Answer:
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
Explanation:
Given that,
Constant speed = 6.38 m/s
Force [tex]F=7.50\times10^{3}\ N[/tex]
Kinetic friction = 0.26
(a). We need to calculate the friction force
Using formula of friction force
[tex]f_{k}=\mu F_{N}[/tex]
Put the value into the formula
[tex]f_{k}=0.26\times7.50\times10^{3}[/tex]
[tex]f_{k}=1950\ N[/tex]
(b). If the only other horizontal force exerted on the sleigh is due to the horses pulling the sleigh,
We need to calculate the magnitude of this force
According to given data,
The same force will be applied to keep constant velocity.
Hence, (a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force.
(a). The kinetic friction force is 1950 N.
(b). The magnitude of force will be equal of friction force
The calculation is as follows;a. The magnitude of the kinetic friction force experienced by the sleigh is
[tex]= 0.76 \times 7.50 \times 10^3[/tex]
= 1950 N
b. It should be equivalent to the friction force.
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The Jamaican Bobsled Team is sliding down a hill in a toboggan at a rate of 5 m/s when he reaches an even steeper slope. If he accelerates at 2 m/s2 for the 5 m slope, how fast is he traveling when he reaches the bottom of the 5 m slope?
Answer:
6.7 m/s
Explanation:
Given:
Δx = 5 m
v₀ = 5 m/s
a = 2 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (5 m/s)² + 2 (2 m/s²) (5 m)
v = 6.7 m/s
Which pair of objects would be most strongly attracted to each other?
A. A positively charged particle and a negatively charged particle
B. Two positively charged particles
C. Two negatively charged particles
D. A negatively charged particle and a neutral particle
Answer:
Its A
Explanation:
Just did the quiz
The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.
Pair of objects that are most strongly attracted:When there is a positively charged particle & the negatively charged particle so due to this it should be strongly attracted. Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.
learn more about particles here: https://brainly.com/question/12734041
Super Mario and Bowser Jr. are racing around a track when Baby Bowser launches a green shell at Mario, bringing him to rest. Bowser Jr. then passes Mario at his top speed of 30 blocks/h, moving down the track in a straight line. Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track. Mario world measure distance using the units of blocks, with 1 block = 0.47 m.
a) What are Mario and Bowser Jr.'s speeds in m/s?
Assuming both Mario and Bowser Jr. race to the finish in a straight line at their top speeds,
b) How long does it take for Mario to catch Bowser Jr.?
c) How far down the track is Mario from the point at which he reaches his top speed?
Answer:
(a). Mario's speed in m/s = 5.2 × 10^-3 m/s.
Bowser Jr.'s speeds in m/s = 3.92 × 10^-3 m/s.
(b). 27001.2 seconds(s)..
(c). 141 metre(m).
Explanation:
So, the following data or parameters or information was given in the question above. These informations are going to help us in solving this question or problem;
=>" Bowser Jr. then passes Mario at his top speed = 30 blocks/h.
=> " Mario quickly accelerates and reaches his top speed of 40 blocks/h in order to catch back up and pass Bowser Jr., but by this point he has opened up a 75 block head start down the track."
=> "Mario world measure distance using the units of blocks, with 1 block = 0.47 m"
Therefore, the solution is given below;
(1). For the first part, we are to determine or calculate Mario and Bowser Jr.'s speeds in m/s.
Therefore, Mario's speed in m/s = 40 × 0.47) ÷ 3600 = 5.2 × 10^-3 m/s.
Also, Bowser Jr.'s speeds in m/s = ( 30 × 0.47) ÷ 3600 = 3.92 × 10^3 m/s.
(2). So, the next thing to do now.is to determine or calculate how long it took for Mario to catch Bowser Jr.
Thus, the time it took for Mario to catch Bowser Jr. Can be related as below;
[ ( 5.2 × 10^-3 m/s) - (3.92 × 10^-3 m/s) × (time,t taken for Mario to catch Bowser Jr.) = 75 × 0.47.
Therefore, the time it took for Mario to catch Bowser Jr. = 27001.2 seconds.
(3). Now, we calculate How far down the track Mario from the point at which he reaches his top speed.
The distance = 5.2 × 10^-3 m/s × 27001.2m = 141m
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
Can someone please explain how to find the acceleration of the hanging mass?
Answer:
Acceleration = m/s²
Explanation:
T= Newtons compared to the weight W = Newtons for the hanging mass. If the weight of the hanging mass is less than the frictional resistance force acting on the mass on the table, then the acceleration will be zero.
waht is science
wjwissbsskdldmndndnd
Answer:
the intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.
