The normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is 102.41 N.
Centripetal force of the Ferris wheelThe normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is calculated as follows;
Fn = Fc + mg
Fn = mω²r + mg
where;
ω is the angular speed = 1 rev/30 s = 2π/30 s = 0.21 rad/sFn = 10(0.21²) 10 + 10(9.8)
Fn = 102.41 N
Thus, the normal force of the ferris wheel on a 10kg toddler at the very bottom of the ferris wheel is 102.41 N.
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2. A tennis ball machine launches balls horizontally with an initial speed of 5.3 m/s, from a height of 1.2 m.
a) What will the time of flight be for a tennis ball launched by the ball machine? (3)
b) What will the range of the tennis ball be? (2)
c) What will be the final velocity of the ball with which it reaches the ground? (3)
(a) The time of flight be for a tennis ball launched by the ball machine is 0.19 s.
(b) The range of the tennis ball be is 1.01 m.
(c) The final velocity of the ball with which it reaches the ground is 7.16 m/s.
Time of flight of tennis ballThe time of flight of the tennis ball is calculated as follows;
h = vt + ¹/₂gt²
1.2 = 5.3t + 0.5(9.8)t²
1.2 = 5.3t + 4.9t²
4.9t² + 5.3t - 1.2 = 0
a = 4.9, b = 5.3, c = 1.2
solve using quadratic formula
t = 0.19 s
Thus, the time of flight be for a tennis ball launched by the ball machine is 0.19 s.
Range of the tennis ballThe range of the tennis ball is calculated as follows;
R = vt
R = 5.3 x 0.19
R = 1.01 m
Final velocity of the ballThe final velocity of the ball with which it reaches the ground is calculated as follows;
vf = vo + gt
vf = 5.3 + 9.8(0.19)
vf = 7.16 m/s
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A 4500 kg Aston Martin traveling at 102 m/s has to stop short because some ducklings
hazard onto the road. The Aston Martin was able to stop in 1.77 seconds. How much
force was placed on the car?
Answer:
-259322.03N
Explanation:
[tex]F=m*(\frac{v}{t})\\ F=4500kg*(\frac{0-102m/s}{1.77s} )\\F=-259322.033898\\\\[/tex]
The ________ of an object is the same on the earth as it is on the moon.
A. mass
B. kilogram
C. newton
D. acceleration of gravity
E. weight
Answer:
mass
Explanation:
The mass of an object is the same on the earth as it is on the moon.
Mass always remains constant, it never changes with respect to place.
23) What is the magnitude of the electric field intensity at a point where a proton experiences an
electrostatic force of magnitude 2.30 x 10-25 newton?
Answer: I am so sorry i hadn't learned this Yet~!
Explanation:
A cable with 19.0 N of tension pulls straight up on a 1.50 kg block that is initially at rest. What is the block's speed after being lifted 2.00 m ? Solve this problem using work and energy
The final speed of the block, after being lifted 2.00 m is 3.39 m/s
What is speed?Speed can be defined as the rate of change in the distance of a body.
To calculate the speed of the block after being lifted 2.00 m, first, we need to calculate the acceleration of the block using the formula below
Formula:
T-mg = ma......... Equation 1Where:
T = Tension in the cablem = mass of the cablea = accelerationg = acceleration due to gravityRestructuring the formula above,
a = (T-mg)/m............... Equation 2From the question,
Given:
T = 19 Nm = 1.5 kgg = 9.8 m/s²Substitute these values into equation 2
a = [(19)-(1.5×9.8)]/1.5a = 4.3/1.5a = 2.87 m/s²Finally, to calculate the speed of the block, we use the formula below.
v² = u²+2as.......... Equation 3Where:
v = Final speedu = initial speeda = accelerations = distanceFrom the question,
Given:
u = 0 m/sa = 2.87 m/s²s = 2.00 mSubstitute these values into equation 3
v² = 0²+(2×2×2.87)v² = 11.48v = √11.48v = 3.39 m/sHence, The final speed of the block, after being lifted 2.00 m is 3.39 m/s.
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a
Which of these is a chemical
change?
A. water boiling
B. salt disolving
C. paper burning
Answer:
Burning coal and boiling water are both chemical changes. Burning coal is a chemical change, and boiling water is a physical change. Burning coal is a physical change, and boiling water is a chemical change.
Explanation:
Which process describes the warm ocean that transfers heat to the air above it?
O conduction
O convection
O reflection
O radiation
Answer:
convection is the answer
The process that describes the warm ocean that transfers heat to the air above it is known as Convection. Thus, the correct option for this question is B.
