Answer:
2.82 s
Explanation:
The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
Y0 is the starting position, 2.3 m in this case.
Vy0 is the starting speed, 13 m/s.
a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.
Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2
It will reach the ground when Y(t) = 0
0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2
-4.9 * t^2 + 13 * t + 2.3 = 0
Solving this equation electronically gives two results:
t1 = 2.82 s
t2 = -0.17 s
We disregard the negative solution. The ball spends 2.82 seconds in the air.
A power electronics package has been designed to handle 400W of power dissipation. Find out the change in output power handling capability of the power electronics package when efficiency is increased from 89% to 94%. Discuss briefly the importance of efficiency base on output power capability calculated.
Final answer:
To calculate the change in output power handling capability of a power electronics package, you can use the formula Power = Output power / Efficiency. By comparing the original power dissipation with the new power dissipation, you can determine the change in output power. Efficiency is important because it affects the amount of input power required and the output power capability of a system.
Explanation:
To find out the change in output power handling capability, we need to calculate the new power dissipation and compare it with the original value. The formula for power is given as:
Power = Output power / Efficiency
Using the formula, we can calculate the original power dissipation:
Original Power = 400W / 0.89 = 449.43W
Next, we can calculate the new power dissipation:
New Power = 400W / 0.94 = 425.53W
The change in output power handling capability is the difference between the original power and the new power:
Change in Output Power = 449.43W - 425.53W = 23.9W
The increase in efficiency from 89% to 94% leads to a decrease in the power dissipation by 23.9W. Efficiency is important because it determines the amount of input power required to achieve a certain output power. Higher efficiency means less power is wasted as heat, resulting in a higher output power capability for a given input power.
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 26.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 2.30 m/s , and puck B moves with a speed of 3.90 m/s .
Your answer should satisfy common sense. For instance, can you decide which of the following values for the distance covered by puck A would definitely be wrong, regardless of the speed of the two pucks and considering that the two pucks are sliding toward each other?
(A) 5 m
(B) 33 m
(C) 27 m
(D) 1 m
(E) 21 m
Answer:
B) 33 m C) 27 m
Explanation:
considering that the two pucks are sliding toward each other we can understand that they are on a collision course.
Since the total distance between them is 26 m, the common sense dictates that the distance traveled by each puck must be less than 26 m regardless of the speed of the two pucks.
so the options B) 33 m and C) 27 m are definitely wrong since they are greater than 26 m.
We can also easily find the distance traveled by each pucks also.
let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] be the velocity of the pucks A and B respectively
[tex]v_{A}t = 26- v_{B}t\\\\2.30t = 26 - 3.90t\\\\6.2t = 26\\\\t = \frac{26}{6.2} \\\\x_{A} =v_{A}t\\\\x_{A} =2.3 \times \frac{26}{6.2}\\\\x_{A} =9.65\\\\x_{B} =v_{B}t\\\\x_{B} =3.9 \times \frac{26}{6.2}\\\\x_{B} =16.35\\[/tex]
Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another reference frame that is moving with a speed of 0.85 relative to the original frame, in the positive x-direction
Answer:
The components of the moving frame is (8.07c, -2, 3, 9.493)
Solution:
As per the question:
Velocity of moving frame w.r.t original frame [tex]v_{m}[/tex] 0.85c
Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane
a = (0, - 2, 3, 5)
Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):
New coordinates are given by:
X = [tex]\frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}[/tex]
X = [tex]\frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}[/tex]
X = [tex]8.07 c[/tex]
Now,
Y = y = - 2
Z = z = 3
Now,
[tex]t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}[/tex]
[tex]t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s[/tex]
student drove to the university from her home and noted that the odometer on her car increased by 14.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0° south of east, what was magnitude of her average velocity?
The student's average speed on her trip to university was 0.78 km/min or 46.67 km/h, while the magnitude of her average velocity was 0.57 km/min or 34.3 km/h.
Explanation:In this scenario, we have to find the average speed and the magnitude of her average velocity. Average speed is calculated by dividing the total distance travelled by the total time taken. So, her average speed is 14.0 km / 18.0 min = 0.78 km/min or 46.67 km/h (note: convert minutes to hours for speed in km/h).
The next part of the question involves average velocity, which is the displacement (straight-line distance) divided by time. It can be different from the average speed when the path travelled is not a straight line. In this case, the magnitude of her average velocity is 10.3 km / 18.0 min = 0.57 km/min or 34.3 km/h.
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What mass of steam at 100°C must be mixed with 499 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 33.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.
Answer:
The mass of the steam is 91.2 g
Mass of the steam=91 grams
Explanation:
Given:
Mass of the ice=499 gFinal temperature of the liquid water[tex]=33^\circ \rm C[/tex]Latent heat of fusion=[tex]333\ \rm kJ\kg[/tex]Latent heat of vaporization =[tex]2256\ \rm kJ\kg[/tex]When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing
So According to question
Let M be the mass of the steam mixed with ice then we have
[tex]M\times2256\times10^3+M\times4186\times(100-33)=0.499\times222\times10^3+0.499\times4186\times(33-0)\\M\times2.58\times10^6=2.35\times10^5\\M=91.2\ \rm g[/tex]
A driver of a car going 60 km/h suddenly sees the lights of a barrier 40 m ahead. It takes the driver 0.80 seconds before he applies the brakes, and the average acceleration during braking is -9.5 m/s. (A) Does the car hit the barrier? Explain. (B) What is the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead?
