Answer:
The man's total displacement is equal to 0.
Explanation:
Given that,
A man ran a 5 mile race. The race looped around a city park and back to the starting line.
We need to find the total displacement of the man.
We know that,
Displacement = shortest path covered
Also,
Displacement = final position - initial position
As it reaches back to its starting line, it means, the displacement is equal to 0.
Hence, the man's total displacement is equal to 0.
A pendulum is moving 2.0 m/s at the bottom of its swing. How high vertically will it go before it begins to swing back? Group of answer choices 0.4 m 0.1 m 0.2 m 0.8 m 1.0 m
Answer:
h = 0.2 m
Explanation:
Given that,
A pendulum is moving 2.0 m/s at the bottom of its swing.
We need to find the height high it swing back. Let the height is h.
Using the conservation of energy such that,
[tex]\dfrac{1}{2}mv^2=mgh\\\\h=\dfrac{v^2}{2g}[/tex]
Put all the values,
[tex]h=\dfrac{(2)^2}{2\times 10}\\\\h=0.2\ m[/tex]
So, it will reach to a height of 0.2 m.
a 10 kg box is pulled across an ice rink for a distance of 50m. it is pulled with a constant force of 10 N on a rope angled at 60° to the horizontal. How much work is done on the box? (A) 50 ) (B) 100 ) (C) 250 ) (D) 500)
Answer:
(C) 250
Explanation:
Given the following data;
Mass = 10 kg
Distance = 50 m
Force applied = 10 N
Angle = 60°
To find the work done on the box, we would first of all find the horizontal component of the force applied.
[tex] Horizontal force, Fx = mgCosd[/tex]
Where;
Fx represents the horizontal force.
m is the mass of an object.
g is the acceleration due to gravity.
d is the angle of inclination (theta).
mg = weight = 10 N
Substituting into the equation, we have;
[tex] Fx = 10 * Cos60 [/tex]
[tex] Fx = 125 * 0.5 [/tex]
Fx = 5 Newton
Next, we find the work done.
Mathematically, workdone is given by the formula;
[tex] Workdone = force * distance[/tex]
Substituting into the formula, we have;
[tex] Workdone = 5 * 50 [/tex]
Workdone = 250 Nm
A child has a mass of 30 kg on Earth. If th gravity on Moon is one sixth that of the Earth what is the weight of the child on the Moon? Gravity on Earth = 10 N^-1
Answer:
the baby would still be 30 kg on the moon because that’s mass, not weight.
However, it will feel like 5 kg, because the acceleration is 1/6 and the force is 1/6 and the actual weight (product of mass and acceleration will be 1/6).
The child has a mass of 30 kg on earth and has one-sixth gravity on the moon will have a weight of 50 kg on the moon.
What is Gravity?Gravity is a fundamental interaction that causes all objects with mass or energy to attract one another. The weakest of the four fundamental interactions is gravity, by far. The Moon's gravity causes sublunar tides in the oceans, just as gravity on Earth imparts weight to physical objects.
Given:
The mass of the child (m) = 30 kg,
The gravity on the moon (G) = 1/6 of the earth.
To calculate the weight of the child on the moon, use the formula given below,
F = m × a
Here
[tex]F[/tex] is the weight, [tex]m[/tex] is the mass and [tex]a[/tex] is the gravity on earth, gravity on earth [tex]10\ m/s^2[/tex].
Substitute the values
[tex]F=30*10/6[/tex]
[tex]F = 50\ kg[/tex]
Therefore, the weight of a child on the moon is 50 kg.
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what are scalar quantities and list their examples
Explanation:
these are quantities that have magnitude but no direction
Examples include : speed, mass, volume, density, everything, time
A massive truck collides with a less massive car. The force experienced by the truck is ____________ (less than, equal to, greater than) the force experienced by the car. The resulting acceleration of the truck is ____________ (less than, equal to, greater than) the resulting acceleration of the car.
Please answer ASAP
Answer:
equal to; less than
Explanation:
newtons 3rd law says that all forces have equal and opposite reactions so it would be equal to for the first one. but the acceleration depends on the mass of the object and because the truck is more massive, the acceleration would be slower, or less than.
a string 2m long used to whirl a 200gm stone in horizontal circle at a speed of 2m/s . find the tension in the string
Answer:
Explanation:
First off, we need the mass in kg and it's in g so we have to convert it. Then we can do the problem. 200 g = .200 kg. Moving on.
