat transfer can be calculated using the formula Q = k·A·(T1 - T2)·t / L, where Q is the heat transferred, k is the thermal conductivity of the material, A is the cross-sectional area, T1 and T2 are the temperatures at both ends of the material respectively, t is the time of heat transfer, and L is the length of the material.
In this case, the steel bar is square, so its cross-sectional area A is calculated by squaring the side length w, so A = w². So, the exact formulation to calculate the heat transferred to the cold end of the bar as a function of time will be Q= 14.6 J/(s⋅m⋅°C)⋅(0.18 m)²⋅(88°C - T2)·t / 1.7 m.
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The heat transferred to the cold end of the steel bar is calculated by finding the rate of heat transfer and multiplying it by the time. The rate of heat transfer is approximately 23.5729 W.
To find the heat transferred to the cold end of the bar as a function of time, we use the formula for the rate of heat transfer through a material.
Given data:
Side length of the square bar, [tex]w = 0.18 m[/tex]Thermal conductivity of steel,[tex]k = 14.6 J/(s.m.^oC)[/tex]Length of the bar,[tex]L = 1.7 m[/tex]Temperature at the hot end,[tex]T_1 = 88^oC[/tex]Temperature at the cold end, [tex]T_2 = 0^oC[/tex] (since it rests on ice)Cross-sectional area, [tex]A = w^2 = (0.18 m)^2 = 0.0324 m^2[/tex]The rate of heat transfer Q/t can be calculated using the formula:
[tex]Q/t = k * A * (T_1 - T_2) / L[/tex]
Plugging in the values:
[tex]Q/t = 14.6 J/(s.m.^oC) * 0.0324 m^2 * (88^oC - 0^oC) / 1.7 m[/tex]
[tex]Q/t = 14.6 * 0.0324 * 88 / 1.7[/tex]
[tex]Q/t = 23.5729 W[/tex]
Therefore, the heat transferred to the cold end of the bar as a function of time is:
[tex]Q = (23.5729 W) * t[/tex]
The heat transferred to the cold end of a square steel bar over time can be calculated by finding the rate of heat transfer and multiplying it by the time. The rate of heat transfer for the given data is approximately 23.5729 W.
Roller coaster loops are rarely perfectly circular. Instead, they are tightly curved at the top, where the cars are moving more slowly, and they have a gentler curve at the bottom, where the cars are moving much faster. Explain why the loops are designed this way.
Answer:
[tex]a_{c}=v^{2}/R[/tex]
The radius of curvature changes so that centripetal acceleration is similar along the entire roller coaster.
Explanation:
We know that the centripetal acceleration is directly proportional to the tangential velocity and inversely proportional to the radius of curvature:
[tex]a_{c}=v^{2}/R[/tex]
By energy conservation (and common sense), we know that the speed at the top of the roller coaster is smaller. Therefore if the roller coaster has similar accelerations (therefore also similar normal forces) at the top and at the bottom, it is necessary that the difference in speed be compensated with the radius of curvature, i.e. smaller radius at the top than at the bottom.
The car has a constant deceleration of 4.20 m/s^2. If its initial velocity was 24.0 m/s, how long does it take to come to a stop? Answer in s
Answer:
The time is 5.71 sec.
Explanation:
Given that,
Acceleration [tex]a= -4.20 m/s^2[/tex]
Initial velocity = 24.0 m/s
We need to calculate the time
Using equation of motion
v = u+at[/tex]
Where, v = final velocity
u = inital velocity
t = time
a = acceleration
Put the value into the formula
[tex]0 =24.0 +(-4.20)\times t[/tex]
[tex]t = \dfrac{-24.0}{-4.20}[/tex]
[tex]t=5.71\ sec[/tex]
Hence, The time is 5.71 sec.
The x-component of vector R is Rx = −23.2 units and its y-component is Ry = 21.4 units. What is its direction? Give the direction as an angle measured counterclockwise from the +x-direction.
The direction of vector R is [tex]138.47^o[/tex]counterclockwise from the +x-direction and the angle is measured in counterclockwise direction.
For the direction of vector R, which is the angle measured counterclockwise from the +x-direction, we utilize trigonometry.