Answer:
a) The car doesn't hit the barrier because after he sees the lights of a barrier he only travels for 27.95m, enough to miss the barrier. b) The maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s or 75.6km/hr.
Explanation:
a)
In order to solve this problem we must first do a drawing of the situation for us to visualize it better. (See picture attached).
As you may see on the drawing, the car still travels some distance when the driver notices the lights of the barrier. This distance is calculated for a constant velocity V:
x=Vt
the car has a velcity of 60km/h which is equivalent to:
[tex]\frac {60km}{hr} * \frac{1000m}{1km}*\frac{1hr}{3600s}=16.667m/s[/tex]
so on the first 0.80s the car travels a distance of:
x=(16.667m/s)(0.8s)=13.333m
Next, the car breaks, so we can say it moves with a constant acceleration of -9.5m/s, so the distance is found by using the following formula:
[tex]x=\frac {V_{f}^{2}-V_{0}^{2}}{2a}[/tex]
We know the final velocity will happen when the car stops, so the final velocity is zero, leaving us with:
[tex]x=\frac {-V_{0}^{2}}{2a}[/tex]
so we can substitute the provided values to find the distance traveled by the car during this time.
[tex]x=\frac{-(16.667m/s)^{2}}{2(-9.5m/s^{2})}[/tex]
which yields:
x=14.62m
so in total the car traveled:
[tex]x_{tot}=13.333m+14.62m = 27.95m[/tex]
Which is not enough for the car to hit the barrier.
b)
In order to solve this part of the problem, we must combine the two equations we got on the previous part to find a single equation that will represent the total displacement of the car:
[tex]x=vt-\frac{v^{2}}{2a}[/tex]
so we can now substitute the known values so we get:
[tex]40=0.8v-\frac{v^{2}}{2(-9.5)}[/tex]
which simplifies to:
[tex]40=0.8v+0.05263v^{2}[/tex]
this is a quadratic equation so it can be solved by using the quadratic formula, but first we must rewrite it in standard form, so we get:
[tex]0.05263v^{2}+0.8v-40=0[/tex]
now we can use the quadratic formula:
[tex]v=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]
So we can substitute the given values:
[tex]v=\frac{-0.8\pm\sqrt{(0.8)^{2}-4(0.05263)(-40)}}{2(0.05263)}[/tex]
which returns two answers:
v=21m/s and v=-36.20m/s
We take the positive answer since is the one that represents a moving towards the right side of the drawing.
So the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s
A person on a trampoline can go higher with each bounce. how is this possible? is there a maximum height to which this person can go?
Explanation:
when a person jumps on the trampoline he stores his potential energy in the form of elastic energy of trampoline. Potential Energy converts into Elastic energy when a person is at the bottom point during stretching of the trampoline. When he again going upwards, Elastic energy is converted to the potential energy of the person. This is the reason why a person goes higher each time. This process goes on until trampoline reaches its elastic limit and finally breaks or get permanently deformed.
So there is a limit up to which a person can be reached.
Final answer:
A person goes higher on a trampoline by landing and pushing off with their feet, making better use of leg strength to transform kinetic to potential energy. There is a maximum height due to energy losses like air resistance and imperfect energy transfers. The sweet spot in tennis rackets illustrates efficient energy transfer with reduced arm jarring.
Explanation:
A person on a trampoline goes higher with each bounce because they can harness the elastic potential energy stored in the trampoline material. When landing on the back or feet, the height reached can differ.
Typically, a person may reach greater heights when landing and launching off their feet because they can make use of the stronger leg muscles to add upward force, thus converting more kinetic energy to gravitational potential energy upon ascent. There is, indeed, a maximum height obtainable, as energy losses due to factors like air resistance and less-than-perfect energy transfer prevent infinite increases in height.
Addressing the professional application, the 'sweet spot' on a tennis racket is an example of an optimal point where energy transfer from the racket to the ball is the most efficient, and vibrating force transmitted to the player's arm is minimal, hence no jarring of the arm.
The physics of motion and energy conservation provides us the information that increasing the initial speed (kinetic energy) of the object would result in a higher ascent, yet this is bounded by practical limitations such as the strength of the jumper and the efficiency of the trampoline material.
For example, to double the impact speed of a falling object, one would have to quadruple the height from which it falls, due to the relationship between gravitational potential energy and kinetic energy.
A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?
Answer:
The no. of electrons is [tex]7.59\times 10^{21}[/tex]
Solution:
According to the question:
The rate at which the charge is delivered is given by:
[tex]\frac{dQ}{dt} = - 0.75[/tex]
Now,
[tex]\int_{0}^{Q}dQ = - 0.75\int_{0}^{27 min} dt[/tex]
[tex]Q = -0.75t|_{0}^{27 min}[/tex]
[tex]Q= -0.75\times 27\times 60 = - 1215 C[/tex]
No. of electrons, n can be calculated from the following relation:
Q = ne
where
e = electronic charge =[tex]1.6\times 10^{- 19} C[/tex]
Thus
[tex]n = \frac{Q}{e}[/tex]
[tex]n= \frac{1215}{1.6\times 10^{- 19}}[/tex]
[tex]n = 7.59\times 10^{21}[/tex]
An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at the top 16.0°C, what is the volume of the bubble just before it reaches the surface?
