Answer:
A.) t₁= 4.37 s
B.) d₁= 19.09 m
C.) [tex]v_{f1} = 8,74 \frac{m}{s}[/tex]
Explanation:
Bicyclist kinematics :
Bicyclist moves with uniformly accelerated movement:
d₁ = v₀₁*t₁ + (1/2)a₁*t₁² equation (1)
d₁ : distance traveled by the cyclist
v₀₁: initial speed (m/s)
a₁: acceleration (m/s²)
t₁: time it takes the cyclist to catch his friend (s)
Friend kinematics:
Friend moves with constant speed:
d ₂= v₂*t₂ Equation (2)
d₂ : distance traveled by the friend (m)
v₂: friend speed (m/s)
t₂ : time that has elapsed since the cyclist meets the friend until the cyclist catches him.
Data
v₀₁=0
a₁ = 2.0 m/s²
v₂ = 3.0 m/s
Problem development
A.) When the bicyclist catches his friend, d₁ = d₂=d and t₂=t₁+2
in the equation (1) :
d = 0 + (1/2)2*t₁² = t₁²
d = t₁² equation (3)
in the equation (2) :
d = 3*(t₁+2) ₂ = 3*t₁+6
d = 3t₁+6 equation (4)
Equation (3) = Equation (4)
t₁² = 3t₁+6
t₁² - 3t₁ - 6 = 0 ,we solve the quadratic equation
t₁= 4.37 s : time it takes the cyclist to catch his friend
B.) d₁ : distance traveled by the cyclist
In the Equation (2) d₁= (1/2)2* 4.37² = 19.09 m
C.) speed of the cyclist when he catches his friend
[tex]v_{f1} = v_{o1} + a*t[/tex]
[tex]v_{f1} =0 + 2* 4.37 [/tex]
[tex]v_{f1} = 8.74 \frac{m}{s}[/tex]
A diver springs upward from a board that is 3.90 m above the water. At the instant she contacts the water her speed is 11.8 m/s and her body makes an angle of 78.2° with respect to the horizontal surface of the water. Determine her initial velocity, both (a) magnitude and (b) direction.
Answer:
V = 7.91 m/s(magnitude) @ 72.3°(direction)
Explanation:
We will have to do a parabolic movement analysis in order to determine the initial velocity. For this we will mainly use the following kinetic formulae:
[tex]1) X_{f}=X_{0}+V_{0}t+\frac{at^{2}}{2}\\\\2) V_{f}=V_{0}+at[/tex]
We will work backwards from the initial information: a final velocity of 11.5 m/s at a 78.2° angle. From this we can extract the final horizontal and vertical velocity:
[tex]V_{fx}=11.8\frac{m}{s}*Cos(78.2)=2.41\frac{m}{s}\\\\V_{fy}=11.8\frac{m}{s}*Sin(78.2)=11.55\frac{m}{s}[/tex]
We will use [tex]V_{fy}[/tex] with a negative value (since the diver is moving in a downwards direction). That way we can find the time elapsed between the highest point of the trayectory to instants before she contacts the water. At the maximum height of the movement, the velocity in the vertical direction is zero. Therefore we can rewrite equation 2) and solve for t:
[tex]-11.55\frac{m}{s}=0\frac{m}{s} - 9.8\frac{m}{s^{2}}t\\\\t= \frac{-11.55\frac{m}{s}}{- 9.8\frac{m}{s^{2}}}=1.17s[/tex]
This t that we found represents the time from the maximum height to the surface of the water. Now we can use equartion 1) to determine the maximum height. We must remember a starting point of 3.9m above the water:
[tex]0 = h+3.9m+0\frac{m}{s}*t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\ \\h=\frac{9.8\frac{m}{s^{2}}1.17s^{2}}{2}-3.9m=2.9m\\\\[/tex]
Since we now know what is the maximum height we can determine the time that it took for the diver to reach that point using equation 1) and 2):
[tex]2.9m =V_{0y}t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\0\frac{m}{s}= V_{0y}t-9.8\frac{m}{s^{2}}t[/tex]
We will solve for [tex]V_{ox}[/tex] in equation 2) and replace it in equation 1):
[tex]V_{0y} = 9.8\frac{m}{s^{2}}t\\\\2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}[/tex]
We solve for t (time from diving board to maximum height):
[tex]2.9m =(9.8\frac{m}{s^{2}}t)t-\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\2.9m=\frac{9.8\frac{m}{s^{2}}t^{2}}{2}\\\\5.8m=9.8\frac{m}{s^{2}}t^{2}\\\\t=\sqrt{ \frac{5.8m}{9.8\frac{m}{s^{2}}}}=0.77s[/tex]
Finally we can replace our t value in [tex]V_{0y}t=9.8\frac{m}{s^{2}}t[/tex] to obtain:
[tex]V_{0y}t=9.8\frac{m}{s^{2}}t=9.8\frac{m}{s^{2}}*0.77s=7.54\frac{m}{s}[/tex]
With this we have the information we need in order to determine the initial velocity of the diver. The magnitude will be calculated from the initial horizontal and vertical velocities (consider final and initial horizontal velocity to be equal in PARABOLIC MOVEMENTS):
[tex]V_{0}=\sqrt{V_{0x}^{2}+V_{0y}^{2}}=\sqrt{(2.41\frac{m}{s})^{2}+(7.54\frac{m}{s})^{2}}=7.91\frac{m}{s}[/tex]
And the direction is determined by an arctangent:
[tex]tan\theta=\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}}\\\\\theta=arctan(\frac{7.54\frac{m}{s}}{2.41\frac{m}{s}})=72.3[/tex]
Final answer:
To find the diver's initial velocity, use equations involving vertical and horizontal components considering her final speed and displacement. The magnitude is 15.93 m/s, and the direction is horizontal.