Given:
x-component of vector R [tex](R_x) = -23.2\ units[/tex]
y-component of vector R [tex](R_y) = 21.4\ units[/tex]
The angle is determined using the arctangent function:
[tex]\theta = tan^{-1}(R_y / R_x)[/tex]
Substituting the given values in the equation:
[tex]\theta = tan^{-1}(21.4 / -23.2).[/tex]
Calculating using a calculator:
[tex]\theta=-41.53^o[/tex]
Since angles are measured counterclockwise from the positive x-direction, adding 180° to the calculated angle gives the direction in that context. The final angle is calculated as:
[tex]\theta= -41.53^o + 180^o \\\theta= 138.47^o\\[/tex]
Therefore, the direction of vector R is [tex]138.47^o[/tex]counterclockwise from the +x-direction.
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The angle of vector R relative to the positive x-axis can be calculated using the arctangent function with the given x and y components, modified to reflect the vector's presence in quadrant II of the Cartesian plane.
Explanation:To determine the direction of vector R given its x and y components, we can use the arctangent function to find the angle of the vector relative to the positive x-axis. Since the x-component (Rx) is -23.2 units and the y-component (Ry) is 21.4 units, we calculate the angle θ using the formula θ = tan⁻¹(Δy/Δx). Plugging in our values, we get θ = tan⁻¹(21.4 / -23.2). Remember to adjust the angle based on the signs of Rx and Ry since tan⁻¹ only provides results for quadrants I and IV, and our vector lies in quadrant II. This angle will be measured counterclockwise from the positive x-direction.
In attempting to
pass thepuck to a teammate, a hockeyplayer
gives it an initialspeedof 1.7
m/s. However, this speed is inadequate tocompensate for
the kinetic frictionbetween the puck and theice. As
a result, the puck travels only half the
distancebetweenthe players before slidingto a
halt. What minimum initial speed should the puck
havebeengiven so that it reached
theteammate, assuming that the same force of
kineticfrictionacted on the puck
everywherebetween the two players?
Answer:
2.04 m/s
Explanation:
Given:
[tex]u[/tex] = initial inadequate speed of the puck = 1.7 m/s[tex]v[/tex] = final velocity of the puck while reaching half the distance of the targeted teammate = 0 m/sAssumptions:
[tex]m[/tex] = mass of the puck[tex]U[/tex] = minimum initial speed of the puck so that it reaches the target[tex]x[/tex] = distance of the targeted teammate [tex]f_k[/tex] = kinetic friction between the puck and the iceWork-energy theorem: For the various forces acting on an object, the work done by all the forces brings a change in kinetic energy of an object which is equal to the total work done.
For the initial case, the puck travels half the distance of the target teammate. In this case, the change in kinetic energy of the puck will be equal to the work done by the friction.
[tex]\therefore \dfrac{1}{2}m(v^2-u^2)=f_k\dfrac{x}{2}\\\Rightarrow \dfrac{1}{2}m(0^2-1.7^2)=f_k\dfrac{x}{2}[/tex]...........eqn(1)
Now, again using the work energy theorem for the puck to reach the targeted teammate, the change in kinetic energy of the puck will be equal to the work done by the kinetic friction.
[tex]\therefore \dfrac{1}{2}m(v^2-U^2)=f_k\dfrac{x}{2}\\\Rightarrow \dfrac{1}{2}m(0^2-U^2)=f_kx[/tex]...........eqn(2)
On dividing equation (1) by (2), we have
[tex]\dfrac{-1.7^2}{-U^2}=\dfrac{1}{2}\\\Rightarrow \dfrac{1.7^2}{U^2}=\dfrac{1}{2}\\\Rightarrow U^2= 2\times 1.7^2\\\Rightarrow U^2 = 5.78\\\Rightarrow U=\pm \sqrt{5.78}\\\Rightarrow U=\pm 2.04\\\textrm{Since the speed is always positive.}\\\therefore U = 2.04\ m/s[/tex]
Hence, the puck must be kicked with a minimum initial speed of 2.04 m/s so that it reaches the teammate.