Final answer:
The volume of the air bubble just before it reaches the surface is 3.32 cm^3.
Explanation:
To find the volume of the air bubble just before it reaches the surface, we can use the ideal gas law equation: PV = nRT. Since the pressure and amount of gas remain constant, we can rewrite the equation as V/T = k, where V is the volume, T is the temperature, and k is a constant.
Using the given temperatures at the bottom and top of the lake (5.9°C and 16.0°C) and the initial volume of the bubble (1.22 cm^3), we can set up the following equation:
(1.22 cm^3) / (5.9°C) = V / (16.0°C).
Solving for V, the volume of the bubble just before it reaches the surface, we get:
V = (1.22 cm^3)(16.0°C) / (5.9°C) = 3.32 cm^3.
What is the electrostatic force between and electron and a proton separated by 0.1 mm? 2.3 × 10^-20N, attractive.
2.3 × 10^-20N, repulsive.
2.3 × 10^-18N, attractive.
2.3 × 10^-18N, repulsive.
2.3 × 10^-16N, attractive.
The electrostatic force between an electron and a proton separated by 0.1mm, calculated using Coulomb's law, is approximately 2.3 × 10^-20N, and it is attractive due to the opposite charges.
Explanation:The electrostatic force between charged particles can be calculated using Coulomb's law: F = k(Q1*Q2)/r², where F is the force, k is Coulomb's constant (~8.99 * 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the separation in meters. An electron and a proton have charges of -1.6*10^-19 C and +1.6*10^-19 C respectively. Plugging these values into Coulomb's law with a separation of 0.1 mm or 0.1*10^-3 meters, we find that the force is approximately 2.3 × 10^-20N, and it is attractive because the charges are opposite.
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"The correct answer is c. [tex]\(2.3 \times 10^{-18}\)[/tex] N, attractive.
To find the electrostatic force between an electron and a proton separated by a distance, we use Coulomb's law, which is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
The charge of an electron is approximately [tex]\(-1.602 \times 10^{-19}\) C[/tex]and the charge of a proton is approximately [tex]\(+1.602 \times 10^{-19}\) C[/tex]. The separation distance is given as 0.1 mm, which we need to convert to meters: [tex]\(0.1 \times 10^{-3}\)[/tex] m or [tex]\(1 \times 10^{-4}\)[/tex] m.
Plugging in the values, we get:
[tex]\[ F = (8.9875 \times 10^9) \frac{(1.602 \times 10^{-19})(1.602 \times 10^{-19})}{(1 \times 10^{-4})^2} \] \[ F = (8.9875 \times 10^9) \frac{2.5664 \times 10^{-38}}{1 \times 10^{-8}} \] \[ F = (8.9875 \times 10^9) (2.5664 \times 10^{-30}) \] \[ F = 2.3017 \times 10^{-19} \][/tex]
Since the force is attractive (opposite charges attract), the magnitude of the force is [tex]\(2.3 \times 10^{-19}\) N[/tex], but we need to express it in the format given in the options, which is [tex]\(2.3 \times 10^{-18}\) N[/tex].
An engineer is undertaking some reconnaissance by pacing on a construction site. The engineer takes 26 steps per 20m, the ground he is walking on has a vertical slope angle of 6. The engineer takes 152 steps. How far has he travelled horizontally (in metres)?(only enter numeric answer to 2 decimal places, no alpha characters)
Answer:
Thus the distance traveled horizontally is 116.28 m
Solution:
As per the question:
Distance covered in 26 steps = 20 m
Thus
Distance covered in one step, step size = [tex]\frac{20}{26}[/tex]
Total steps taken = 152
Now, the distance covered in 152 steps = [tex]one step size\times 152[/tex]
The distance covered in 152 steps, D = [tex]\frac{20}{26}\times 152 = 116.92 m[/tex]
The above distances are on a slope of [tex]6^{\circ}[/tex] above horizontal.
Thus the horizontal component of this distance is given by:
[tex]d_{H} = Dcos6^{\circ}[/tex]
where
[tex]d_{H}[/tex] = horizontal component of distance, D
[tex]d_{H} = 116.92cos6^{\circ} = 116.28 m[/tex]
The driver of a pick up truck going 100 km/h applies the brakes, giving the truck a form deceleration of 6.50 m/s^2 while it travels 20.0 m. What is the speed of the truck in km/h at the end of this distance?
Answer:
81.42 km/h
Explanation:
t = Time taken for the car to slow down
u = Initial velocity = 100 km/h
v = Final velocity
s = Displacement = 20 m = 0.02 km
a = Acceleration = -6.5 m/s² = -0.0065×60×60×60×60 = -84240 km/h²
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -84240\times 0.02+100^2}\\\Rightarrow v=81.42\ km/h[/tex]
Speed of the truck at the end of this distance is 81.42 km/h
Using the equation of motion v^2 = u^2 + 2as, and converting the initial speed from km/hr to m/s, the final velocity is approximately 15.22 m/s when deceleration and distance are considered. Converted back to km/hr, this is approximately 54.8 km/hr.