The equation used to find the tension in the string is the same one we use to find the centripetal force, because the tension is what is upplying the centripetal force needed to keep the stone moving in a circular manner. The formula for that is
[tex]F_c=T=\frac{mv^2}{r}[/tex] where ma is the mass in kg, v is the velocity in m/s, and r is the radius of the circle about which the stone spins in meters.
Filling in:
[tex]T=\frac{(.200)(2)^2}{2}[/tex] which gives us
T = .4 N
In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.71 km mark at a time of 25.0 min . If he accelerated for 60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 7.71 km mark was the same as his overall average speed up to that time.
Answer:
0.18 m/s²
Explanation:
The total time taken to cover a 10 km race is 27 min, 43.6 seconds.
At 25 min, Pre was at 7.71 km mark. Therefore the average speed = 7.71 km / 25 min = 7710 m / (25 * 60) s = 5.14 m/s
The distance remaining = 10 km - 7.71 km = 2.29 km = 2290 m
The remaining time = 27 min, 43.6 seconds - 25 min = 2 min 43.6 second = 163.6 seconds
She accelerates for 60 seconds, therefore the distance covered (S) during the acceleration (a) is:
S₁ = 5.14(60) + 0.5a(60)² = 308.4 + 1800a
She maintains the speed for the remaining distance (S₂). The remaining time = 163.6 seconds - 60 seconds = 103.6 seconds. The final speed after the acceleration = (5.14 + 60a) m/s
S₂ = (5.14 + 60a)* 103.6 = 532.5 + 6216a
S₁ + S₂ = 2290 m
(308.4 + 1800a) + (532.5 + 6216a) = 2290
8016a + 840.9 = 2290
8016a = 1449.1
a = 0.18 m/s²
An 84% efficient single pulley is used to lift a 230 kg piano 3.5 m. How much work must be input?
Answer:
35%
Explanation:
win
The amount of work required to lift the 230 kg to a height of 3.5 m at 84% efficiency is 9391.67 J
What is efficiency?This can be defined as the capacity to achieve a given task with little or no waste. Mathematically, it can be expressed as
Efficiency = (output / input) × 100
How to determine the work output Mass (m) = 230 KgHeight (h) = 3.5 mAcceleration due to gravity (g) = 9.8 m/s²Work output =?Work output = mgh
Work output = 230 × 9.8 × 3.5
Work output = 7889 J
How to determine the work input Output = 7889 JEfficiency = 84% Input =?Efficiency = (output / input) × 100
84% = 7889 / input
Cross multiply
84% × input = 7889
Divide both side by 84%
Input = 7889 / 84%
Input = 7889 / 0.84
Input = 9391.67 J
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what is electricity?write any two defects of a simple cell.
Answer:
Electricity is the set of physical phenomena associated with the presence and motion of matter that has a property of electric charge. Electricity is related to magnetism, both being part of the phenomenon of electromagnetism, as described by Maxwell's equations.
A bat is flying at 0.6 m/s when it spots an appetizing insect. The bat accelerated at a rate of 1.2m/s2 for 1.9
seconds. How fast is the bat traveling at the time?
Answer:
v_f = 2.9m/s
Explanation:
v_f = v_I + a(t)
v_f = 0.6m/s + 1.2m/s²(1.9s)
v_f = 2.88m/s
v_f = 2.9m/s
what the balance electrons for calcium
Answer:
its two valence electrons
Explanation:
Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions.
UR WELCOME!! :)
What is moment? Write down the law of moment.A long spanner is used to unscrew the tight nut.Why?
Answer:
Moment is the product of force and its perpendicular distance from a point along its line of action.
The net moments on clockwise and anticlockwise is zero at a point.
The net force applied on a spanner handle is equal to the moment of the rotation
Which is NOT a characteristic of a solid?
a) All solids have a definite shape.
b) All solids have molecules that are loosely packed together.
C) All solids have mass.
d) All solids take up space
Answer:
b
Explanation:
what is the momentum of an object weighing 7.5 kg moving at 1.6 m/s?
