Answer:
(a) emf = 0.507 V
(b) emf = 0.0507 V
(c) emf = 0.00234 V
Explanation:
Given;
number of turns of the coil, N = 40 turns
diameter of the coil, d = 11 cm
radius of the coil, r = 5.5 cm = 0.055 m
magnitude of the magnetic field, B = 0.4 T
The magnitude of the induced emf is calculated as;
[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]
(a) when the time, t = 0.3 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]
(b) when the time, t = 3.0 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]
(c) when the time, t = 65 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]
What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s
Answer:
[tex]K=0.023J[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=0.15[/tex]
Velocity [tex]v=0.5m/s[/tex]
Angular Velocity [tex]\omega=8.4rad/s[/tex]
Generally the equation for Kinetic Energy is mathematically given by
[tex]K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)[/tex]
[tex]K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))[/tex]
[tex]K=0.023J[/tex]
Which person ha the most freedom to make his or her own lifestyle decisions?
A. Frieda is 10 years old and lives with her grandmother
B. Vladimir is 3 years old and attends preschool
C. Quincy is 16 years old and lives in a dormitory
D. Lucinda is 32 years old and has two children
Answer:
C Quincy
Explanation:
Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.
I hope this helps!
Answer:
The correct answer is C. Quincy is 16 years old and lives in a dormitory
Explanation:
Quincy being 16 may be able to get a job and hold a membership at a gym even if not at a gym he still has the most freedom period with Lucinda having children and Frieda and Vladimir being to young to make the choices of exercising on their own.
Please tell me if I'm wrong so I may give you the correct answer!
Happy Holidays!!
A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period
Answer: [tex]P=5573.43\ W[/tex]
Explanation:
Given
Mass of the elevator is [tex]M=650\ kg\\\[/tex]
Time period of ascension [tex]t=3\ s[/tex]
cruising speed [tex]v=1.75\ m/s[/tex]
Distance moved by elevator during this time
Suppose Elevator starts from rest
[tex]\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s[/tex]
Distance moved
[tex]\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m[/tex]
Gain in Potential Energy is
[tex]\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N[/tex]
Average power during this period is
[tex]\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W[/tex]
Answer:
The power is 331.7 W.
Explanation:
mass, m = 650 kg
time, t= 3 s
initial velocity, u = 0 m/s
final velocity, v = 1.75 m/s
(a) The power is defined as the rate of doing work.
Work is given by the change in kinetic energy.
W = 0.5 m (V^2 - u^2)
W = 0.5 x 650 x 1.75 x 1.75 = 995.3 J
The power is given by
P = W/t = 995.3/3 = 331.7 W
a small rock with a mass 0.20 kg is attachted to the lower wns od a long string and moves in a horixontal circle with radiu 0.6 m. what angle does the string make with vertical
An object carries a charge of -8.5 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the first to the second object so that both objects have the same charge?
Answer:
Approximately 2*10^13 electrons must be transferred
Explanation:
Below is the given information:
First object carries charge = -8.5 µC
Number of electrons in 1st = 8.5 x 10^-6/(1.6 x 10^-19) = 5.3125 x 10^13
Second object carries a charge = -2.0 µC
The number of electrons in 2nd = 2*10^-6/(1.6*10^-19) = 1.25 x 10^13
so, approximately 2 x 10^13 electrons must be transferred
Convert the unit of 0.0063 milliseconds into microseconds. (Answer in scientific notation)
Answer:
microsecond = 1 × 10-6 seconds
1 second = 1 × 100 seconds
1 microsecond = (1 / 1) × 10-6 × 10-0 seconds
1 microsecond = (1) × 10-6-0 seconds
1 microsecond = (1) × 10-6 seconds
1 microsecond = 1 × 1.0E-6 seconds
1 microsecond = 1.0E-6 seconds,just do like it
Answer:
6.3 x [tex]10^{-3}[/tex]
Explanation:
1 millisecond = 1000 microseconds
0.0063 millisecond = 6.3 microseconds
Which is the best description of a molecule?
A-) a molecule of an element is composed of at least two types of atoms.