What are some characteristics of heat?Some characteristics of heat are as follows:
It has no mass, shape, color, odor, volume, and weight.It is one of the invisible forms of energy.It stimulates the presence of warmth.It gets remarkably transferred from one body to another.This form of energy has the capacity to flow in each and every direction.The process of convection transfers heat with the help of the migration of a fluid (liquid or gas) between areas of different temperatures. It is an approach that signifies the transfer of heat between two bodies by involving currents of moving gas or fluid.
Therefore, convection is the process that describes the warm ocean that transfers heat to the air above it. Thus, the correct option for this question is B.
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#SPJ2
list out the use of simple machine
Explanation:
simple machine can multiplayer of speed and force
What is environmental?
Answer:
Environmental means relating to or caused by the surroundings in which someone lives or something exists. It protects against environmental hazards such as wind and sun. The form the human family takes is a response to environmental pressures.
Explanation:
I HOPE IT HELPS!!
THANX!!!
STAY SAFE!!!!
A capacitor of cylindrical shape as shown in the red outline, few cm long carries a uniformly distributed charge of 7.2 uC per meter of length. By constructing a suitable Gaussian surface around the wire, Find the magnitude and direction of the electric field at points
a)5.5m
b)2.5m
perpendicular from the center of the wire.
(a) The magnitude of the electric field at point 5.5m is 2.35 x 10⁴ N/C.
(b) The magnitude of the electric field at point 2.5m is 5.18 x 10⁴ N/C.
Electric field at a point on the Gaussian surface
The magnitude of the electric field at a point on the cylindrical Gaussian surface is calculated as follows;
E = λ/2πε₀r
where;
λ is linear charge densityε₀ is permitivity of free spacer is the position of the chargeAt a distance of 5.5 m[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 5.5} \\\\E = 2.35 \times 10^4 \ N/C[/tex]
At a distance of 2.5 m[tex]E = \frac{\lambda}{2\pi \varepsilon _0 r} \\\\E = \frac{7.2 \times 10^{-6}}{2\pi \times 8.85 \times 10^{-12} \times 2.5} \\\\E = 5.18 \times 10^4 \ N/C[/tex]
Thus, the magnitude of the electric field at points of 5.5m is 2.35 x 10⁴ N/C, and the magnitude of the electric field at points of 2.5m is 5.18 x 10⁴ N/C.
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Determine
i. the total capacitance for the circuit
ii. the total charge stored in the circuit
iii. the charge stored in C9 (3μF)
(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
Total capacitance of the circuit
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
C1 and C2 are in series[tex]\frac{1}{C_{12}} = \frac{1}{C_1 } + \frac{1}{C_2} \\\\\frac{1}{C_{12}} = \frac{1}{4 } + \frac{1}{4} \\\\\frac{1}{C_{12}} = \frac{1}{2} \\\\C_{12} = 2 \ \mu F[/tex]
C1 and C2 are parallel to C3[tex]C_{123} = C_{12} + C_3\\\\C_{123} = 2\ \mu F + 2\ \mu F \\\\C_{123} = 4 \ \mu F[/tex]
C(123) is series to C5 and C6[tex]\frac{1}{C_{t} } = \frac{1}{C_{123}} + \frac{1}{C_5} + \frac{1}{C_6} \\\\\frac{1}{C_{t} } = \frac{1}{4} + \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{t} } = \frac{12}{24} \\\\\frac{1}{C_{t} } = \frac{1}{2} \\\\C_t = 2 \ \mu F[/tex]
C7 and C8 are in series[tex]\frac{1}{C_{78}} = \frac{1}{6} + \frac{1}{6} \\\\\frac{1}{C_{78}} = \frac{2}{6} \\\\\frac{1}{C_{78}} =\frac{1}{3} \\\\C_{78} = 3 \ \mu F[/tex]
Total capaciatnce of the circuitCt + C(78) = 2 μF + 3 μF = 5 μF
Total charge stored in the circuitThe total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
Charge stored in 3μF capacitorQ = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
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The force of gravity on an object gives the object its _______.
A. mass
B. kilogram
C. acceleration of gravity
D. weight
Answer:
Weight
Explanation:
The force of gravity on a an object gives the object its weight.
A spring-loaded ballistic cart measuring 0.68 kg is in contact with a second 0.80 kg
cart. The carts are initially at rest on a level surface. The spring is released and the
lighter cart is observed to move at +0.52 m/s afterward. What is the velocity of the
other cart?
Answer:
wait in comments.................
7. A 2.0 kg block, starting from rest, is pushed by a
constant force along a frictionless track. The
position of the block as a function of time is
recorded in the data above. The final momentum
of the block is
(A) 0.8 kgm/s
(B) 1.2 kgm/s
(C) 1.6 kgm/s
(D) 3.2 kgm/s
Answer:
(A) 0.8 kgm/s
Explanation:
because of the even ground it would only slow down
A student in the Biomechanics class has decided that she would like to make her arms
stronger. She has a mass of 63 kg, She chooses to complete some elbow flexion exercises
using a kettlebell. For this problem, consider the hand and forearm to be a single segment.