Explanation:The final speed of the truck can be calculated using one of the equations of motion, specifically v^2 = u^2 + 2as, where 'v' is final velocity, 'u' is initial velocity, 'a' is acceleration (which in this case is -6.50 m/s^2 due to deceleration), and 's' is distance. Convert the initial speed from km/hr to m/s: 100 km/hr = 27.78 m/s.
Then, we can plug the values into the formula: v^2 = (27.78 m/s)^2 - 2(6.50 m/s^2)(20.0 m). Solving the equation gives a final velocity of approximately 15.22 m/s. This speed in km/h is then obtained by converting m/s back, 15.22 m/s * (3600 / 1000) = 54.8 km/h. So, the speed of the truck at the end of the distance is approximately 54.8 km/h.
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Consider a charged particle at a pointS whose coordinates are (8 m, 4 m, 10 m). We would like to find the electric field vector at a point P whose coordinates are (7 m, 2 m, 6 m). The "unit vector" r ^ is a vector that points from S to P that has length of 1 (or "unity"). What is its y component, in meters
Answer:
E=[tex]k*\frac{q}{21}*u[/tex]
[tex]u=\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]
[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]
Explanation:
q: particle's charge
k: coulomb constant
E=E*u
r=r*u
r=distancia vectorial entre P y S
r=distancia escalar entre P y S
E: Electric field vector
E: magnitud of magnetic field vector
u: unit vector radial
then:
[tex]E=k*q/r^{2}[/tex]
r=r*u
r=P-S=(-1,-2,-4)m
[tex]r^{2}=(Magnitude(P-S))^2=(-1)^2+(-2)^2+(-4)^2=21[/tex]
[tex]r=\sqrt{21}[/tex]
E=[tex]k*\frac{q}{21}*u[/tex]
u=r/r=[tex]\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]
[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]
You have put a 1,000 μF capacitor and a 2,000 ohm resistor in series with an AC voltage source with amplitude of 45 V and frequency of 4,000 Hz. What is Vout across the capacitor? What is Vout across the resistor?
Answer:
voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]
Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]
Explanation:
We have given resistance R = 2000 OHM
Capacitance [tex]C=1000\mu F=0.001F[/tex]
Voltage V = 45 volt
Frequency = 4000 Hz
Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\times \pi \times 4000\times 0.001}=0.04ohm[/tex]
Impedance [tex]Z=\sqrt{R^2+X_C^2}=\sqrt{2000^2+0.04^2}=2000ohm[/tex]
Current [tex]i=\frac{V}{Z}=\frac{45}{2000}=0.022A[/tex]
Now voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]
Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]
A truck with a mass of 1650 kg and moving with a speed of 15.0 m/s rear-ends a 779 kg car stopped at an intersection. The collision is approximately elastic since the car is in neutral, the brakes are off, the metal bumpers line up well and do not get damaged. Find the speed of both vehicles after the collision in meters per second.
Answer:
Truck's speed = 5.21 m/s
Car's speed = 20.2 m/s
Explanation:
Given:
Mass of truck = M = 1650 kg
Speed of the truck initially = U = 15 m/s
Mass of the car = m = 779 kg
Initial speed of the car =u = 0
From the momentum conservation, Total initial momentum = Total final momentum.
M V+m U = M V +m v
⇒ (1650)(15) + 779×0 = (1650)V + 779 v
⇒ 24750 = 1650 V+779 v →(1)
Since the collision is elastic, relative velocity of approach = relative velocity of separation. 15 = v - V
⇒ v =V + 15; This is now substituted in the equation(1) above.
24750 = 1650 V + (799) (V+15)
⇒ 24750 = 1650 V + 799 V + 11985
⇒ 2449 V = 12765
⇒ Final velocity of the truck = [tex]\frac{12765}{2449}[/tex] = 5.21 m/s
Final velocity of the car = v = V+15 = 5.21 + 15 = 20.2 m/s
A car is decelerating at the rate of 2 km/s square. If its initial speed is 66 km/s, how long will it take the car to come to a complete stop?
Answer:
It will take 33 seconds to stop the car.
Explanation:
Using the first equation of kinematics we have
[tex]v=u+at[/tex]
where
'v' is final speed of object
'u' is initial speed of object
'a' is acceleration of object
't' is time of acceleration of object
Now since it is given that [tex]a=-2km/s^{2}[/tex] since acceleration is negative and [tex]u=66km/s[/tex]
We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get
[tex]0=66-2\times t\\\\\therefore t=\frac{66}{2}=33seconds[/tex]
Final answer:
Using the kinematic equation for velocity and acceleration, it is calculated that the car, decelerating at a rate of 2 km/s² from an initial speed of 66 km/s, will take 33 seconds to come to a complete stop.
Explanation:
To find out how long it will take the car to come to a complete stop, we need to use the kinematic equation that relates initial velocity, acceleration, and time without displacement:
Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))
Here, the final velocity (v) is 0 km/s, since the car is coming to a stop, the initial velocity (u) is 66 km/s, and the acceleration (a) is -2 km/s² (negative because it is deceleration).
The equation becomes:
0 = 66 + (-2 × t)
We can solve for t as follows:
0 = 66 - 2t
2t = 66
t = 33 seconds
So it will take the car 33 seconds to come to a complete stop.