Answer:
[tex]momentum = mass \times velocity \\ = 7.5 \times 1.6 \\ = 12 \: kg {ms}^{ - 1} [/tex]
Mass is 7.5 kg
Velocity is 1.6 m/s
we know that,
momentum = mass × velocity
or; momentum =7. 5 kg × 1. 6 m/s
or; momentum = 12 kgms^1
Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π ΔL /√(L g)
Answer:
ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]
Explanation:
In a simple harmonic motion, specifically in the simple pendulum, the angular velocity
w = [tex]\sqrt{\frac{g}{L} }[/tex]
angular velocity and period are related
w = 2π / T
we substitute
2π / T = \sqrt{\frac{g}{L} }
T = [tex]2\pi \ \sqrt{\frac{L}{g} }[/tex]
In this exercise indicate that for a long Lo the period is To, then and increase the long
L = L₀ + ΔL
we substitute
T = [tex]2\pi \ \sqrt{\frac{L + \Delta L}{g} }[/tex]
T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }[/tex]
in general the length increments are small ΔL/L «1, let's use a series expansion
[tex]\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...[/tex]
we keep the linear term, let's substitute
T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )[/tex]
if we do
T = T₀ + ΔT
T₀ + ΔT = [tex]2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}[/tex]
T₀ + ΔT = T₀ + [tex]\pi \sqrt{\frac{1}{Lg} } \ \Delta L[/tex]
ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]
a ball of mass 0.2 kg is dropped from a height of 20m on impact to the ground it loses in 30 joule of energy calculate the height it reaches on rebound
Answer:
new PE at top=original PE - energy lost
mgh=.2(9.8)20-30
Explanation:
mgh=.2(9.8)20-30 = 9.2
A small part of a current loop consisting of a very, very large triangular conductor in the x-y plane is shown below, with the current i=39.30 A.
12 10 4 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 x (cm)
Calculate the magnitude of the B-field at point 'p'.
Answer: hello your question is poorly written attached below is the well written question
answer: 4.9 * 10^-5 T
Explanation:
Current ( i ) = 39.30 A
Determine the magnitude of the B-field at Point P
Mag of B-field at Point P = μo*I / 2π*r --- ( 1 )
where : I = 39.30 , μo = 4π * 10^-7 , r = 8*10^(-2)
Input values into equation 1
( 10^-7 * 39.30 ) / ( 8*10^-2 )
∴ Mag of B-field = 4.9 * 10^-5 T
A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?
Answer:
592000 J
Explanation:
We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:
1 Pa = 1 Kg/ms²
Therefore,
3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²
Next, we shall determine the workdone.
Workdone is given by the following equation:
Workdone (Wd) = pressure (P) × change in volume (ΔV)
Wd = PΔV
With the above formula, the work done can be obtained as follow:
Pressure (P) = 3.7×10⁵ Kg/ms²
Change in volume (ΔV) = 1.6 m³
Workdone (Wd) =?
Wd = PΔV
Wd = 3.7×10⁵ × 1.6
Wd = 592000 Kgm²/s²
Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:
1 Kgm²/s² = 1 J
Therefore,
592000 Kgm²/s² = 592000 J
Therefore, the Workdone is 592000 J.
Can someone please help with part c)
Energy released ,
[tex] = 3.789 \times {10}^{ - 12} [/tex] J
9) What is the temperature of 5 moles of nitrogen at 1 atm in a 2 liter container?*
- 4.88 K
- 0.0328 K
- 0.41 K.
- 0.0164 K
Answer:
4.88 K.
Explanation:
From the question given above, the following data were obtained:
Number of mole (n) = 5 moles
Pressure (P) = 1 atm
Volume (V) = 2 L
Gas constant (R) = 0.082 atm.L/Kmol
Temperature (T) =?
The temperature of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
1 × 2 = 5 × 0.082 × T
2 = 0.41 × T
Divide both side by 0.41
T = 2 / 0.41
T = 4.88 K
Therefore, the temperature of the gas is 4.88 K.
The rock falls from the distance of 15 m before it hits the water. Calculate its kinetic energy just before hitting the water. Show your working
Answer:
[tex]k.e = \frac{1}{2} m {v}^{2} \\ {v}^{2} = {u}^{2} + 2gs \\ {v}^{2} = 0 + (2 \times 9.8 \times 15) \\ v = 17.1 \: m {s}^{ - 1} \\ k.e = \frac{1}{2} \times m \times {17.1}^{2} \\ = 147m \: joules \\ m \: is \: mass[/tex]
Kinetic energy of the rock just before hitting the water is 147m joules where m is the mass of the object.