The distance from her elbow to her wrist is 22.86 cm.
The force from the kettlebell is applied to her hand, which is 30.48 cm from her elbow joint.
She knows that the moment arm of the elbow extensor muscles about the elbow axis is
Answer:
what is heat and transfer
Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?
Answer:
Gravitational force is the force that attracts objects towards each other. Two factors that affect the gravitational force between objects are the mass of the two objects and the distance between
Explanation:
Gravity is what pulls us towards the Earth if we were to jump into the air, so it is the force that pulls things towards other things. The bigger the objects are the more gravity they have, so a planet has more gravity than say, an apple. Distance between objects also makes their gravity change, so the Earth's pull on the moon is different than the Earth's pull on the sun.
Hopefully this helps- let me know if you have any questions!
Describe gravitational force in your own words. Which two factors affect the gravitational force between two objects?
Answer :-Gravitational Force :- It is the attractive force between the two body having some mass.Example of gravitational force :- attraction between the human and earth . earth pulls the body towards it with the help of this This force helps the celestial bodies to revolve in there orbits Factors effecting Gravitational force:- Mass :- gravitational force directly proportional on the masses of the body between which it is acting [tex]gravitational \: force ∝m{ \tiny1}m{ \tiny2}[/tex]Distance between the masses :- Gravitational force is inversely proportional to the square of the distance between the masses [tex]gravitational \: force∝ \frac{1}{ {r}^{2} } [/tex]Formula of the Gravitational force :-[tex]F = \frac{G m{ \tiny1}m{ \tiny2} }{ {r}^{2} } [/tex]
where
F is gravitational force G is gravitational constant ( universal constant ) m 1 is mass of first body m2 is mass of second body r is the distance between the two bodiesA block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed 4.20 m/s makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 8.40 block recoils with a speed of 0.400 m/s. In the figure; the blocks are in contact for 0.200 s.
For A block of mass m = 8.40 kg, moving on a horizontal frictionless surface with a speed of 4.20 m/s is mathematically given as
F = 193.2N
What is the magnitude of the average force on the 8.40-kg block, while the two blocks are in contact, is closest to?
Generally, the equation for the magnitude of the average force mathematically given as
F = m(v1+v2)/t
F = 8.40(4.2+O.4)/t
F = 193.2N
In conclusion magnitude of the average force is
F = 193.2N
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Suppose a grower sprays (2.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water when it freezes?
There is no temperature change which drives heat flow, thus no heat will be released by the water.
Heat released by the water when it freezesThe heat released by the water when it freezes is calculated as follows;
Q = mcΔФ
where;
m is mass of waterc is specific heat capacity of waterΔФ is change in temperature = Фf - ФiInitial temperature of water, Фi = 0 °C
when water freezes, the final temperature, Фf = 0 °C
Q = 22 x 4200 x (0 - 0)
Q = 0
Since there is no temperature change which drives heat flow, thus no heat will be released by the water.
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Why would it be impractical to wire a home with a circuit in which all loads were connected in series.
For residences, series circuits are impracticable. Parallel circuits are practical because each appliance is controlled by its own switch, which prevents the other appliances from being turned off. A) The appliances are all turned on. The conductor wire is carrying a considerable quantity of current (arrow).
An electron has a mass of 9.1x10^-31 kg and a charge
of -1.6x10^-19 C. Suppose you could isolate one electron in
a perfect vacuum and then create an electric field to pull
and edi 107. upward on the electron. How strong would the field have to
be to counteract the electron's weight? (In other words,
di Delfine bu how strong would the field have to be to put the electron in
a state of force equilibrium?)
is one of these the question
Why was the government in giving support to addressing and researching the HIV epidemic ?
Mention the objective of the Experiment?
Answer:
I don't understand your question ❓,the object.....of what experiment
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
As a conservation biologist for the Chesapeake Bay, you and your
colleagues have been conducting a research study that tracks the decrease
in the bald eagle population over the past few years.
What evidence can you find for the decrease in the bald eagle population?
An unbalanced force acting on an object results in ______.
A. inertia
B. friction
C. acceleration
D. faster
E. slower
Answer:
friction
Explanation:
What’s the weight of a box w/ a mass of 150kg on earth?