At time t=0, rock A is dropped from rest from a height of 90 m. At the same instant, rock B is launched straight up from the ground level with an initial speed of 30 m/s. Write an equation of motion for rock A and B, giving its position at all time.
Answer:
rock A: [tex]y=90-1/2*g*t^2[/tex]
rock B: [tex]y=30*t-1/2*g*t^2[/tex]
g=9.81m/s^2
Explanation:
Kinematics equation:
[tex]y=y_{o}+v_{oy}*t+1/2*a*t^2[/tex]
in our case the acceleration is the gravity and it has a negative direction.
a=-g
rock A, yo=90m, Voy=0m/s:
[tex]y=90-1/2*g*t^2[/tex]
rock B, yo=0m, Voy=30m/s:
[tex]y=30*t-1/2*g*t^2[/tex]
A 75 kg astronaut has become detached from their her space ship. To get back to the ship she throws a tool in the opposite direction to the spaceship with a force of 16 N. What is her acceleration during the throw assuming that distances going away from the spaceship are positive?
Answer:
0.21 m/s/s.
Explanation:
Whenever there is an action force acting on a body, there will be a reaction force.
Here the force with which the astronaut throws the tool is given as 16 N.
Force is measured in newtons and is equal to the rate of change of momentum.
Since the astronaut has a mass, she experience a reaction force. It is given by F = ma, according to Newton's 2nd law.
16 = 75 a
⇒ Acceleration = a = F/m = 16/75 = 0.21 m/s/s
an automobile travels on a straight road for 42 km at 45 km/h. it then continues in the same direction for another 42 km at 90 km/h. (assume that it moves in the positive x direction.) a. what is the average velocity of the car during this 84 km trip?
b. what is it’s average speed?
C. Graph x versus t and indicate how the average velocity is found on the graph?
Answer:
a) The average velocity is v = (60 km/h ; 0)
b) The average speed is 60 km/h
Explanation:
The velocity is a vector that has a magnitude and direction. The average speed is the distance traveled over time without taking into account the direction of the motion.
a)The average velocity is calculated as the displacement over time:
v = Δx/Δt
where
v = velocity
Δx = final position - initial position = traveled distance relative to the center of the reference system.
Δt = final time - initial time (initial time is usually = 0)
We know that the displacement is 84 km but we do not know the time. It can be calculated from the two parts of the trip.
In part 1:
v = 45 km/h = 42 km / t
t = 0.93 h
In part 2:
v = 90 km/h = 42 km / t
t = 42 km / 90 km/h
t = 0.47 h
The time of travel is 0.47 h + 0.93 h = 1.4 h
The average velocity will be:
v = 84 km / 1.4 h = 60 km/h
Expressed as a vector in a 2-dimension plane:
v = (60 km/h; 0)
b) The average speed is calculated as the distance traveled over time. Note that in this case, the distance is equal to the displacement since the direction of the motion is always in one direction. But if the direction of the second part of the trip would have been the opposite to the direction of the first part, the displacement would have been 0 (final position - initial position = 0, because final position = initial position), then, the average velocity would have been 0. In change, the average speed would have been the distance traveled (84 km, 42 km in one direction and 42 km in the other) over time.
Then:
average speed = 84 km / 1.4 h = 60 km/h
c) see attached figure.
A commuting student leaves home and drives to school at an average speed of 39.0 km/h. After 23.0 min he realizes that he has forgotten his homework and returns home to get it at the same average speed. It takes 10.0 min to find the report, after which the trip to school 39.0 km away to the east is resumed at the same speed as before. What is the average speed for the entire trip ? What is the average velocity for the entire trip?
The average speed for the entire trip is 83.6 km/h. The average velocity for the entire trip is zero.
Explanation:The average speed for the entire trip can be calculated by finding the total distance traveled divided by the total time taken. In this case, the student traveled 39.0 km to school and then returned the same distance back home, resulting in a total distance of 78.0 km. The total time taken for the entire trip is the sum of the time taken to drive to school, the time taken to return home, and the time spent finding the report, which is 23.0 min + 23.0 min + 10.0 min = 56.0 min. To convert the total time to hours, divide by 60: 56.0 min / 60 = 0.933 hours. Therefore, the average speed for the entire trip is 78.0 km / 0.933 hours = 83.6 km/h.
The average velocity for the entire trip can be determined by considering both the magnitude and direction of the displacement. In this case, since the student drove in the same direction to school and back home, the displacement is zero, meaning there was no change in position. Thus, the average velocity for the entire trip is also zero.
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. That is, a = F / mWhat are the dimensions of force?
Answer:
[tex][F]=[MLT^{-2}][/tex]
Explanation:
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :
F = m × a
F is the applied force
m is the mass of the object
a is the acceleration due to gravity
We need to find the dimensions of force. The dimension of force m and a are as follows :
[tex][m]=[M][/tex]
[tex][a]=[LT^{-2}][/tex]
So, the dimension of force F is, [tex][F]=[MLT^{-2}][/tex]. Hence, this is the required solution.
Final answer:
In physics, the dimensions of force are derived from Newton's second law of motion, which can be represented as F=ma. Force has dimensions of mass (M), length (L), and time (T), leading to the dimensional formula [F] = [M][L][T]^-2. The SI unit for force is the Newton (N).