Explanation:
To determine the diver's initial velocity, we first find the vertical component of her velocity using the equation vf=vi+at. Here, vf = 11.8 m/s, vi is the vertical component of the initial velocity, a = -9.81 m/s^2, and t = time in the air. Solving for vi, we get vi = 15.93 m/s. Then, to find the horizontal component of the initial velocity, we use the equation dx=vit+1/2at^2. Since the diver contacts the water at the highest point, her vertical displacement is 3.9m, and we can find the horizontal component of the initial velocity to be 0 m/s. Therefore, her initial velocity magnitude is 15.93 m/s, and her initial velocity direction is horizontal.
The position vector of an object of mass 0.5 kg is given by r 2.0t2i+ 3.0tj m where t represents time. Which ONE of the following statements is FALSE when a time of t 2.0 s has elapsed. a) The object has moved a distance of 10 m. b) The momentum of the object is 41+1.5/ kg.m.s c) The kinetic energy of the object is 18 J d) The force on the object acts perpendicular to the direction of the unit vectorj e) The magnitude of the force acting on the object is 4 N
Answer:
Statement E is false.
Explanation:
Statement a says that the object would have moved a distance of 10 m.
At t = 0
x0 = 0
y0 = 0
At t= 2
x2 = 2 * 2^2 = 8
y2 = 3 * 2 = 6
[tex]d2 = \sqrt{x2^2 + y2^2}[/tex]
[tex]d2 = \sqrt{8^2 + 6^2} = 10 m[/tex]
Statement A is true.
Statement B says that the momentum of the object would be 4*i+1.5*j
To test this we need to know the speed.
The speed is the first derivative by time:
v = 4*t*i + 3*j
At t = 2 the speed is
v2 = 8*i + 3*j
The momentum is
P = m * v
P2 = 0.5 * (8*i + 3*j) = (4*i+1.5*j)
This statement is true.
Statement C says that the kinetic energy of the object would be 18 J.
The magnitude of the speed is
[tex]v = \sqrt{8^2 + 3^2} = 8.54 m/s[/tex]
The kinetic energy is
Ek = 1/2 * m * v^2
Ek = 1/2 * 0.5 * 8.54^2 = 18 J
Statement C is true
Statement D says that the force acts perpenducular to the direction of the unit vector j.
The acceleration is the second derivative respect of time
a = 4*i + 0*j
Since there is no acceleration in the direction of j we can conclude that the force is perpendicular to the direction of j.
Statement D is true.
Statement E says that the force acting on the object is of 4 N
f = m * a
f = 0.5 * 4 = 2 N
Statement E is false.
What is the magnitude of the electric field of a proton at a distance of 50 micrometers? _____________ (in units of N/C)
Answer:
Electric field at distance of 50 micrometer due to a proton is 0.576 N/C
Explanation:
We have given distance where we have to find the electric field [tex]r=50\mu m=50\times 10^{-6}m[/tex]
We know that charge on proton [tex]q=1.6\times 10^{-16}C[/tex]
Electric field due to point charge is given by [tex]E=\frac{Kq}{r^2}[/tex], here K is a constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So electric field [tex]E=\frac{9\times 10^9\times 1.6\times 10^{-16}}{(50\times 10^{-6})^2}=0.576N/C[/tex]
A 7.0-kg bowling ball experiences a net force of 5.0 N.
Whatwill be its acceleration?
Answer:
It will have an acceleration of [tex]0.71 m/s^{2}[/tex].
Explanation:
Newton’s second law is defined as:
[tex]F = ma[/tex] (1)
Equation 1 can be rewritten for the case of the acceleration:
[tex]a = \frac{F}{m}[/tex]
The force and the acceleration are directly proportional (if one increase the other will increase too).
[tex]a = \frac{5.0 N}{7.0 Kg}[/tex]
But 1 N is equivalent to [tex]1 Kg.m/s^{2}[/tex]
[tex]a = \frac{5.0 Kg.m/s^{2}}{7.0 Kg}[/tex]
[tex]a = 0.71 m/s^{2}[/tex]
So a body will have an acceleration as a consequence of the action of a force.
The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 539-nm photons. What is the energy gap between the ground state and excited state in the laser material?
Answer:
The answer is 2.3 eV
Explanation:
As energy should be conserved, we have to understand that the energy gap between the ground state and excited state is equal to the energy of the emitted photons.
The wavelength of the emitter photons is
[tex]\lambda = 539\ nm = 539 \times 10^{-9}\ m[/tex],
then their energy is
[tex]E = \frac{h c}{\lambda} = 36.855 \times 10^{-20}\ J[/tex]
where h is the Planck constant,
[tex]h = 6.626 \times 10^{-34}\ Js[/tex]
and c is the speed of light,
[tex]c = 2.998 \times 10^{8}\ m/s[/tex].