Explanation:
The equation that governs the period of a pendulum’s swinging. T=2π√L/g
Where T is the period, L is the length of the pendulum and g is a constant, equal to 9.8 m/s2. The symbol g is a measure of the strength of Earth’s gravity, and has a different value on other planets and moons.
On our Moon, the strength of earth’s gravity is only 1/6th of the normal value. If a pendulum on Earth has a period of 4.9 seconds, what is the period of that same pendulum on the moon?
Answer:
The period of that same pendulum on the moon is 12.0 seconds.
Explanation:
To determine the period of that same pendulum on the moon,
First, we will determine the value of g (which is a measure of the strength of Earth's gravity) on the Moon. Let the value of g on the Moon be [tex]g_{M}[/tex].
From the question, the strength of earth’s gravity is only 1/6th of the normal value. The normal value of g is 9.8 m/s²
∴ [tex]g_{M}[/tex] = [tex]\frac{1}{6} \times 9.8 m/s^{2}[/tex]
[tex]g_{M}[/tex] = 1.63 m/s²
From the question, T=2π√L/g
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
We can write that,
[tex]T_{E} = 2\pi \sqrt{\frac{L}{g_{E} } }[/tex] .......... (1)
Where [tex]T_{E}[/tex] is the period of the pendulum on Earth and [tex]g_{E}[/tex] is the measure of the strength of Earth's gravity
and
[tex]T_{M} = 2\pi \sqrt{\frac{L}{g_{M} } }[/tex] .......... (2)
Where [tex]T_{M}[/tex] is the period of the pendulum on Moon and [tex]g_{M}[/tex] is the measure of the strength of Earth's gravity on the Moon.
Since we are to determine the period of the same pendulum on the moon, then, [tex]2\pi[/tex] and [tex]L[/tex] are constants.
Dividing equation (1) by (2), we get
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
From the question,
[tex]T_{E} = 4.9secs[/tex]
[tex]g_{E}[/tex] = 9.8 m/s²
[tex]g_{M}[/tex] = 1.63 m/s²
[tex]T_{M}[/tex] = ??
From,
[tex]\frac{T_{E} }{T_{M} } = \sqrt{\frac{g_{M} }{g_{E} } }[/tex]
[tex]\frac{4.9}{T_{M} } = \sqrt{\frac{1.63}{9.8} }[/tex]
[tex]\frac{4.9}{T_{M} } = 0.40783[/tex]
[tex]T_{M} =\frac{4.9}{0.40783 }[/tex]
[tex]T_{M} = 12.01 secs[/tex]
∴ [tex]T_{M} = 12.0secs[/tex]
Hence, the period of that same pendulum on the moon is 12.0 seconds.
Answer:
The period of that same pendulum on the moon is 12.0 s
Explanation:
Given;
period of a pendulum’s swinging, T=2π√L/g
the strength of earth’s gravity on moon, g₂ = ¹/₆(g₁)
period of pendulum on Earth, T₁ = 4.9 s
period of pendulum on moon, T₂ = ?
The length of the pendulum is constant, make it the subject of the formula;
[tex]T = 2\pi \sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}}\\\\(\frac{T}{2\pi} )^2 =\frac{L}{g}\\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\ L = \frac{gT^2}{4\pi^2}\\\\L_1 = L_2\\\\\frac{g_1T_1^2}{4\pi^2}= \frac{g_2T_2^2}{4\pi^2}\\\\g_1T_1^2 = g_2T_2^2\\\\T_2^2 = \frac{g_1T_1^2}{g_2} \\\\T_2 = \sqrt{\frac{g_1T_1^2}{g_2}}\\\\ T_2 = \sqrt{\frac{g_1T_1^2}{g_1/6}}\\\\ T_2 = \sqrt{\frac{6*g_1T_1^2}{g_1}}\\\\T_2 = \sqrt{6T_1^2}\\\\ T_2 = T_1\sqrt{6} \\\\T_2 = (4.9)\sqrt{6}\\\\ T_2 = 12.0 \ s[/tex]
Therefore, the period of that same pendulum on the moon is 12.0 s
A car travels at a velocity of 80 m/s, accelerates than stops in 250 second. What is the acceleration of the truck?
(I’ll give brainliest please someone help me on this and explain it with work I don’t get it)
A disc at rest without slipping, rolls down a hill of height (3×9.8)m.What is its speed when it reaches at the bottom?