5. The oldest rocks in the South Atlantic Ocean, immediately adjacent to the African and South American continental shelves, are 120,000,000 years old (time). Calculate the average rate of seafloorspreading for the South Atlantic Ocean over its entire existence (Hint: Use formula: velocity = distance/time): ______________km/yr (velocity). Now convert that to: _______________ cm/yr:
Answer:
The continental drift speed is 6.18 cm/yr
Explanation:
In order to calculate the drift speed of the continents, we can assume that it is constant for what the relationship meets
v = d / t
Where v is the speed, t the time and d the distance
To find the distance we use the closest points that are in the South Atlantic, after reviewing a world map, these points are Brazil and Namibia that has an approximate distance of 7415 km
To start the calculation let's reduce the magnitude
d = 7415 km (1000m/1km) (100cm/1m)
d = 7.415 10⁸ 8 cm
t = 120000000 years
t = 1.2 10⁸ year
With these values we calculate the average speed
v = 7.415 10 3 / 1.2 108 [km / yr.]
v = 6.18 10-5 Km / yr
v = 7.415 10 8 / 1.2 10 8 [cm / yr]
v = 6.18 cm / yr.
The continental drift speed is 6.18 cm/yr
Calculate the least velocity of projection required to give
amissile a horizontal displacement of 500m if the angle
ofprojection is 24 degrees?
Answer:81.24 m/s
Explanation:
Given
Horizontal displacement([tex]R_x[/tex])=500
Angle of projection[tex]=24 ^{\circ}[/tex]
Let u be the launching velocity
and horizontal range is given by
[tex]R_x=\frac{u^2sin2\theta }{g}[/tex]
[tex]500=\frac{u^2sin48}{9.81}[/tex]
[tex]u^2=\frac{500\times 9.81}{0.7431}[/tex]
[tex]u^2=6600.32854[/tex]
[tex]u=\sqrt{6600.32854}=81.24 m/s[/tex]
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.6 m/s . Two seconds later the bicyclist hops on his bike and accelerates at 2.0 m/s2 until he catches his friend. A.) How much time does it take until he catches his friend (after his friend passes him)? t=___s B.) How far has he traveled in this time? x= ____ m
C.) What is his speed when he catches up? v=____ m/s
Answer:
A) t = 7.0 s
B) x = 25 m
C) v = 10 m/s
Explanation:
The equations for the position and velocity of an object traveling in a straight line is given by the following expressions:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
v = velocity at time t
A)When both friends meet, their position is the same:
x bicyclist = x friend
x0 + v0 · t + 1/2 · a · t² = x0 + v · t
If we place the center of the frame of reference at the point when the bicyclist starts following his friend, the initial position of the bicyclist will be 0, and the initial position of the friend will be his position after 2 s:
Position of the friend after 2 s:
x = v · t
x = 3.6 m/s · 2 s = 7.2 m
Then:
1/2 · a · t² = x0 + v · t v0 of the bicyclist is 0 because he starts from rest.
1/2 · 2.0 m/s² · t² = 7.2 m + 3.6 m/s · t
1 m/s² · t² - 3.6 m/s · t - 7.2 m = 0
Solving the quadratic equation:
t = 5.0 s
It takes the bicyclist (5.0 s + 2.0 s) 7.0 s to catch his friend after he passes him.
B) Using the equation for the position, we can calculate the traveled distance. We can use the equation for the position of the friend, who traveled over 7.0 s.
x = v · t
x = 3.6 m/s · 7.0 s = 25 m
(we would have obtained the same result if we would have used the equation for the position of the bicyclist)
C) Using the equation of velocity:
v = a · t
v = 2.0 m/s² · 5.0 s = 10 m/s
A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A is the mass number of the nucleus. Estimate the den- sity of the nucleus of 127I (which has a nuclear mass of 2.1 × 10222 g) in grams per cubic centimeter. Compare with the density of solid iodine, 4.93 g cm23.