If we want our answer in electron-volts instead of joules...
[tex]1 J = 6.242 \times 10^{18}\ eV[/tex]
then
[tex]E = 36.855 \times 10^{-20} J = 2.3\ eV[/tex].
The energy gap between the ground state and excited state in the laser material is approximately 3.69 x 10^-19 Joules. This was calculated using the energy equation of photons, considering the given wavelength of 539 nm.
Explanation:The energy gap between the ground state and the excited state in a laser material can be calculated using the energy equation of photons, E=hc/λ, where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength (539 x 10^-9 m in this case). Upon calculation, we find the energy to be about 3.69 x 10^-19 Joules. Hence, the energy gap between the ground state and the excited state is 3.69 x 10^-19 Joules.
This calculation entails quantum physics because it is based on the principle that when an electron in an atom moves from an excited state to a ground state, it emits energy in the form of a photon. This emitted photon has a certain wavelength and frequency, in this case, 539 nm.
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Consider the scenario where a person jumps off from the edge of a 1 m high platform and lands on the ground Suppose his initial jumping speed was 3 m/s. For how long was this person in the air?
Answer:
For 0.24 sec the person was in the air.
Explanation:
Given that,
Height = 1 m
Initial velocity = 3 m/s
We need to calculate the time
Using equation of motion
[tex]s=ut+\dfrac{1}{2}gt^2[/tex]
Where, u = initial velocity
s = height
Put the value into the formula
[tex]1 =3\times t+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]4.9t^2+3t-1=0[/tex]
[tex]t = 0.24\ sec[/tex]
On neglecting the negative value of time
Hence, For 0.24 sec the person was in the air.
The kilowatt-hour is a unit of speed
power
force
momentum
energy
Answer:
Answer:
energy
Explanation:
The capacity to do work is called energy.
There are several types of energy. Some of the types of energy are given below:
1. Kinetic energy
2. Potential energy
3. Solar energy
4. thermal energy
5. Sound energy
6. Nuclear energy, etc.
The SI unit of energy is Joule.
Some other units of energy are
erg, eV, etc.
1 J = 10^7 erg
1 eV = 1.6 x 10^-19 J
The commercial unit of energy is kilowatt hour.
1 Kilo watt hour = 3.6 x 10^6 J
Explanation:
An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the electron be moving when it is 3.00 cm from the proton?
Answer:
electron moving at 91.82 m/s
Explanation:
given data
distance d1 = 6 cm
distance d2 = 3 cm
to find out
how fast will the electron be moving
solution
we know potential energy formula that is
potential energy = [tex]k\frac{q1q2}{r}[/tex]
here k is 9× [tex]10^{9}[/tex] N-m²/C²
m mass of electron is 9.11 × [tex]10^{-31}[/tex] kg
and q = 1.60 × [tex]10^{-19}[/tex] C
we consider here initial potential energy = U1
U1 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.06^2}[/tex]
U1 = 3.84 × [tex]10^{-27}[/tex] J
and final potential energy = U2
U2 = [tex]9*10^{9}\frac{1.60*10^{-19}*1.60*10^{-19}}{0.03}[/tex]
U2 = 7.68 × [tex]10^{-27}[/tex] J
and speed of electron = v
so we will apply here conservation of energy
0.5×m×v² = U2 - U1 ................1
so
0.5×9.11× [tex]10^{-31}[/tex] ×v² = 3.84 × [tex]10^{-27}[/tex]
v = 91.82 m/s
so electron moving at 91.82 m/s
Answer:
The speed of the electron is 91.86 m/s.
Explanation:
Given that,
Distance of electron from proton = 6.00 cm
Distance of proton = 3.00 cm
We need to calculate the initial potential energy
[tex]U_{1}=\dfrac{kq_{1}q_{2}}{r}[/tex]
Put the value into the formula
[tex]U_{i}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{6.00\times10^{-2}}[/tex]
[tex]U_{i}=3.84\times10^{-27}\ J[/tex]
The final potential energy
[tex]U_{f}=\dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{3.00\times10^{-2}}[/tex]
[tex]U_{f}=7.68\times10^{-27}\ J[/tex]
We need to calculate the speed of the electron
Using conservation of energy
[tex]\dfrac{1}{2}mv^2=U_{f}-U_{i}[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times9.1\times10^{-31}\times v^2=7.68\times10^{-27}-3.84\times10^{-27}[/tex]
[tex]v=\sqrt{\dfrac{2\times3.84\times10^{-27}}{9.1\times10^{-31}}}[/tex]
[tex]v=91.86\ m/s[/tex]
Hence, The speed of the electron is 91.86 m/s.
A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the horizontal and the sand is moved without slipping at the rate of 2 m/s. The sand is collected in a big drum 5 m below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
[tex]y = yo+ V_y t +\frac{1}{2}gt^2[/tex]
[tex]0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2[/tex]
solving for t
t = 1.075 sec
for horizontal motion
[tex]x = V_x t[/tex]
x = 2cos18*1.075
x = 2.044 m
The motion of sand is due to the movement of conveyor belt. The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.
What is equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The second equation of the motion for distance can be given as,
[tex]y=ut+\dfrac{2}{2}gt^2[/tex]
Here, [tex]u[/tex] is the initial body, [tex]g[/tex] is the acceleration of the body due to gravity and [tex]t[/tex] is the time taken by it.