Answer:
The speed as it reaches the bottom is 24m/s
Explanation:
Given parameters:
Height of hill = (3 x 9.8)m = 29.4m
Unknown:
Final speed as it reaches the bottom = ?
Solution:
To solve this problem, we apply;
v² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity = 9.8m/s²
h is the height
note, the initial speed of the body is 0;
Input the parameters and solve;
v² = 0² + 2 x 9.8 x 29.4
v² = 576.24
v = √576.24 = 24m/s
The speed as it reaches the bottom is 24m/s
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
A sound wave can be considered as a displacement wave or a pressure wave? What phase difference exists between the displacement and the pressure of a sound wave?
A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
This question is incomplete, the complete question is;
A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
Answer: the time constant of the damped oscillation is 47.44s
Explanation:
Given that;
t = 5.0s
Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao
EXPRESSION for amplitude is A(t) = Ao e^-t / T
t is time while T is time constant
so
0.9Ao = Ao e^-t / T
0.9 = e^ -t/T
So we take the natural log of both the sides
ln (0.9) = -t/T
-0.1054 = -t/T
0.1054 = t/T
WE now substitute our value of t
0.1054 = t/T
0.1054T = 5.0
T = 5 / 0.1054
T = 47.44s
therefore the time constant of the damped oscillation is 47.44s
Question 16 1 pts Jessie feels pressured by his parents to get a job. This is an example of the law of?
readiness
disuse
effect
belonging
A spring gun is able to launch a 7.0 gram marble to a vertical height of 22 mmeasured from the compressed point of the marble). If the spring iscompressed 8.0 cm from its relaxed state, what will be the spring constant
Answer: Spring constant = 472N/m
Explanation:
The change in gravitational potential energy by the spring is given as = mgh
where m= 7.0 g = 7 X 10 -3kg
g= 9.8m/s
h= 22m
Gravitational potential energy= mgh
= 7.0 x 10^-3 X 9.8 x 22 = 1.5092 J
Remember that change in gravitational potential energy by the spring =elastic potential energy
Therefore, Potential energy P. E = 1/2 K x²
where K= CONSTANT
x= 8.0
2 X 1.5092 J / (8.0 X 10^-2)²= 471.625 ROUNDED TO 472 N/m
Assume that the particle has initial speed viviv_i. Find its final kinetic energy KfKfK_f in terms of viviv_i, MMM, FFF, and DDD.
Answer:
KE= 1/2mv²
Explanation:
The kinetic energy of a body is the energy possessed by virtue of the body in motion
Given the parameters
m which is the mass of the body
v which is the velocity of the body too
K.E = kinetic energy
The expression for the kinetic energy of a body is given as
KE= 1/2mv²
Which object will require the greatest amount of force to change its motion?
A. A 148 kg object moving 131 m/s
B. A 153 kg object moving 127 m/s
C. A 160 kg object moving 126 m/s
O D. A 162 kg object moving 124 m/s
Answer: D 160kg object moving 126 m/s
Explanation:
An object having a mass of 162 kg and moving with a velocity of 124 m/sec will require the greatest amount of force to change its motion. The correct option is D.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
If the object has to stop, the final velocity must be zero. If the time is constant, the amount of force only depends on the mass and the velocity at which the body is moving.
The amount of force on the object depends on the momentum of the body.
The momentum of the body is;
P = mv
Object D will require the greatest amount of force to change its motion. Because the momentum of the body for option D is the greatest.
Hence, the correct option is D.
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A water balloon launcher with a mass of 2.2 kg is suspended on a wire. It fires a 0.85 kg balloon to the north at a velocity of 13.0 m/s. What is the resulting velocity of the launcher if the net force on the launcher is equal to the reaction force?
Answer:
5.0 m/s south
Hope this Helps!
Answer:
5.0 m/s south
Explanation:
1. A speed boat is racing across a lake at 25 meters per
second when its motor burns out. It then slowly
comes to a stop over the next 45 seconds. What was its
acceleration?
v = u + a t
where u = initial velocity (25 m/s), v = final velocity (0), a = acceleration, and t = time (45 s). So
0 = 25 m/s + a (45 s)
a = (-25 m/s) / (45 s)
a ≈ -0.56 m/s²
A man walking at 3.56 m/s accelerates at 2.50 m/s2 for 9.28 s. How far does he get?
141 m
26.8 m
44.6 m
248 m
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
A single-turn circular loop of radius 9.4 cm is to produce a field at its center that will just cancel the earth's field of magnitude 0.7 G directed at 70degrees below the horizontal north direction. Find the current in the loop.