Answer:
Density of 127 I = [tex]\rm 1.79\times 10^{14}\ g/cm^3.[/tex]
Also, [tex]\rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]
Explanation:
Given, the radius of a nucleus is given as
[tex]\rm r=kA^{1/3}[/tex].
where,
[tex]\rm k = 1.3\times 10^{-13} cm.[/tex]A is the mass number of the nucleus.The density of the nucleus is defined as the mass of the nucleus M per unit volume V.
[tex]\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.[/tex]
For the nucleus 127 I,
Mass, M = [tex]\rm 2.1\times 10^{-22}\ g.[/tex]
Mass number, A = 127.
Therefore, the density of the 127 I nucleus is given by
[tex]\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.[/tex]
On comparing with the density of the solid iodine,
[tex]\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.[/tex]
A compass in a magnetic field will line up __________.
(A) with the north pole pointing perpendicular to the magnetic field
(B) with the north pole pointing opposite the direction of the magnetic field
(C) with the north pole pointing in the direction of the magnetic field
Answer:
Option c
Explanation:
Magnetic field lines form loops starting from north pole to south pole outside the magnet and from south pole to north pole inside the magnet.
Thus the field is such that it is directed outwards from the North pole and directed inwards to the South pole of the magnet.
A compass in a magnetic field will will comply with the magnet's North pole directing towards the magnetic field.
The equation for bouyancy force on a fully submerged object of volume V and mass M is given by: OF) = PwVg OF) = Pwg/V OF, = Mg
Answer:
[tex]OF=\rho _wVg[/tex]
Explanation:
The bouyancy force an object feels (OF) when submerged ina fluid is always the weight of the liquid the object displaces. This weight will be the mass of fluid displaced ([tex]m_d[/tex]) multiplied by the acceleration of gravity g. The mass displaced will be the density of the fluid [tex]\rho_w[/tex] multiplied by the volume of fluid displaced [tex]V_d[/tex]. If the object is fully submerged, then this volume will be the same as the volume of the object V. We write all these steps in equations:
[tex]OF=m_dg=\rho _wV_dg=\rho _wVg[/tex]
A girl is whirling a ball on a string around her head in
ahorizontal plane. She wants to let go at precisely the right
timeso that the ball will hit a target on the other side of the
yard.When should she let go of the string?
Answer:
Answered
Explanation:
The girl whirling the ball should let go off ball when the ball is at a position such that tangent to the circle is in the direction of the target.
the tangent at any point in a circular path indicates the direction of velocity at that point. And the moment when the centripetal force is removed the ball will follow the tangential path at that moment.
Final answer:
The girl should release the ball when it aligns with the target in a straight line from her, following principles of inertia and circular motion. In a scenario where a ball winds around a post, it would speed up due to the conservation of angular momentum, aligning with Michelle's prediction.
Explanation:
The question about when a girl should release a ball while whirling it around her head in a horizontal plane to hit a target involves understanding circular motion and projectile motion principles in physics. According to the laws of physics, the ball will continue to move tangentially to the circle at the point of release because of inertia. Thus, she should let go of the string when the ball is directly in line with the target, assuming no air resistance and that the path follows a straight line in the horizontal direction.
The conservation of angular momentum and the conservation of energy are relevant in situations where objects are in circular motion, like a ball tied to a string being whirled around. For an object in circular motion, when the radius of the motion decreases (as in the ball winding around a post), it will speed up because angular momentum is conserved. This implies Michelle’s view on the ball having to speed up as it approaches the post is correct, contrasting Astrid’s expectation of constant speed due to energy conservation.
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is σ = 1.99 x 10^-7 C/m^2, and the plate separation is 1.69 x 10^-2 m. How fast is the electron moving just before it reaches the positive plate?
Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area[tex] = \sigma = 1.99*10^{-7} C/m^2[/tex]
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is[tex] E = \frac{\sigma}{\epsilon}[/tex]
force acting on charge is F = q E
Work done by charge q id[tex] \Delta X =\frac{ q\sigma \Delta x}{\epsilon}[/tex]
this work done is converted into kinectic enerrgy
[tex]\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}[/tex]
solving for v
[tex]v = \sqrt{\frac{2q\Delta x}{\epsilon m}[/tex]
[tex]\epsilon = 8.85*10^{-12} Nm2/C2[/tex]
[tex]v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}[/tex]
v = 1.15*10^{7} m/s
What must be 'n' of a sphere surrounded by water so that the parallel rays that affect one of its faces converge on the second vertex of the sphere?