Given information-
The conveyor is tilted at an angle of 18° above the horizontal.
The Sand is moved without slipping at the rate of 2 m/s.
The sand is collected in a big drum 5 m below the end of the conveyor belt.
The horizontal component of the velocity is given as,
[tex]v_y=2\cos 18[/tex]
The vertical component of the velocity is given as,
[tex]v_y=2\sin18[/tex]
Put the value in the above equation as,
[tex]y-y_0=v_yt+\dfrac{1}{2}gt^2[/tex]
[tex]0-5=2\sin18 (t)+\dfrac{1}{2}\times9.8\tiems t^2\\t=1.075\rm sec[/tex]
The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is,
[tex]d=v_xt\\d=2\cos18\times1.075\\d=2.044\rm m[/tex]
Thus, the horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.
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Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied by 4.25×10^−6 C charges. Express your answer to three significant figures and include the appropriate units.
Ans:
12500 N/C
Explanation:
Side of square, a = 2.42 m
q = 4.25 x 10^-6 C
The formula for the electric field is given by
[tex]E = \frac{Kq}{r^2}[/tex]
where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges
According to the diagram
BD = [tex]\sqrt{2}\times a[/tex]
where, a be the side of the square
So, Electric field at B due to charge at A
[tex]E_{A}=\frac{Kq}{a^2}[/tex]
[tex]E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]
EA = 6531.32 N/C
Electric field at B due to charge at C
[tex]E_{C}=\frac{Kq}{a^2}[/tex]
[tex]E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]
Ec = 6531.32 N/C
Electric field at B due to charge at D
[tex]E_{D}=\frac{Kq}{2a^2}[/tex]
[tex]E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}[/tex]
ED = 3265.66 N/C
Now resolve the components along X axis and Y axis
Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C
Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C
The resultant electric field at B is given by
[tex]E=\sqrt{E_{x}^{2}+E_{y}^{2}}[/tex]
[tex]E=\sqrt{8840.5^{2}+8840.5^{2}}[/tex]
E = 12500 N/C
Explanation:
Answer:
[tex]1.250\times 10^4\ N/C.[/tex]
Explanation:
Given:
Charge on each corner of the square, [tex]q=4.25\times 10^{-6}\ C.[/tex] Length of the side of the square, [tex]a = 2.42\ m.[/tex]According to the Coulomb's law, the strength of the electric field at a point due to a charge [tex]q[/tex] at a point [tex]r[/tex] distance away is given by
[tex] E = \dfrac{kq}{r^2}[/tex]
where,
[tex]k[/tex] = Coulomb's constant = [tex]9\times 10^9\ Nm^2/C^2[/tex].
The direction of the electric field is along the line joining the point an d the charge.
The electric field at the point P due to charge at A is given by
[tex]E_A = \dfrac{kq}{a^2}[/tex]
Since, this electric field is along positive x axis direction, therefore,
[tex]\vec E_A = \dfrac{kq}{a^2}\ \hat i.[/tex]
[tex]\hat i[/tex] is the unit vector along the positive x-axis direction.
The electric field at the point P due to charge at B is given by
[tex]E_B = \dfrac{kq}{a^2}[/tex]
Since, this electric field is along negative y axis direction, therefore,
[tex]\vec E_B = \dfrac{kq}{a^2}\ (-\hat j).[/tex]
[tex]\hat j[/tex] is the unit vector along the positive y-axis direction.
The electric field at the point P due to charge at C is given by
[tex]E_C = \dfrac{kq}{r^2}[/tex]
where, [tex]r=\sqrt{a^2+a^2}=a\sqrt 2[/tex].
Since, this electric field is along the direction, which is making an angle of [tex]45^\circ[/tex] below the positive x-axis direction, therefore, the direction of this electric field is given by [tex]\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)[/tex].
[tex]\vec E_C = \dfrac{kq}{2a^2}\ (\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j))\\=\dfrac{kq}{2a^2}\ (\dfrac{1}{\sqrt 2}\hat i-\dfrac{1}{\sqrt 2}\hat j)\\=\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\[/tex]
Thus, the total electric field at the point P is given by
[tex]\vec E = \vec E_A+\vec E_B +\vec E_C\\=\dfrac{kq}{a^2}\hat i+\dfrac{kq}{a^2}(-\hat j)+\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\=\left ( \dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )\hat i+\left (\dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )(-\hat j)\\=\dfrac{kq}{a^2}\left [\left ( 1+\dfrac{1}{2\sqrt 2} \right )\hat i+\left (1+\dfrac{1}{2\sqrt 2} \right )(-\hat j)\right ]\\=\dfrac{kq}{a^2}(1.353\hat i-1.353\hat j)[/tex]
[tex]=\dfrac{(9\times 10^9)\times (4.25\times 10^{-6})}{2.42^2}\times (1.353\hat i-1.353\hat j)\\=(8.837\times 10^3\hat i-8.837\times 10^3\hat j)\ N/C.[/tex]
The magnitude of the electric field at the given point due to all the three charges is given by
[tex]E=\sqrt{(8.837\times 10^3)^2+(-8.837\times 10^3)^2}=1.250\times 10^4\ N/C.[/tex]
During constant linear acceleration, what changes uniformly? A. acceleration B. distance C. displacement D. velocity E. all of these
Answer:D-velocity
Explanation:
Given
Constant linear acceleration i.e. acceleration is constant
a=constant
and we know [tex]\frac{\mathrm{d}V}{\mathrm{d} t}=a[/tex]
Thus velocity is changing uniformly to give constant acceleration
While displacement and distance are function of time thus they are not changing uniformly.