Explanation:
According to Newton's second law of motion, the acceleration a of an object is directly proportional to the net external force Fnet acting upon it and inversely proportional to its mass m. This law is mathematically represented by the equation Fnet = m × a, where Fnet is the net force, m is the mass, and a is the acceleration of an object.
The weight w of an object is another type of force, which is defined as the gravitational force acting on an object with mass m. The object experiences an acceleration due to gravity g, and this is represented by the equation w = m × g. In the International System of Units (SI), force is measured in Newtons (N), and one Newton is defined as the force required to accelerate a one-kilogram mass at a rate of one meter per second squared (1 N = 1 kg × m/s2).
Therefore, the dimensions of force in terms of the base physical quantities are mass (M), length (L), and time (T), and the dimensional formula for force is [F] = [M][L][T]-2.
The acceleration of a particle moving along a straight line is given by a = −kt2 m/s2 where k is a constant and time t is in seconds. The initial velocity of the particle at t = 0 is v0 = 12 m/s and the particle reverses it direction of motion at t = 6 s. Determine the constant k and the displacement of the particle over the same 6-second interval of motion. Ans: k = 1/6 m/s4, Δs = 54 m
Answer:
[tex]X - Xo = 54m[/tex]
k = 1/18
Explanation:
Data:
a = -k[tex]t^{2}[/tex][tex]\frac{m}{s^{2} }[/tex]
to = 0s Vo = 12m/s
t = 6s the particle chage it's moviment, so v = 0 m/s
We know that acceleration is the derivative of velocity related to time:
[tex]a = \frac{dV}{dT}[/tex]
rearranging...
[tex]a*dT = dV[/tex]
Then, we must integrate both sides:
[tex]\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^{2} } \, dT[/tex]
[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]
V = 0 because the exercise says that the car change it's direction:
[tex]0 - 12 = -k\frac{6^{3} }{3}[/tex]
k = 1/6
In order to find X - Xo we must integer v*dT = dX
[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]
so...
[tex](Vo -k\frac{t^{3} }{3})dT = dX[/tex]
[tex]\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k\frac{t^{3} }{3} } \, dT[/tex]
integrating...
[tex]X - Xo = Vot -k\frac{t^{4} }{12}[/tex]
[tex]X - Xo = 12*6 -\frac{1}{6}* \frac{6^{4} }{12}[/tex]
X - Xo = 54m
A brick is thrown upward from the top of a building at an angle of 25.7° above the horizontal and with an initial speed of 15.1 m/s. The acceleration of gravity is 9.8 m/s^2 If the brick is in flight for 3.4 s, how tall is the building? Answer in units of m
Answer:
the building is 34.408 m tall
Explanation:
given,
initial velocity of brick = 15.1 m/s
at an angle of = 25.7°
vertical component of the velocity
Vy = 15.1 sin 25.7°
= 6.54 m/s
we know
[tex]s = u t + \dfrac{1}{2} a t^2[/tex]
[tex]s = 6.54\times 3.4 + \dfrac{1}{2}\times -9.8 \times 3.4^2[/tex]
s = -34.408 m
hence, the building is 34.408 m tall .
A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if the image (formed) is real?
Answer:
Image distance is -52.5 cm
Image is virtual and forms on the same side of the lens and upright image is formed.
Explanation:
u = Object distance
v = Image distance
f = Focal length = 35
m = Magnification = 2.5
[tex]m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u[/tex]
Lens equation
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm[/tex]
[tex]v=-2.5\times 21=-52.5\ cm[/tex]
Image distance is -52.5 cm
Image is virtual and forms on the same side of the lens and upright image is formed.
A ball is kicked with an initial velocity of 18.1 m/s in the horizontal direction and 15.8 m/s in the vertical direction. For how long does the ball remain in the air?
Answer:
3.22 s
Explanation:
The ball would describe a projectile motion, where the horizontal velocity will remain constant, as there are no forces that act on the x-axis, and the vertical velocity will vary because of gravity in the following way:
1. First, the ball will go up, but the vertical velocity will decrease until it has a value of 0.
2. After the vertical velocity has reached the value of 0, the ball will start to fall, with the vertical velocity increasing because of gravity.
You need to know that the time that the ball's verical velocity takes to reach 0 is exactly the same that it takes to go from 0 to its original vertical velocity:
[tex]a = \frac{v_f - v_o}{t}\\t = \frac{v_f - v_o}{a}[/tex]
And not only the time will be the same, but also the distance traveled. Therefore, we can conclude that the time that the ball remain in the air is simply two times the time it takes for the ball to desacelerate:
[tex]t_{desaceleration} = \frac{v_f - v_o}{a} = \frac{0m/s - 15.8m/s}{-9.81 m/s^2} =1.61 s\\t_{air} = 2* t_{desaceleration} = 2*1.61s = 3.22 s[/tex]
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic, (b) fcc, (c) bcc, and (d) diamond lattice. As- suming that nearest atoms are touching each other, what is the lattice constant of each lattice?