Answer:
The refractive index of the sphere is 2.66
Solution:
The refractive index, [tex]n_{w} = 1.33[/tex] and since the sphere is surrounded by water.
Therefore, according to the question, the parallel rays that affect one of the faces of the sphere converges on the second vortex:
Thus the image distance from the pole of surface 1, v' = 2R
where
R = Radius of the sphere
Now, using the eqn:
[tex]\frac{n_{w}}{v} + \frac{n}{v'} = \frac{n - n_{w}}{R}[/tex]
[tex]0 + \frac{n}{2R} = \frac{n - 1.33}{R}[/tex]
Since, v is taken as infinite
n = 2.66
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding the speed limit?
Answer:
yes driver exceed the car speed
Explanation:
given data
speed of car = 38 m/s
speed limit = 75.0 mi/h
to find out
Is the driver exceeding the speed limit
solution
we know car speed is 38 m/s and limit is 75 mi/hr
so for compare the speed limit we convert limit and make them same
as we know
1 m/s = 2.236 mi/hr
so
car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr
as this car speed is exceed the speed limit that is 75 mi/hr
yes driver exceed the car speed
Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alyssa starts 15 s later and runs at a constant 4.0 m/s. At what time after Jenny's start does Alyssa catch up with Jenny?
Answer:
285 seconds
Explanation:
Jenny speed is 3.8 m/s
Alyssa speed in 4.0 m/s
Alyssa starts after 15 seconds
Find the distance covered by Jenny, when Alyssa starts
Distance=Speed*time
Distance covered by Jenny in 15 seconds= 3.8×15=57m
Relative speed of the two members heading same direction will be;
4.0m/s-3.8m/s=0.2m/s
To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two
Distance=57m, relative speed=0.2m/s t=57/0.2 =285 seconds
=4.75 minutes
Answer:
After 100 seconds Alyssa catch up with Jenny.Explanation:
Jenny's data:[tex]v_{Jenny} =3.8m/s[/tex]
[tex]t_{Jenny}=t[/tex]
[tex]d_{Jenny}=d[/tex]
Alyssa's data:[tex]v_{Alyssa}=4.0m/s[/tex]
[tex]t_{Alyssa}=t-15[/tex], because she has a difference of 15 seconds.
[tex]d_{Alyssa}=d[/tex]
Both move at a constant speed, that means there's no acceleration, their speed is always the same.
Now, the equation of each movement is
[tex]d=3.8t[/tex] and [tex]d=4(t-15)[/tex], then we solve this two.
We replace the first equation into the second one
[tex]3.8t=4(t-15)\\3.8t=4t-20\\20=4t-3.8t\\0.2t=20\\t=\frac{20}{0.2}\\ t=100[/tex]
That means after 100 seconds Alyssa catch up with Jenny.
Question Part Points Submissions Used A pitcher throws a 0.200 kg ball so that its speed is 19.0 m/s and angle is 40.0° below the horizontal when a player bats the ball directly toward the pitcher with velocity 46.0 m/s at 30.0° above the horizontal. Assume +î to be along the line from the batter to the pitcher and +ĵ to be the upward vertical direction. (Express your answers in vector form.) (a) Determine the impulse (in N · s) delivered to the ball. I = N · s (b) If the force on the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases linearly to zero in another 4.00 ms, what is the maximum force (in N) on the ball?