During constant linear acceleration, the following quantity changes uniformly: velocity. Therefore option D is correct.
Constant linear acceleration refers to a situation where an object's acceleration remains constant over a specific period of time. In this case, the rate of change of velocity is uniform, meaning the velocity increases or decreases by the same amount in equal time intervals.
Velocity is the rate of change of displacement and is directly affected by acceleration. During constant linear acceleration, the velocity changes uniformly. It either increases or decreases at a constant rate, depending on the direction of acceleration.
Therefore,
During constant linear acceleration, the following quantity changes uniformly: velocity.
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A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball? (Assume that the player gets the ball at his maximum height.)
Answer:4.95 m/s
Explanation:
Given
Ball rises to a height of 1.25 m
Let u be the velocity of player while leaving ground .
Its final velocity will be zero as he reaches a height of 1.25 m
thus
[tex]u^2=2gh[/tex]
[tex]u=\sqrt{2\times 9.81\times 1.25}[/tex]
[tex]u=\sqrt{24.525}[/tex]
u=4.952 m/s
Final answer:
To reach a height of 1.25 meters, a basketball player must leave the ground with an initial upward velocity of 4.95 m/s, calculated using the kinematic equation for motion under constant acceleration due to gravity.
Explanation:
To find the required initial velocity, we use the kinematic equation which relates distance (d), initial velocity ([tex]v_i[/tex] ), final velocity ([tex]v_f[/tex]), acceleration (a), and time (t). We can ignore time since we are not asked for it and we know that the final velocity at the peak of the jump is 0 m/s because the ball will momentarily stop moving upwards before it starts falling down. We will use the acceleration due to gravity, which is approximately 9.81 m/s².
The equation is:
[tex]v_f^2 = v_i^2 + 2ad[/tex]
Setting [tex]v_f[/tex] to 0 m/s (since the ball stops at the peak), the equation simplifies to:
[tex]v_i[/tex] = √(2ad)
Plugging in the values we get:
[tex]v_i[/tex] = √(2 * 9.81 m/s² * 1.25 m) = √(24.525 m²/s²)
[tex]v_i[/tex] = 4.95 m/s
Therefore, the basketball player must leave the ground with an initial velocity of 4.95 m/s to reach a height of 1.25 meters.
You are taking a picture of a giraffe that is standing far away from you. The image is just too small, so you swap the 60-mm-focal-length lens in your camera for a 960 mm telephoto lens. By what factor does this increase the size of the image?
Explanation:
Given that, you are taking a picture of a giraffe that is standing far away from you. The image is just too small, so you swap the 60-mm-focal-length lens in your camera for a 960 mm telephoto lens such that,
[tex]m_1=960\ mm[/tex]
[tex]m_2=60\ mm[/tex]
We need to find the the factor with which the size of the image increases. It can be calculated by taking ratios of both [tex]m_1\ and\ m_2[/tex]
So, [tex]\dfrac{m_1}{m_2}=\dfrac{960}{60}[/tex]
[tex]\dfrac{m_1}{m_2}=16[/tex]
So, the size of image increases by a factor of 16. Hence, this is the required solution.
A boat moves ar 10.0 m/s relative to
thewater. If the boat is in a river where the current is 2.00 m/s,
howlong does it take the boat to make a complete round trip of 1000
mupsteam followed by a 1000- m trip downstream?
Answer: 208.3 s
Explanation:
Hi!
You need to calculate the velocity of the boat relative to the shore, in each trip.
Relative velocities are transformed according to:
[tex]V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\[/tex]
Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s
In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s
In upstream trip we have:
[tex]V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms[/tex]
In dowstream we have:
[tex]V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms[/tex]
In both cases the distance travelled is 1000m. Then the time it takes the round trip is:
[tex]T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s[/tex]
A hydraulic press must exert a force of 20,000N to crush a wrecked car into a convienent block. If the output piston is eight times the area of the input piston, what input force is required?
Answer:
Explanation:
We shall apply Pascal's law here to find the solution .According to this law if we apply a pressure on a particular area over an in-compressible liquid it gets transferred to other point inside or over the surface of the liquid without getting diminished.
in the present case force is applied on a car to crush it which is enormous in magnitude. This force is 20000 N, which is also called output force Let the surface on which car is placed be A₂. We apply input force F₁ at other point over a small area be it A₁.
According to Pascal law ,
20000/ A₂ = F₁ / A₁
F₁ = 20000 X (A₁ / A₂)
= 20000 X 1/8
= 2500 N.
Abus moves to the cast with a speed 10 ms, and a student in the bus moves with a speed 9 m/s to the cust. What is the velocity of the . relative to the bus A. 19 m/s to the cast B. 19 ms to the west c. 1 m/s to the east D. L us to the west.
Answer : The velocity of the student relative to the bus is 1 m/s to the east.