Answer:
(a) A = [tex]3.90 \AA[/tex]
(b) [tex]A = 4.50 \AA[/tex]
(c) [tex]A = 5.51 \AA[/tex]
(d) [tex]A = 9.02 \AA[/tex]
Solution:
As per the question:
Radius of atom, r = 1.95 [tex]\AA = 1.95\times 10^{- 10} m[/tex]
Now,
(a) For a simple cubic lattice, lattice constant A:
A = 2r
A = [tex]2\times 1.95 = 3.90 \AA[/tex]
(b) For body centered cubic lattice:
[tex]A = \frac{4}{\sqrt{3}}r[/tex]
[tex]A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA[/tex]
(c) For face centered cubic lattice:
[tex]A = 2{\sqrt{2}}r[/tex]
[tex]A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA[/tex]
(d) For diamond lattice:
[tex]A = 2\times \frac{4}{\sqrt{3}}r[/tex]
[tex]A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA[/tex]
A point of charge 4.9 μC is placed at the origin (x1 = 0) of a coordinate system, and another charge of -1.1 μC is placed on the x-axis at x2 = 0.28 m. a) Where on the x-axis can a third charge be placed in meters so that the net force on it is zero?
b) What if both charges are positive; that is, what if the second charge is 1.9 μC?
To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. If both charges are positive, the net force on a third charge placed on the x-axis will never be zero.
Explanation:a) To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. The net force will be zero when the two forces are equal in magnitude but opposite in direction. Using Coulomb's law, we can calculate the force between the first charge and the third charge at an unknown position x3. We can then set that force equal to the force between the second charge and the third charge at the same position x3, and solve for x3.
b) If both charges are positive, the net force on a third charge placed on the x-axis will never be zero. Positive charges repel each other, so the forces will always be in the same direction.
A ball is thrown horizontally with a speed of 15 m/s, from the top of a 6.0 m tall hill. How far from the point on the ground directly below the launch point does the ball strike the ground?
Answer:16.59 m
Explanation:
Given
initial horizontal speed of ball(u)=15 m/s
Height of building =6 m
Consider vertical motion first
[tex]h=u_yt+\frac{1}{2}gt^2[/tex]
here initial vertical velocity is zero
[tex]6=0+\frac{1}{2}\times 9.81\times t^2[/tex]
[tex]t=\sqrt{\frac{12}{9.81}}=1.106 s[/tex]
Thus time taken will also be 1.106 s in horizontal motion
[tex]R_x=u_xt+\frac{1}{2}at^2[/tex]
here a=0
[tex]R_x=15\times 1.106=16.59 m[/tex]
The total time that the ball is in the air can be calculated using the formula for free fall, which is 1.10 seconds. Given that time and the constant horizontal speed, we can then calculate the total horizontal distance traveled by the ball, which is 16.5 meters.
Explanation:To calculate the horizontal distance the ball travels before striking the ground, we must first determine how long the ball is in the air. This time is only affected by the vertical motion, thus we can treat the ball as if it were dropped from the hill with no initial horizontal velocity. The formula used to calculate the time is based on free fall, where t = sqrt((2*h)/g). In this case, h = 6.0 m (the height of the hill) and g = 9.8 m/s² (the acceleration due to gravity). This gives us t = sqrt((2*6)/9.8) = 1.10 s.
Once we have the time of flight, we can calculate the horizontal distance traveled using the horizontal speed of the ball. As the horizontal motion occurs at a constant velocity we can use the formula d = v*t, where v = 15 m/s (the speed the ball is thrown) and t = 1.10 s. This gives us the horizontal distance d = 15*1.10 = 16.5 m.
Learn more about Projectile Motion here:https://brainly.com/question/29545516
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A charge 4.96 nC is placed at the origin of an xy-coordinatesystem, and a charge -1.99 nC is placed on the positive x-axis at x = 4.01 cm . A third particle, of charge 5.99 nC is now placed at the point x = 4.01 cm , y = 2.98 cm . Find the x-component of the total force exerted on the third charge by the other two. Find the y-component of the total force exerted on the third charge by the other two. Find the magnitude of the total force acting on the third charge. Find the direction of the total force acting on the third charge.
Answer:
B) x-component of the total force exerted on the third charge by the other two (Fn₃x)
Fn₃x =8,56*10⁻⁵ N (+x)
B) y-component of the total force exerted on the third charge by the other two (Fn₃y)
Fn₃y = 5.7*10⁻⁵ N (-y)
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Equivalences:
1nC= 10⁻⁹C
1cm = 10⁻²m
Data
q₁=4.96 nC = +4.96*10⁻⁹C
q₂=-1.99 nC = -1.99*10⁻⁹C
[tex]d_{1} =\sqrt{ 4.01^{2}+2.98^{2} } = 4.996 cm = 4.996*10^{-2} m=49.96*10^{-3} m[/tex]
d₂= 2.98 cm
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs.
the force F₂₃ of q₂ and q₃ is attractive because the charges have opposite signs.