What is Kinetic Energy?The kinetic energy of an object is defined as the energy it possesses due to its motion. It is the work required to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains the kinetic energy as long as its momentum does not change.
It is directly proportional to the mass of the object and the square of its velocity:[tex]K.E. = 1/2 mv^2[/tex].
Where, m= mass of the object
v= velocity
So if the units of mass are in kilograms and the units of velocity are meters per second, then kinetic energy has units of kilogram-meter squared per second squared.
For above given information,
[tex]K.E= 1/2 mv^2\\v^2= u^2 + 2gs\\ \\v^2= 0+ (2* 9.8*15)\\v^2= 294 m/s\\So, K.E. = 1/2 m* 294 m/s\\K.E.= 147m joules[/tex]
Thus, Kinetic energy of the rock just before hitting the water is 147m joules where m is the mass of the object.
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Water is being heated on the stove at 90 ⁰C. What is this temperature on the Fahrenheit degrees?
Answer:
194 degrees farenheit! hot
Explanation:
(90°C × 9/5) + 32 = 194°F
Answer:
194
Explanation:
to get from celcius to Fahrenheit
multiply temperature by 2 and add 30
Vector B has components Bx = 6.18 and By = 4.43. Find the magnitude and direction counter clockwise from the positive x-axis.
Answer:
Explanation:
The formula for the magnitude of this resultant vector, which is in Q1, btw, is:
[tex]B_{mag}=\sqrt{6.18^2+4.43^2}[/tex] which gives us, to 3 sig fig's:
7.60 m. Since the resultant vector is in Q1, we don't need to add anything to the angle once we find it to 3 sig fig:
[tex]tan^{-1}(\frac{4.43}{6.18})[/tex] gives us an angle of 35.6 degrees
A student is in class 400 minutes every day for 5 days each week. How many seconds is this?
Answer:
120,000seconds
Explanation:
400 minutes x 5= 2000minutes (2000minutes x 60)seconds =120,000seconds.
btw :
stay safe! :3
Answer:
12000
Explanation:
5 x 400 = 2000, 2000 x 60 = 12000
for the light bulb to produce the aame illuminace with the candle to the screen, the luminous intensity of the bulb is _______ that of the candle?
Answer:
9 times
Explanation:
luminosity decreases to the square of the distance
What's scientific notation
Katie and her sister have entirely different features and traits. Katie has blond hair and brown eyes, but her sister has red hair and blue eyes. What causes these variations in traits such as hair color, eye color, and height?
A. competition
B. offspring
C. alleles
D. fitness
Answer:
C. Alleles
Explanation:
Our genes are responsible for our phenotype (physical traits). Alleles is the only genetic term in there.
A carpenter applies a force of 60N horizontally to push a plane 40 cm along a piece of wood, how much work does she do?
Answer:
W = 24 J
Explanation:
Given that,
Applied force, F = 60 N
Distance moved, d = 40 cm = 0.4 m
We need to find the work done by the carpenter. We know that,
Work done, W = Fd
Put all the values,
W = 60 N × 0.4 m
= 24 J
Hence, the required work done is equal to 24 J.
10 points
Calculate the force of attraction between the moon and the earth, if their masses and distance apart are 10^22Kg, 10^24kg and 6.4x10^6m. Take g = 6.67x10^-11Nm^2Kg^-2
a) 0.63x10^24N
b) 1.63x10^22N
c) 2.63x10^20N
d) 3.63x10^18N
Answer:
Explanation:
You didn't fill in the proper masses which is why you never got an answer to this. But that's ok...I got you. I happen to know what they are! We will use the universal law of gravitation and the gravitational constant to solve this.
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:
[tex]F_g=\frac{(6.67*10^{-11})(5.98*10^{24})(7.36*10^{22})}{(3.84*10^8)^2}[/tex] The denominator is the radius of the earth plus the radius of the moon plus the distance between their surfaces, just FYI.
That gives us that
[tex]F_g=1.99*10^{20}N[/tex] Not sure what your choices entail, but I'd have to say, taking into consideration that maybe your problem didn't figure in the distance between the surfaces, you'd be at choice B.
A cyclist is taking part in the Tour de France, which is a bicycle race that takes place every year.
a Two forces acting on the cyclist are weight and reaction. Name two other forces acting on him
as he cycles along.