Explanation :
Relative speed : It is defined as the speed of a moving object when compared to the another moving object.
There are two condition of relative speed.
(1) It two bodies are moving in same direction then the relative speed will be the difference of two bodies.
[tex]\text{Relative speed}=|V_2-V_1|[/tex]
(2) It two bodies are moving in opposite direction then the relative speed will be the sum of two bodies.
[tex]\text{Relative speed}=V_1+V_2|[/tex]
As per question,
The speed of bus, [tex]V_1[/tex] = 10 m/s to east
The speed of student, [tex]V_2[/tex] = 9 m/s to east
In this problem, the two bodies are moving in same direction then the relative speed will be the difference of two bodies.
[tex]\text{Relative speed}=|V_2-V_1|[/tex]
[tex]\text{Relative speed}=|9-10|m/s[/tex]
[tex]\text{Relative speed}=1m/s[/tex]
Therefore, the velocity of the student relative to the bus is 1 m/s to the east.
A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pH of the solution, but 30 excess electrons is typical. What is the magnitude of the electric force on a protein with this charge in a 1700 N/C electric field?
Express your answer in newtons.
Answer:
Electric force, [tex]F=8.16\times 10^{-15}\ N[/tex]
Explanation:
A protein molecule in an electrophoresis gel has a negative charge. Electric field, E = 1700 N/C
There are 30 excess electrons. The charge on 30 electrons is, q = 30e
[tex]q=30\times 1.6\times 10^{-19}\ C=4.8\times 10^{-18}\ C[/tex]
The electric force in terms of electric field is given by :
[tex]F=q\times E[/tex]
[tex]F=4.8\times 10^{-18}\ C\times 1700\ N/C[/tex]
[tex]F=8.16\times 10^{-15}\ N[/tex]
So, the magnitude of the electric force on a protein is [tex]8.16\times 10^{-15}\ N[/tex]. Hence, this is the required solution.
Answer:
8.16 x 10^-15 N
Explanation:
Number of excess electrons = 30
Electric field, E = 1700 N/C
Charge of one electron, e = 1.6 x 10^-19 C
Total charge, q = charge of one electron x number of electrons
q = 30 x 1.6 x 10^-19 = 48 x 10^-19 C
The relation between the electric field and the force is given by
F = q E
F = 48 x 10^-19 x 1700
F = 8.16 x 10^-15 N
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.66 N. Determine the distance between the charges.ns
Answer:
it is separated by 80 cm distance
Explanation:
As per Coulombs law we know that force between two point charges is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]q_1 = +8.4\mu C[/tex]
[tex]q_2 = +5.6 \mu C[/tex]
force between two charges is given as
[tex]F = 0.66 N[/tex]
now we have
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}[/tex]
[tex]r = 0.8 m[/tex]
so it is separated by 80 cm distance
Final answer:
The distance between the charges is approximately 20.73 cm.
Explanation:
To determine the distance between the charges, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is: F = k × (q1 × q2) / r², where F is the force, k is the Coulomb's constant (8.99 x 10⁹ Nm²/C²), q1 and q2 are the charges of the particles, and r is the distance between them.
In this case, we have the force (F) as 0.66 N, q1 as 8.4 uC, q2 as 5.6 uC, and we need to find the distance (r). Plugging these values into the formula, we get:
0.66 = (8.99 x 10⁹) × ((8.4 x 10⁻⁶) × (5.6 x 10⁻⁶)) / r²
Solving for r, we can rearrange the equation to get:
r² = ((8.99 x 10⁹) × ((8.4 x 10⁻⁶) × (5.6 x 10⁻⁶))) / 0.66
r² = 8.99 x 10⁹ x 8.4 x 10⁻⁶ x 5.6 x 10⁻⁶ / 0.66
r² = 8.99 x 8.4 x 5.6 / 0.66
r² = 429.643
r = √429.643
r ≈ 20.73 cm
A large asteroid of mass 98700 kg is at rest far away from any planets or stars. A much smaller asteroid, of mass 780 kg, is in a circular orbit about the first at a distance of 201 meters as a result of their mutual gravitational attraction. What is the speed of the second asteroid?
Answer:
1.81 x 10^-4 m/s
Explanation:
M = 98700 kg
m = 780 kg
d = 201 m
Let the speed of second asteroid is v.
The gravitational force between the two asteroids is balanced by the centripetal force on the second asteroid.
[tex]\frac{GMm}{d^{2}}=\frac{mv^2}{d}[/tex]
[tex]v=\sqrt{\frac{GM}{d}}[/tex]
Where, G be the universal gravitational constant.
G = 6.67 x 10^-11 Nm^2/kg^2
[tex]v=\sqrt{\frac{6.67 \times 10^{-11}\times 98700}{201}}[/tex]
v = 1.81 x 10^-4 m/s
When a car is on an inclined bank of angle θ and rounding a curve with no friction, what is the centripetal force equal to? a. The weight of the car b. N cos (θ) c. N sin (θ) d. Zero
Answer:
[tex]F_{net}=N\ sin\theta[/tex]
Explanation:
Let a car of m is on an incline bank of angle θ and it is rounding a curve with no friction. We need to find the centripetal force acting on it.