B-) a molecule of a compound is composed of only one type of atom.
C-) a molecule of a compound is composed of at least two types of atoms.
D-) a molecule of an element is composed of exactly two types of atoms.
Una bala de 10 g se dispara contra un bloque de madera de 102 g inicialmente en reposo sobre una superficie horizontal. Después del impacto el bloque se desliza 8 m antes de detenerse. Si el coeficiente de fricción entre el bloque y la superficie es 0,5, ¿Cuál es la velocidad de la bala inmediatamente antes del impacto?
Answer:
una ess abola cola sola answer
A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is
[tex]\implies F_1 < F_2[/tex]
[tex] \implies F_{net} = F_2 - F1[/tex]
[tex]\implies F_{net} = 42 -15[/tex]
[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]
The net force on the 4.0 kg box is 27 N towards EAST.
A tractor of mass 2000kg Pulls a trailer of mass 1500kg. The total frictional force is 3000N and the acceleration of the tractor is 3ms^-2. Calculate;
(a) the force exerted on the tractor by the tow-bar when the acceleration is 3ms^-2
(b) the force exerted when the tractor and trailer are moving at a constant speed of 4m/s
Answer:
a) T = -22796.5 N, b) F = 3000 N
Explanation:
a) For this part we use Newton's second law
Let's set a reference frame with the x-axis in the direction of motion and the y-axis in the vertical direction.
We make a free-body diagram for each body,
the tractor has the bar force (T) and the push force (F) and the friction force (fr₁)
Y axis
N₁ -W₁ = 0
N₁ = M₁ g
X axis
F - T - fr₁ = M₁ a
the Trailer has the bar force (T) and the friction force (fr₂)
Y axis
N₂ - W₂ = 0
N₂ = m₂ g
X axis
T - fr₂ = m₂ a
let's write the system of equations
F - T - fr₁ = M₁ a (1)
T - fr₂ = m₂ a
we solve
F - (fr₁ + fr₂) = (M₁ + m₂) a
indicate that the total friction forces are fr = 3000N
fr = fr₁ + fr₂
F =[tex]\frac{(M_1+m_2) a}{fr}[/tex]
let's calculate
F =[tex]\frac{(2000+1500) \ 3}{3000}[/tex]
F = 3.5 N
The friction force is
fr = μ N
the norm of the system is N = N₁ + N₂
μ = [tex]\frac{fr}{N_1 + N_2}[/tex]
μ = [tex]\frac{3000}{2000+1500}[/tex]
μ = 0.858
with this value we can find the friction force 1 and substitute in equation 1
F - T - μ N₁ = M₁ a
T = F - M₁ (a + μ g)
T = 3.5 - 2000 (3 + 0.858 9.8)
T = -22796.5 N
b) when the system moves with constant velocity the acceleration is zero
F - T - fr₁ = 0
T - fr₂ = 0
we solve
F + (fr₁ + fr₂) = 0
F = fr₁ + fr₂
F = 3000 N
A fast moving vehicle travelling at a speed of 25.4 m/s comes up behind another vehicle which is
travelling at a slower constant speed of 13.6 m/s. If the faster vehicle does not begin braking until it
is 11.4 meters away from the car in front of it, what is the minimum acceleration that the faster car
must exhibit if it is to avoid colliding with the car in front? Assume that both cars are travelling in the
positive direction
Answer:
a = 6.1 m / s²
Explanation:
For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle
Let's find the relative initial velocity of the two vehicles
v₀ = v₀₂ - v₀₁
v₀ = 25.4 - 13.6
v₀ = 11.8 m / s
the fastest vehicle
x = v₀ t + ½ a t²
The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is
x = 11.4 m
let's use the expression
v² = v₀² - 2 a x
how the vehicle stops v = 0
a = v₀² / 2x
a = [tex]\frac{11.8^2}{2 \ 11.4}[/tex]
a = 6.1 m / s²
this velocity is directed to the left
A baseball player (hitter) hits the ball with her bat. The ball flies north at an angle a above ground and with an initial speed v towards a fielder whose job is to catch the ball before it hits the ground. The distance between the fielder and the hitter is D. At the moment when the bat hits the ball, the fielder starts running to the south towards the place where he believes the ball will fall. Assume the fielder runs with a constant speed all the time the ball is in the air. Assume the hand of the fielder catches the ball just before it hits the ground and that the ball is hit by the bat at ground level.