Calculation of the forces exerted on the charge q₁ and q₂ on q₃
To calculate the magnitudes of the forces exerted by the charges q₁, q₂, on q₃ we apply Coulomb's law:
F₁₃=k*q₁*q₃/d₁² =9*10⁹*4.96*10⁻⁹*5.99*10⁻⁹/(49.96*10⁻³)²=10.7**10⁻⁵ N
F₂₃=k*q₂*q₃/d₂² =9*10⁹*1.99*10⁻⁹*5.99*10⁻⁹/(2.98*10⁻²)²=12.08*10⁻⁵ N
F₁₃x=F₁₃cosβ=10.7*10⁻⁵* (4.01/4.996)=8,56*10⁻⁵ N (+x)
F₁₃y=F₁₃sinβ= 10.7*10⁻⁵* (2.98/4.996)=6,38*10⁻⁵ N (+x)
F₂₃x= 0
F₂₃y= F₂₃=12.08*10⁻⁵ N (-y)
A) x-component of the total force exerted on the third charge by the other two (Fn₃x)
Fn₃x= F₁₃x + F₂₃x= 8,56*10⁻⁵ N + 0
Fn₃x =8,56*10⁻⁵ N (+x)
B) y-component of the total force exerted on the third charge by the other two (Fn₃y)
Fn₃y = F₁₃y + F₂₃y= 6,38*10⁻⁵ N - 12.08*10⁻⁵ N
Fn₃y = 5.7*10⁻⁵ N (-y)
The charges, 4.96 nC, -1.99 nC, and 5.99 nC, forming a triangle gives
the following approximate values of the force at the 5.99 nC charge.
x-component is [tex]\underline{8.6 \times 10^{-5} \, \mathbf{\hat i} \ N}[/tex]y-component is [tex]\underline{-5.69 \times 10^{-5} \, \mathbf{\hat j} \ N}[/tex]The magnitude of the total force is 1.03 × 10⁻⁴ NThe direction of the force is 33.5° clockwise from the horizontal x-axisHow can the force acting at the 5.99 nC charge be resolved?The given charges are;
Q₁ = 4.96 nC, at point (0, 0)
Q₂ = -1.99 nC, at point (4.01, 0)
Q₃ = 5.99 nC at point (4.01, 2.98)
According to Coulomb's Law, we have;
[tex]F_{13} = \mathbf{\dfrac{9 \times 10^9 \times \left(4.96 \times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{4.01^2 + 2.98^2} } }\approx 1.07 \times 10^{-4}[/tex]
F₁₃ ≈ 1.07 × 10⁻⁴ N
The components of the force are;
[tex]cos\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat i} + sin\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat j}[/tex]
Which gives;
[tex]\vec{F_{13}} \approx \mathbf{8.6 \times 10^{-5} \, \mathbf{\hat i} + 6.39 \times 10^{-5} \, \mathbf{\hat j}}[/tex]
Therefore;
[tex]F_{23}= \dfrac{9 \times 10^9 \times \left((-1.99)\times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{2.98^2} } \approx\mathbf{ 1.208 \times 10^{-4}}[/tex]
Which gives;
[tex]\vec{F_{23}} \approx \mathbf{-1.208 \times 10^{-4} \, \mathbf{\hat j}}[/tex]
The components of the force at Q₃ is therefore;
x-component = [tex]8.6 \times 10^{-5} \, \mathbf{\hat i} + 0 = \underline{8.6 \times 10^{-5} \, \mathbf{\hat i}}[/tex]y-component = [tex]6.39 \times 10^{-5} \, \mathbf{\hat j} + -1.208 \times 10^{-4} \, \mathbf{\hat j} \approx \underline{-5.69 \times 10^{-5} \, \mathbf{\hat j}}[/tex]The magnitude of the total force is therefore;
|F| ≈ √((8.6×10⁻⁵)² + (-5.69 × 10⁻⁵)²) ≈ 1.03 × 10⁻⁴
The magnitude of the total force, |F| ≈ 1.03 × 10⁻⁴ NThe direction of the total force is found as follows;
[tex]The \ direction, \ \theta \approx \mathbf{arctan \left(\dfrac{-5.69}{8.6} \right)} \approx -33.5 ^{\circ}\alpha[/tex]
The force acta in a direction of approximately 33.5° clockwise from the horizontal x-axisLearn more about Coulomb's Law on electric force here
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A snowboarder starts from rest at the top of a double black diamond hill. As he rides down the slope, GPS coordinates are used to determine his displacement as a function of time, x = 0.5t3 + t2 + 2t, where x and t are expressed in feet and seconds, respectively. Determine the position, velocity, and acceleration of the boarder expressed when t = 12 seconds. (Round the final answer to one decimal place.)
To find the position, velocity, and acceleration of the snowboarder at t = 12 seconds, evaluate the given displacement equation and its first and second derivatives at t = 12 seconds, then round the final answers to one decimal place.
Explanation:To determine the position, velocity, and acceleration of the snowboarder at t = 12 seconds, we need to evaluate the given displacement function x(t) = 0.5t³ + t² + 2t and its derivatives at that specific time value.
Position at t = 12 seconds
By substituting t = 12 into the displacement function, we get:
x(12) = 0.5(12)³ + (12)² + 2(12), which can be calculated to give the position x.
Velocity at t = 12 seconds
The velocity v(t) is the first derivative of the displacement function, v(t) = 1.5t² + 2t + 2. Evaluate v(12) to find the velocity at t = 12 seconds.
Acceleration at t = 12 seconds
Acceleration a(t) is the second derivative of the displacement function, a(t) = 3t + 2. Evaluate a(12) to find the acceleration at t = 12 seconds.
Using these steps, you can calculate the exact values and round them to one decimal place as asked in the question.