The attached free body diagram shows the car on the banked turn. It is clear that,
In vertical direction,
[tex]N\ cos\theta=mg[/tex]
In horizontal direction,
[tex]F_{net}=F_{centripetal}[/tex]
[tex]F_{net}=N\ sin\theta[/tex]
So, the centripetal force is equal to [tex]N\ sin\theta[/tex]. Hence, the correct option is (c).
(a) A 1.00-μF capacitor is connected to a 15.0-V battery. How much energy is stored in the capacitor? ________ μJ (b) Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored?________ μJ
Answer:
(a) [tex]E_{ c} = 112.5 \mu J[/tex]
(b) [tex]E'_{ c} = 18 \mu J[/tex]
Solution:
According to the question:
Capacitance, C = [tex]1.00\mu F = 1.00\times 10^{- 6} F[/tex]
Voltage of the battery, [tex]V_{b} = 15.0 V[/tex]
(a)The Energy stored in the Capacitor is given by:
[tex]E_{c} = \frac{1}{2}CV_{b}^{2}[/tex]
[tex]E_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 15.0^{2}[/tex]
[tex]E_{ c} = 112.5 \mu J[/tex]
(b) When the voltage of the battery is 6.00 V, the the energy stored in the capacitor is given by:
[tex]E'_{c} = \frac{1}{2}CV'_{b}^{2}[/tex]
[tex]E'_{c} = \frac{1}{2}\times 1.00\times 10^{- 6}\times 6.0^{2}[/tex]
[tex]E'_{ c} = 18 \mu J[/tex]
(a) The energy stored in the capacitor is [tex]1.125 \times 10^{-4} \ J[/tex]
(b) The energy stored in the capacitor is [tex]1.8 \times 10^{-5} \ J[/tex]
The given parameters;
charge of the capacitor, q = 1 -μFvoltage across the capacitor, V = 15The energy stored in the capacitor is calculated as follows;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 15^2\\\\E = 1.125 \times 10^{-4} \ J[/tex]
When the battery voltage changes to 6 V, the energy stored in the capacitor is calculated as follows;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} \times (1\times 10^{-6}) \times 6^2\\\\E = 1.8 \times 10^{-5} \ J\\\\[/tex]
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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 5.3 km, but when the snow clears, he discovers that he actually traveled 8.3 km at 50° north of due east. (a) How far and (b) in what direction (south of due west) must he now travel to reach base camp?
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 41.1 m. (a) At what time do they pass each other? s (b) At what location do they pass each other relative to the window?
Answer:1.44 s,10.17 m
Explanation:
Given
two balls are separated by a distance of 41.1 m
One person drops the ball from a height of 41.1 and another launches a ball with velocity of 41.1 m exactly at the same time.
Let the ball launches from ground travels a distance of x m in t sec
For Person on window
[tex]41.1-x=ut+\frac{1}{2}gt^2[/tex]
[tex]41.1-x=0+\frac{1}{2}\times 9.81\times t^2--------1[/tex]
For person at ground
[tex]x=v_ft-\frac{1}{2}gt^2---------2[/tex]
add (1) & (2)
[tex]41.1=v_ft-\frac{1}{2}gt^2+\frac{1}{2}gt^2[/tex]
[tex]41.1=v_ft[/tex]
and [tex]v_f[/tex] is given by
[tex]v_f=\sqrt{2\times 9.81\times 41.1}=28.39 m/s[/tex]
[tex]t=\frac{41.1}{28.39}=1.44 s[/tex]
Substitute value of t in equation 1
[tex]41.1-x=0+\frac{1}{2}\times 9.81\times 1.44^2[/tex]
41.1-x=10.171 m
Thus the two ball meet at distance of 10.17 m below the window.
A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a bronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.
Answer:
153.6 kN
Explanation:
The elastic constant k of the block is
k = E * A/l
k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m
0.12% of the original length is:
0.0012 * 0.25 m = 0.0003 m
Hooke's law:
F = x * k
Where x is the change in length
F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)
The compressive load will generate a stress of
σ = F / A
F = σ * A
F = 80*10^6 * 0.048 * 0.04 = 153.6 kN
The smallest admisible load is 153.6 kN
In this exercise we have to use the knowledge of force to calculate that:
[tex]153.6 kN[/tex]
Then from the formula of the elastic constant we get that:
[tex]k = E * A/l\\k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m[/tex]
With that, now using Hooke's law we find that:
[tex]F = x * k\\F = 0.0003 * 729.6*10^6 = 218.88 kN[/tex]
The compressive load will generate a stress of
[tex]\sigma = F / A\\F = \sigma * A\\F = 80*10^6 * 0.048 * 0.04 = 153.6 kN[/tex]
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A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer reading of the car in km/hr?
Answer:
17.64 km/h
Explanation:
mass of car, m = 1000 kg
Kinetic energy of car, K = 1.2 x 10^4 J
Let the speed of car is v.
Use the formula for kinetic energy.
[tex]K = \frac{1}{2}mv^{2}[/tex]
By substituting the values
[tex]1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}[/tex]
v = 4.9 m/s
Now convert metre per second into km / h
We know that
1 km = 1000 m
1 h = 3600 second
So, [tex]v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )[/tex]
v = 17.64 km/h
Thus, the reading of speedometer is 17.64 km/h.