[tex]\text{Weight ,} W = mg = 150 \times 9.8 =1470N[/tex]
Explanation:
f=ma
where force is weight. we know that a is the acceleration of gravity which is -9.8 m/s^2
so f is
-1471.5 N
a=5i+4j-6k ,b=-2i+2j+3k ,c=4i+3j+2k. find the vector perpendicular to a and c
Answer:
Explanation:
You can use the cross product. Let the vector that perpendicular to a and c is [tex]\vec{d}[/tex], so:
[tex]\vec{d}=\vec{a}\times\vec{c}=\left|\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\5&4&-6\\4&3&2\end{array}\right] \right|=(8+18)\hat{i}-\hat{j}(10+24)+\hat{k}(15-16)=26\hat{i}-34\hat{j}-\hat{k}[/tex]
To check that c is perpendicular with a and b, do the dot product between c and a and also c and b and if the result is zero, you're true.
[tex]\vec{d}.\vec{a}=(26*5)-(34*4)+(6)=0[/tex] (c perpendicular to a)
[tex]\vec{d}.\vec{c}=(4*26)-(34*3)-(2*1)=0[/tex] (d perpendicular to c)
Terry is walking down the street at 3 m/s. If he
has a mass of 70 kg, what is his momentum?
[tex]\text{Given that, mass m = 70 kg and velocity v = 3 m/s}\\\\\text{Momentum,}~ p = mv = 70 \times 3 = 210~ kg ~ms^{-1}[/tex]
A rope is wrapped around a pulley many times. The pulley can be modeled as a solid disk of radius R and mass M, and a mass mA hangs vertically from the pulley. The mass is released from rest. show answer Incorrect Answer 25% Part (a) What is the magnitude of the tangential acceleration of the hanging mass?
The magnitude of the tangential acceleration of the hanging mass is 2mg/MR
Tangential acceleration of the hanging massThe tangential acceleration of the hanging mass around the pulley is determined from the principle of conservation of angular momentum as shown below;
τ = Iα
Where;
I is the moment of inertiaα is the angular velocity[tex]\alpha = \frac{\tau}{I} \\\\\alpha = \frac{mgR}{3/2MR^2} \\\\\alpha = \frac{2mgR}{3MR^2} \\\\\alpha = \frac{2mg}{3MR}[/tex]
Where;
m is the hanging massM is the mass of solid diskThe tangential acceleration is calculated as follows;
[tex]a = \alpha R\\\\a = \frac{2mg}{3MR} \times R\\\\a = \frac{2mg}{3M}[/tex]
Thus, the magnitude of the tangential acceleration of the hanging mass is 2mg/MR
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Find the temperature
Answer:
-------1
Explanation:
beacuse that is what i know
The magnetic field perpendicular to a single wire loop of diameter 10.0 cm decreases from 0.50 T to zero. The wire is made of copper and has a diameter of 2.0 mm and length 1.0 cm. How much charge moves through the wire while the field is changing?
I know how to do the calculations, but can someone please explain what is the 10cm diameter and 2mm diameter? Why is there one wire and two diameters? I understand this problem mathematically but not conceptually.
Hi there!
We can begin by using Lenz's Law:
[tex]\epsilon = -N\frac{d\Phi _B}{dt}[/tex]
N = Number of Loops
Ф = Magnetic Flux (Wb)
t = time (s)
Also, we can rewrite this as:
[tex]\epsilon = -NA\frac{dB}{dt}[/tex]
A = Area (m²)
Since the area is constant, we can take it out of the derivative.
This is a single wire loop, so N = 1.
Now, we can develop an expression for the induced emf.
We can begin by solving for the area:
[tex]A = \pi r^2 \\\\d = r/2 r = 0.05cm \\\\A = \pi (0.05^2) = 0.007854 m^2[/tex]
We can also express dB/dt as:
[tex]\frac{dB}{dt} = \frac{\Delta B}{t} = \frac{0-0.5}{t} = \frac{-0.5}{t}[/tex]
Now, we can create an equation.
[tex]\epsilon = -(1)(0.007854)\frac{-0.5}{t} = \frac{0.003927}{t}[/tex]
To solve the system, we must now develop an expression for current given an emf and resistance.
Begin by calculating the resistance of the copper wire:
[tex]R = \frac{\rho L}{A}[/tex]
ρ = Resistivity of copper (1.72 * 10⁻⁸ Ωm)
L = Length of wire (0.01 m)
A = cross section area (m²)
Solve:
[tex]R = \frac{(1.72*10^{-8})(0.01)}{\pi (0.001^2)} = 5.475 * 10^{-5} \Omega m[/tex]
Now, we can use the following relation (Ohm's Law):
[tex]\epsilon = iR\\\\\epsilon = \frac{Q}{t}R[/tex]
*Since current is equivalent to Q/t.
Plug in the value of R and set the two equations equal to each other.
[tex]\frac{Q}{t}(5.475 * 10^{-5}) = \frac{0.003927}{t}[/tex]
Cancel out 't'.
[tex]Q (5.475 * 10^{-5}) = 0.003927 \\\\Q = \frac{0.003927}{5.475*10^{-5}} = \boxed{71.73 C}[/tex]