How fast must the fielder run to be able to catch the ball?
Answer:
imma try and still get wrong
The fielder's required speed is (D + (v * t * cos(a))) / t.
To catch the ball, the fielder must run at a speed that allows them to cover the horizontal distance D between themselves and the hitter, as well as the distance the ball travels horizontally before hitting the ground. The time it takes for the ball to hit the ground can be determined based on its initial vertical velocity and the acceleration due to gravity.
Let's call the time the ball is in the air "t." The horizontal distance the ball travels is given by D + (v * t * cos(a)), where "v" is the initial speed of the ball and "a" is the angle of the ball's flight above ground. The fielder's running speed should equal this horizontal distance divided by "t" to ensure they reach the ball's landing point just in time to catch it. Thus, the fielder's required speed is (D + (v * t * cos(a))) / t.
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A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? What is the amplitude?
The period and the amplitude of the weight suspended from spring are 0.33 seconds and 10 centimeters, respectively.
1) The period is given by:
[tex] T = \frac{1}{f} [/tex]
Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
[tex] T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s [/tex]
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
[tex] A = \frac{20 cm}{2} = 10 cm [/tex]
Therefore, the amplitude is 10 cm.
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I hope it helps you!
Urgent please help !!!!!!!
Answer:
resultant vector =0
Explanation:
because it is connected head to tail in a closed figure
please help me .finish this paper
Solution-1:-
[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]
Solution:-2
[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]
Solution:-3
[tex]\boxed{\sf 320m}[/tex]
Solution:-4
[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]
Solution:-5
[tex]\boxed{\sf Mg_3N_2}[/tex]
Solution:-6
[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]
Solution:-7
[tex]\boxed{\sf {}^{}_{}N}[/tex]
Solution:-8
[tex]\boxed{\sf Galactuse}[/tex]
Solution:-9
[tex]\boxed{\sf Y-X}[/tex]
Solution:-10
[tex]\boxed{\sf Cell\:wall}[/tex]
The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?
a. WA = WB
b. WA = 2WB.
c. WA=4WB
d. WB= 2WA
Answer:
Option (c).
Explanation:
Let the mass of each cart is m and the force is F.
Time for cart A is 2t and for cart B is t.
Work done is given by the
W= force x displacement
As the distance is given by
S= u t +0.5 at^2
So, when the time is doubled the distance is four times.
So, WA = F x 4 S
WB = F x S
WA= 4 WB
A 2 nC point charge is embedded at the center of a nonconducting sphere of radius = 4.8 cm which has a charge of 1 nC distributed uniformly throughout its volume. What is the magnitude of the electric field at a point that is 2.4 cm from the center of the sphere? Write your answer in terms of kN/C.
Answer: [tex]3.32\times 10^4\ kN/C[/tex]
Explanation:
Given
Charge at the center of the sphere is [tex]q=2\ nC[/tex]
Charge distributed over the entire sphere [tex]Q=1\ nC[/tex]
Radius of sphere [tex]r=4.8\ cm[/tex]
Using Guass law
[tex]\Rightarrow \oint \vec{E}\cdot \vec{dA}=\dfrac{q_{enc}}{\epsilon_o}\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+\dfrac{Q}{4\pi R^2}\times 4\pi r^2)\\\\\Rightarrow E\cdot 4\pi r^2=\dfrac{1}{\epsilon}(q+Q\dfrac{r^2}{R^2})\\\\\Rightarrow E\cdot (2.4\times 10^{-2})^2=9\times 10^9(2+1\cdot \dfrac{1}{4})\times 10^{-9}\\\\\Rightarrow E=3.32\times 10^4\ kN/C[/tex]
A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.
a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______
Answer:
(a) 31.44 m/s (b) 164.74 m/s²
Explanation:
Given that,
The diameter of a disk, d = 12 cm
Radius, r = 6 cm
Angular speed = 5.24 rad/s
(a) Linear speed,
[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]
(b) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]
A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper
Answer:
The spring constant of the spring is 10.3 N/m.