Answer:
The impulse is (10.88 i^ + 7.04 j^) N s
maximum force on the ball is (4.53 10 2 i^ + 2.93 102 j ^) N
Explanation:
In a problem of impulse and shocks we must use the impulse equation
I = dp = pf-p₀ (1)
p = m V
With we have vector quantities, let's decompose the velocities on the x and y axes
V₀ = -19 m / s
θ₀ = 40.0º
Vf = 46.0 m / s
θf = 30.0º
Note that since the positive direction of the x-axis is from the batter to the pitcher, the initial velocity is negative and the angle of 40º is measured from the axis so it is in the third quadrant
Vcx = Vo cos θ
Voy = Vo sin θ
Vox= -19 cos (40) = -14.6 m/s
Voy = -19 sin (40) = -12.2 m/s
Vfx = 46 cos 30 = 39.8 m/s
Vfy = 46 sin 30 = 23.0 m/s
a) We already have all the data, substitute and calculate the impulse for each axis
Ix = pfx -pfy
Ix = m ( vfx -Vox)
Ix = 0.200 ( 39.8 – (-14.6))
Ix = 10.88 N s
Iy = m (Vfy -Voy)
Iy = 0.200 ( 23.0- (-12.2))
Iy= 7.04 N s
In vector form it remains
I = (10.88 i^ + 7.04 j^) N s
b) As we have the value of the impulse in each axis we can use the expression that relates the impulse to the average force and your application time, so we must calculate the average force in each interval.
I = Fpro Δt
In the first interval
Fpro = (Fm + Fo) / 2
With the Fpro the average value of the force, Fm the maximum value and Fo the minimum value, which in this case is zero
Fpro = (Fm +0) / 2
In the second interval the force is constant
Fpro = Fm
In the third interval
Fpro = (0 + Fm) / 2
Let's replace and calculate
I = Fpro1 t1 +Fpro2 t2 +Fpro3 t3
I = Fm/2 4 10⁻³ + Fm 20 10⁻³+ Fm/2 4 10⁻³
I = Fm 24 10⁻³ N s
Fm = I / 24 10⁻³
Fm = (10.88 i^ + 7.04 j^) / 24 10⁻³
Fm = (4.53 10² i^ + 2.93 10² j ^) N
maximum force on the ball is (4.53 10 2 i^ + 2.93 102 j ^) N
The impulse delivered to the ball is (-10.87 î + 7.04 ĵ) N·s. The maximum force on the ball is -452.92 N, based on the given force-time relationship.
Solution:
(a) To find the impulse delivered to the ball, we can use vector components of velocity and the formula for impulse.
Initial velocity vector of the ball:
vix = 19.0 m/s × cos(40.0°) = 14.55 m/s
viy = 19.0 m/s × sin(-40.0°) = -12.22 m/s
Final velocity vector of the ball:
vfx = -46.0 m/s × cos(30.0°) = -39.78 m/s
vfy = 46.0 m/s × sin(30.0°) = 23.00 m/s
Change in velocity vector:
Δvx = vfx - vix = -39.78 m/s - 14.55 m/s = -54.33 m/s
Δvy = vfy - viy = 23.00 m/s - (-12.22 m/s) = 35.22 m/s
Impulse vector (I = m × Δv):
Ix = 0.200 kg × (-54.33 m/s) = -10.87 N·s
Iy = 0.200 kg × (35.22 m/s) = 7.04 N·s
Total impulse vector:
I = (-10.87 î + 7.04 ĵ) N·s
(b) To determine the maximum force (Fmax) on the ball, consider the force-time relationship given:
Force increases linearly for 4.00 ms
Force holds constant for 20.0 ms
Force decreases linearly to zero in another 4.00 ms
The total duration of the force application is 28 ms (4 + 20 + 4 = 28 ms). The impulse is the area under the force-time graph, which is:
Impulse (I) = (1/2 × Fmax × 4.00 ms) + (Fmax × 20.0 ms) + (1/2 × Fmax × 4.00 ms)
-10.87 N·s = (0.002 × Fmax + 0.020 × Fmax + 0.002 × Fmax)
-10.87 N·s = 0.024 × Fmax
Fmax = -10.87 N·s / 0.024 s = -452.92 N
The negative sign indicates the direction opposite to the assumed positive direction.
The flight path of a jet aircraft as it takes off is defined by the parmetric equations x=1.25 t2 and y=0.03 t3, where t is the time after take-off, measured in seconds, and x and y are given in meters. At t=40 s (just before it starts to level off), determine at this instant (a) the horizontal distance it is from the airport, (b) its altitude, (c) its speed and (d) the magnitude of its acceleration.