To find the speedometer reading, use the kinetic energy and mass to calculate the speed in meters per second and then convert it to kilometers per hour. The calculated speed is approximately 88 km/hr.
Explanation:To determine the speedometer reading of the car in km/hr based on the kinetic energy given, we use the kinetic energy formula KE = ½ mv². We rearrange this formula to solve for speed (v): v = √(2KE/m).
Plugging in the provided kinetic energy of 1.2 x 10´ J and the car's mass of 1000 kg, we get:
v = √(2 × 1.2 x 10´ J / 1000 kg)
v = √(24 x 10´ J/kg)
v ≈ 24.49 m/s
To convert from m/s to km/hr, we multiply by 3.6:
v = 24.49 m/s × 3.6 km/hr per m/s ≈ 88.16 km/hr
Therefore, the speedometer would read approximately 88 km/hr.
What is the approximate uncertainty in the area of a circle of radius 4.5×10^4 cm? Express your answer using one significant figure.
To find the uncertainty in the area of a circle with a radius containing two significant figures, use the formula A = πr² and round the result to two significant figures, taking into account the rules of rounding and the implications for uncertainty.
Explanation:The question pertains to the calculation of the uncertainty in the area of a circle when provided with the radius which has a certain number of significant figures. Given the radius (r) of a circle as 4.5×104 cm, which contains two significant figures, the area (A) can be calculated using the formula A = πr². However, since π is a constant with an infinite number of significant digits and the radius has only two, the answer must be limited to two significant figures to reflect the precision of the given measurement. Consequently, the area will be calculated and then rounded to two significant figures to give an approximate value of the uncertainty.
Assuming the calculated area is A, we must consider the rounding rules to determine the uncertainty. When we round to two significant figures, this may imply an uncertainty, as the final digit in the rounded number can vary up or down by one. This is an important concept to understand, as rounding can lead to a loss of information and increase the relative uncertainty of the final answer.
The heat in an internal combustion engine raises the internal temperatures to 300 deg C. If the outside temperature is 25 deg C, what is the maximum efficiency of this engine? And is it possible to attin this efficiency?
Answer:48 %
Explanation:
Given
outside temperature [tex]=25^{\circ}\approx 298 K[/tex]
Internal temperature[tex]=300^{\circ}\approx 573 K[/tex]
Maximum efficiency is carnot efficiency which is given by
[tex]\eta =1-\frac{T_L}{T_H}[/tex]
[tex]\eta =1-\frac{298}{573}[/tex]
[tex]\eta =\frac{275}{573}=0.4799[/tex]
i.e. [tex]47.99 %\approx 48 %[/tex]
No it is not possible to achieve this efficiency because carnot engine is the ideal engine so practically it is not possible but theoretically it is possible
Why is it easier to use a potentiometer in a circuit rather than two separte resistors in series with one another?
Answer:
Because by potentiometer we can drop voltage according to our requirement but in set of two series resistors we can not do this.
Explanation:
As we know that the potentiometer can act as a variable resistor but resistors have constant value.
In any circuit if we insert resistors they will drop a constant value if we insert a potentiometer we will drop the value of voltage by according to our will.
Therefore, by the above discussion it can be say that the potentiometer is better than two resesters in series.
Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.20 s as it moves a distance of 1.19 m from the bottom to the top of the window. How long does it take before the ball reappears?
Firstly , let's consider the motion of the ball as it travelled up past the window. Given that the distance covered by the ball is 1.19 m in 0.20 s, we can calculate the initial velocity of the ball using the straightforward kinematic equation for velocity, which is v = d/t. Here, 'd' is the distance travelled by the ball, and 't' is the time taken to cover that distance.
Plugging the values into the equation:
vi = d/t
vi = 1.19 m / 0.20 s
vi = 5.95 m/s
Therefore, the initial velocity of the ball is approximately 5.95 m/s.
Next, we want to determine how long it would take for the ball to reach its peak and then reappear. The acceleration due to gravity (g) is 9.81 m/s^2 and acts downwards. Meaning that, as the ball ascends, it slows down until it reaches its peak, where its velocity becomes zero.
To find the time at peak, we can use the equation of motion v = vi + at, where 'v' is the final velocity, 'vi' is the initial velocity, 'a' is the acceleration, and 't' is time. At the peak, the final velocity 'v' becomes 0, thus:
t_peak = vi / g
t_peak = 5.95 m/s / 9.81 m/s²
t_peak = 0.61 s
Therefore, the time it takes to reach the peak is approximately 0.61 seconds.
As we're dealing with symmetrical motion (upwards and downwards motion are equal), the time for the ball to descend from its peak would be identical to the time it took to ascend, which means:
Time for the ball to reappear, t_reappear = 2 * t_peak
t_reappear = 2 * 0.61 s
t_reappear = 1.22 s
Hence, it should take approximately 1.22 seconds before the ball reappears.
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Alternative sources of energy include : geothermal
hydrogen
wind
solar cells All of the above
Answer:
All of the above
Explanation:
The correct answer is option E (All of the above)
All the options are alternative source of energy.
The option given are not the traditional way (energy production from coal) of extracting energy as the loss of natural resources does not take place in these source of energy.
energy extracted from wind, solar light , hydrogen ,etc are the source of energy the alternative source of production of energy because do not exploit the natural resources, reduce the carbon emission and energy produced by them is clean energy.