Explanation:
Given that,
Mass of a bungee jumper, m = 59 kg
The period of oscillation, T = 0.25 min = 15 sec
We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :
[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]
Where
k is the spring constant
[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]
So, the spring constant of the spring is 10.3 N/m.
A thin stream of water flows vertically downward. The stream bends toward a positively charged object when it is placed near it. The positively charged object is then removed. What will happen to the same stream of water when a negatively charged object is placed nearit
Answer:
the water jet and the negative object attract
Explanation:
Let's analyze the situation. Water is a good conductor of electricity, so when an object with a positive charge is brought closer, a charge of the opposite sign is created, which is why the two objects attract each other. The charge created comes from the ground
When claiming the object, the charge is reduced to zero or the charge goes to earth since water is a good conductor of electricity, therefore when approaching an object with a negative charge, more charges go to earth and the ring is left with a positive charge and attracts it to the object.
In short, the water jet and the negative object attract
what is the prefix notation of 0.0000738?
Answer:
The scientific notation of 738 is 7.38 x 1002.
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN
Answer:
1.621 kN
Explanation:
Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).
The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).
So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N
So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N
The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)
= √(R² + 0²) (since R' = 0)
= √R²
= R
= 1620.82 N
= 1.62082 kN
≅ 1.621 kN
So, the sum of these two forces on the barge is 1.621 kN
12. A car travels in a straight line with an average velocity of 80 km/h for 2.5h and then an average velocity of 40 km/h from 1.5 h (a) what is the total displacement for the 4 h trip? (b) What is the average velocity for the total trip?
Answer:
a. Total displacement = 140 km/h
b. Average velocity = 35 km/h
Explanation:
Given the following data;
Average velocity A = 80 km/h
Time A = 2.5 hours
Average velocity B = 40 km/h
Time B = 1.5 hours
a. To find the total displacement for the 4 h trip;
Total time = Time A + Time B
Total time = 2.5 + 1.5
Total time = 4 hours
Next, we would determine the displacement at each velocity.
Mathematically, displacement is given by the formula;
Displacement = velocity * time
Substituting into the formula, we have;
Displacement A = 80 * 2.5
Displacement A = 200 km/h
Displacement B = 40 * 1.5
Displacement B = 60 km/h
Total displacement = Displacement A - Displacement B
Total displacement = 200 - 60
Total displacement = 140 km/h
b. To find the average velocity for the total trip;
Mathematically, the average velocity of an object is given by the formula;
[tex] Average \; velocity = \frac {total \; displacement}{total \; time} [/tex]
Substituting into the formula, we have;
[tex] Average \; velocity = \frac {140}{4} [/tex]
Average velocity = 35 km/h
It is important to keep terms electrical potential and electrical potential energy straight. Let's say a charge q is placed at a point where the electrical potential is V. If I double the charge to 2q, the electrical potential of the point will _________ and the electrical potential energy of the charge will ___________ 1. stay the same; double 2. double; stay the same 3. stay the same; stay the same 4. double; double
Answer:
1. The electric potential would stay the same.
2. The electric potential energy would double
Explanation:
1. Since the electric potential at the point is V and is not dependent on the charge q, the electric potential at the point still remains the same even if the charge q is doubled to 2q.
So, the electric potential remains the same when the charge is doubled to 2q.
2. Since potential energy U = qV where q = charge and V = electric potential. So when the charge is q and electric potential, V, the electric potential energy at the point is U = qV.
When the charge is doubled to 2q and the electic potential remains the same, the electric potential energy is now U' = 2qV = 2U (since U = qV).
So, the electric potential energy at the point is doubled when the charge is doubled.