To answer this question, you substitute t = 40 s into the given parametric equations to find the horizontal distance from the airport and the altitude. Then, you take the derivative of both equations to find the speed, and the second derivative to find the acceleration.
Explanation:To solve this problems, you will need to use the given parametric equations. The horizontal distance from the airport (a) is given by x = 1.25 t^2. At t = 40 s, you can simply substitute the value of t into the equation to find x. (b) The altitude of the jet is represented by y = 0.03 t^3.
Similarly, substitute t = 40 s into this equation to find y. (c) The speed of the jet can be found by calculating the derivative of both x and y with respect to t and then using these to find the magnitude of the velocity vector. (d) The acceleration of the jet can be found by taking the second derivative of both x and y with respect to t and again using these to find the magnitude of the acceleration vector.
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What is the force of gravity between two 15,000 kg cement trucks that are parked 3 meters apart? How does this force compare with the weight of one of the trucks?
Answer:
[tex]\frac{W}{F} = 8.8\times10^7[/tex]
Explanation:
According to newton's law of gravitation
[tex]F=G\frac{m_1\times m_2}{r^2}[/tex]
here m_1=m_2=15000 kg
r= 3 meters and G= 6.67[tex]6.67\times10^{-11}[/tex]
putting values we get
[tex]F= 6.67\times10^{-11}\frac{15000^2}{3^2}[/tex]
solving the above equation we get
Force of gravitation F= [tex]1.6675\times 10^{-3}[/tex] newton
weight of the one of the truck W = mg= 15000×9.81 N
=147000 N
therefore [tex]\frac{W}{F} = \frac{147000}{1.66\times10^{-3}}[/tex]
=8.8×10^{7)
Which of the following is not an appropriate category of childrens book to include in the early childhood classroom?
A. Chapter Book
B. Picture Book
C. Concept Book
D. Counting book
Answer:
chapter book
Explanation:
Suppose you first walk 25.1 m in a direction 15.4º west of north and then 38.8 m in a direction 23.1º south of west. How far are you from your starting point?
Answer:
43.3 m
Explanation:
d1 = 25.1 m in 15.4° west of north
d2 = 38.8 m in 23.1° south of west
Write the displacements in vector form
[tex]\overrightarrow{d_{1}}=25.1\left ( -Sin15.4\widehat{i}+Cos15.4\widehat{j} \right )=-6.67\widehat{i}+24.2\widehat{j}[/tex]
[tex]\overrightarrow{d_{2}}=38.8\left ( -Cos23.1\widehat{i}-Sin23.1\widehat{j} \right )=-35.69\widehat{i}-15.22\widehat{j}[/tex]
The resultant displacement is given by
[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}[/tex]
[tex]\overrightarrow{d}}=\left ( -6.67-35.69 \right )\widehat{i}+\left ( 24.2-15.22 \right )\widehat{j}[/tex]
[tex]\overrightarrow{d}}=\left ( -42.36 \right )\widehat{i}+\left ( 8.98 \right )\widehat{j}[/tex]
The magnitude of the resultant displacement is given by
[tex]d=\sqrt{8.98^{2}+\left ( -42.36 \right )^{2}}=43.3 m[/tex]
Thus, you are 43.3 m far from your starting point.
A rock is thrown straight up and passes by a window. The window is 1.7m tall, and the rock takes 0.22 seconds to pass from the bottom of the window to the top. How far above the top of the window will the rock rise?
Answer:
The rock will rise 2.3 m above the top of the window
Explanation:
The equations for the position and velocity of the rock are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
y = height of the rock at time t
v0 = initial velocity
y0 = initial height
g = acceleration due to gravity
t = time
v = velocity at time t
If we place the center of the frame of reference at the bottom of the window, then, y0 = 0 and at t = 0.22 s, y = 1.7 m. With this data, we can calculate v0:
1.7 m = 0.22 s · v0 - 1/2 · 9.8 m/s² · (0.22 s)²
Solving for v0:
v0 = 8.8 m/s
Now that we have the initial velocity, we can calculate the time at which the rock reaches its maximum height, knowing that at that point its velocity is 0.
Then:
v = v0 + g · t
0 = 8.8 m/s - 9.8 m/s² · t
-8.8 m/s / -9.8 m/s² = t
t = 0.90 s
Now, we can calculate the max height of the rock:
y = y0 + v0 · t + 1/2 · g · t²
y = 8.8 m/s · 0.90 s - 1/2 · 9.8 m/s² · (0.90 s)²
y = 4.0 m
Then the rock will rise (4.0 m - 1.7 m) 2.3 m above the top of the window
A stone tied to the end of a string is whirled around in
avertical circle of radius R. Find the critical speed below
whichthe spring would become slack at the highest point.
Answer:
v = √rg.
Explanation:
The Minimum speed of the stone that can have to the stone when it is rotated in a vertical circle is √rg.
Mathematical Proof ⇒
at the top point on the circle we have
T + mg = m v²/r
We know that minimum speed will be at the place when its tension will be zero.
∴ v² = rg
⇒ v = √rg.
So, the minimum speed or the critical speed is given as v = √rg.
Answer:
[tex]v=\sqrt{rg}[/tex]
Explanation:
radius of circle = R
Let T be the tension in the string.
At highest point A, the tension is equal to or more than zero, so that it completes the vertical circle. tension and weight is balanced by the centripetal force.
According to diagram,
[tex]T + mg = \frac{mv^{2}}{R}[/tex]
T ≥ 0
So, [tex]mg = \frac{mv^{2}}{R}[/tex]
Where, v be the speed at the highest point, which is called the critical speed.
[tex]v=\sqrt{rg}[/tex]
Thus, the critical speed at the highest point to complete the vertical circle is [tex]v=\sqrt{rg}[/tex].
Vector C has a magnitude of 21.0 m and points in the −y‑ direction. Vectors A and B both have positive y‑ components, and make angles of α=43.4° and β=27.7° with the positive and negative x- axis, respectively. If the vector sum A+B+C=0 , what are the magnitudes of A and B?
The magnitudes of vectors A and B can be determined by resolving each vector into its x and y components, setting up equations for their sum, and solving the equations simultaneously.
Explanation:To find the magnitudes of vectors A and B when their sum with vector C results in a zero vector (A + B + C = 0), we can break down each vector into its x and y components and then solve the component equations separately.
Since vector C points in the negative y-direction and has a magnitude of 21.0 m, its components are Cx = 0 m and Cy = -21.0 m.
Vector A has a positive y-component and makes an angle of α = 43.4° with the positive x-axis. Thus, the components of A are Ax = A cos(α) and Ay = A sin(α).
Vector B also has a positive y-component and makes an angle of β = 27.7° with the negative x-axis, which is the same as 180° - β with the positive x-axis. So, B's components are Bx = -B cos(180° - β) = -B cos(β) and By = B sin(180° - β) = B sin(β).
The equations for the x and y components of the vector sum being zero are:
Substituting the component expressions in, we get:
These are two equations with two unknowns (the magnitudes A and B), which can be solved simultaneously to find the values of A and B. The solution involves elementary algebra and the use of trigonometric identities.
In a movie, a monster climbs to the top of a building 30 m above the ground and hurls a boulder downward with a speed of 25 m/s at an angle of 45° below the horizontal. How far from the base of the building does the boulder land?
Final answer:
The boulder will land approximately 26.34 meters from the base of the building.
Explanation:
To determine the horizontal displacement of the boulder, we need to analyze the horizontal and vertical components of its motion separately.
First, we can determine the time it takes for the boulder to hit the ground using the vertical component of its motion. The initial vertical velocity can be found by multiplying the initial velocity (25 m/s) by the sin of the launch angle (45°). We can then use the equation h = v0yt + 0.5gt2 to solve for the time, where h is the vertical displacement (30 m - 0 m = 30 m), v0y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s2), and t is the time. Solving for t, we get t ≈ 1.87 s.
Next, we can determine the horizontal displacement by multiplying the horizontal component of the initial velocity (25 m/s cos 45°) by the time of flight (1.87 s). Multiplying the values, we get approximately 26.34 m. Therefore, the boulder lands approximately 26.34 m from the base of the building.