Answer:
The current in the loop is 10.5 A.
Explanation:
Given that,
Radius = 9.4 cm
Magnetic field = 0.7 G
Angle = 70°
We know that,
The magnetic field due to the current in a loop is
[tex]B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
The magnetic field due to the current is equal to the magnetic field of earth.
[tex]B_{E}=B_{I}=\dfrac{\mu_{0}NI}{2r}[/tex]
We need to calculate the current in the loop
Using formula of magnetic field
[tex]B=\dfrac{\mu_{0}NI}{2r}[/tex]
[tex]I=\dfrac{2rB}{\mu_{0}N}[/tex]
Put the value into the formula
[tex]I=\dfrac{2\times9.4\times10^{-2}\times0.7\times10^{-4}}{4\pi\times10^{-7}\times1}[/tex]
[tex]I=10.5\ A[/tex]
Hence, The current in the loop is 10.5 A.
Can we cure aging? Support your answer?
Answer: yes it can
Explanation:
Answer:
Yes you can. You need to eat healthy and have the proper diet and always exercise and build muscles, this can slow down aging, and most importantly, we must always be calm. However there is no way to stop aging for that’s part of life.
Explanation:
I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object?
Answer:
The net force on an object is the total force applied on the object after adding up all the forces
In the given diagram,
we can see that the 2 forces of 4N and 4N will cancel each other out since they are equal and in the opposite direction
Now, we are left with a force of 2N and 10N,
the net force will be the difference of these forces:
Net force = 10N - 2N
Net force = 8N downwards
Another way to do it:
The two 4N forces will be cancelled out,
and we are left with a 2N and a 10N force
(notice how we cancelled equal and opposite forces for the 4N)
We can divide the 10N force into (2N + 8N)
Since the 2N forces are equal and opposite, they will be cancelled out
and we will be left with a net force of 8N downwards
HELP PLS!!
In a full/angled projectile, Vty is equal to the inverse of viy
O True
O False
Please help me! What is Ohm's law?
Ohm's law shows the relationship between voltage, current, and the resistance of a energy bond. The formula for Ohm's law is:
voltage = current x resistance
This formula tells you that current and resistance is the voltage of an energy bond.
Hope this helps you!
Answer: What is Ohm’s Law?
Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.
Ohm's Law (E = IR) is as fundamentally important as Einstein's Relativity equation (E = mc²) is to physicists.
E = I × R
It means voltage = current × resistance, or volts = amps × ohms, or V = A × Ω.
Resistance cannot be measured in an operating circuit, so Ohm's Law is especially useful when it needs to be calculated. Rather than shutting off the circuit to measure resistance, a technician can determine R using the above variation of Ohm's Law.
Plz mark brainliest:)
Which statement best describes the direction of force shown by the magnetic field lines around a bar magnet?
A. Away from both the magnet's north and south pole
B. From the magnet's north to its south pole
C. From the magnet's south pole to its north pole
D. Toward both the magnet's north and south pole
Answer: I'm pretty sure the answer is B
Explanation: I'll check after the test
:Okay yes the answer is 100% B
The direction of the magnetic field shown by the magnetic field lines around a magnet is from north to its south pole of magnet. Option B is correct.
Magnetic Field Lines:
These are the imaginary lines that are used to represent the strength of the magnetic field.
The magnetic field lines starts from the North pole and ends up in the South pole.The strength of the magnetic field is directly proportional to the number of lines per unit area.The strength of the magnetic field is the highest near the north pole.Therefore, the direction of the magnetic field shown by the magnetic field lines around a magnet is from north to its south pole of magnet.
To know more about Magnetic Field Lines,
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A soccer ball accelerates from rest and rolls 6.5m down a hill in 3.1 s. It then bumps into a tree. What is the speed of the ball just before it hits the tree.
Answer:
2.096m/s
Explanation:
The speed of this soccer ball can be calculated using the formula;
Speed = distance/time
According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;
Speed = 6.5/3.1
Speed of the ball = 2.096m/s
Therefore, the speed of the ball before hitting the tree is 2.096m/s
When electrical energy is used what type of energy is also produced and considered to be waste energy?
Radiant energy
Thermal energy
Mechanical energy
Nuclear energy
Answer:
Thermal Energy
Explanation:
Define reflection of sound?
Explanation:
When sound travels in a given "medium", it would touch the surface of another "medium" and will bounce back in some other direction, this occurrence is called the reflection of sound.