For a solid uniformly charged sphere of radius R, calculate the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere. a) 9/8 b) infinity c) 2.0 d) 8.0 e) 8/9
Answer:
e
Explanation:
From the given information:
Suppose Q = total charge of the sphere.
here, the electric field outside the sphere at distance R/2 can be expressed as:
[tex]E_1 = \dfrac{1}{4 \pi \varepsilon _o}* \dfrac{Q}{(R + \dfrac{R}{2})^2}[/tex]
where:
[tex]k = \dfrac{1}{4 \pi \varepsilon _o}[/tex]
[tex]E_1 = \dfrac{kQ}{(\dfrac{3R}{2})^2}[/tex]
[tex]E_1 = \dfrac{4kQ}{9R^2}[/tex]
For the electric field inside the sphere, we have:
[tex]E_2 = \dfrac{kQr}{R^3}[/tex]
here:
r = distance of the point from the center = R/2
R = radius of the sphere
∴
[tex]E_2 = \dfrac{kQ * \dfrac{R}{2}}{R^3}[/tex]
[tex]E_2 = \dfrac{kQ }{2R^2}[/tex]
As such, the ratio of the electric field outside the sphere to the one inside is:
[tex]\dfrac{E_1}{E_2} = \dfrac{ \dfrac{4kQ}{9R^2}}{ \dfrac{kQ }{2R^2}}[/tex]
[tex]\dfrac{E_1}{E_2} = \dfrac{4kQ}{9R^2} \times \dfrac{ 2R^2 }{kQ}[/tex]
[tex]\mathbf{\dfrac{E_1}{E_2} = \dfrac{8}{9}}[/tex]
For a solid uniformly charged sphere of radius R, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is - e) 8/9
Electric field due to a sphere at a distance R/2 outside the sphere can be calculated by assuming the sphere as a point massfield [tex]E2= \frac{k(Q)}{(3R/2)^2}[/tex]
= [tex]\frac{4}{9} \frac{k(Q)}{R^2}[/tex]
The electric field at a distance R/2 inside the sphere E1= k(Q)(r)/R^3 where r is the distance from the center [tex]E1= \frac{k(Q)}{(3R/2)^3}[/tex]=
Thus, E2/E1
= (4/9)/(1/2)
= 8/9
Thus, the electric field at a distance R/2 outside the sphere, divided by the electric field at a distance R/2 inside the sphere is - 8/9
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A hungry monkey is sitting at the top of a tree 69 m above ground level. A person standing on the ground wants to feed the monkey. He uses a tee-shirt cannon to launch bananas at the monkey. If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the bananas, how should the person aim the banana cannon
Answer:
Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.
If the cannon throws the banana with the same force the monkey falls
(m.g=Fz <=> m.9,81N/kg=...N).
Then the throw will slow down because of the gravitational pull.
Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.
The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.
If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.
So to do so you need to throw the bananas with a speed of at least 9,81m.s
Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.
I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.
kind regards
Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.
Required:
Rank from smallest to largest.
Answer:
Momentum of object A = Momentum of object C < momentum of B.
Explanation:
The momentum of an object is equal to the product of mass and velocity.
Object A has a mass m and a speed v. Its momentum is :
p = mv
Object B has a mass m/2 and a speed 4v. Its momentum is :
p = (m/2)×4v = 2mv
Object C has a mass 3m and a speed v/3. Its momentum is :
p = (3m)×(v/3) = mv
So,
Momentum of object A = Momentum of object C < momentum of B.
I need the help
Please. I’m terrible at physics
Answer:
Explanation:
so opposite and equal , right? forces are. soooo..
528+52= 580 N is the force that is being exerted up on the scale
TRUE or FALSE: The acceleration of projectile is 0 m/s/s at the peak of the trajectory. Identify the evidence which supports your answer.
The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration ( i.e. statement in the question is False )
Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.
In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.
Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration
Learn more : https://brainly.com/question/24658194
Does the zero electric field intensity in a given region imply zero potential?
Answer:
No, just because the electric field is zero at a particular point, it does not necessarily mean that the electric potential is zero at that point. ... At the midpoint between the charges, the electric field due to the charges is zero, but the electric potential due to the charges at that same point is non-zero.
Explanation:
