Answer:
Gravity
because it's factorised by mass of a body.
For other forces, they deal with charges of negligible mass and weights
Answer:
Gravity
Explanation:
A construction laborer holds a 20 kg sheet of wallboard 3 m above the floor for 4 seconds. During these 4 seconds how much power was expended on the wallboard
Answer:
zero
Explanation:
no distance has moved while holding the sheet...so no distance means no workdone..no workdone means no power...
a 45 kg boy sits on a horse on a carousel 5.0 m from the center of the circle. he makes a revolution every 8.0 s.
calculate his speed.
what is centripetal force acting on the boy?
Answer:
2.9m/s and 140N
Explanation:
For speed convert rev/s to m/s
(Rev/8s)=(2π/8)(5m) = 3.9m/s
For Centripetal Force
Fc = (ac)(m) = (v^2/r)(m) = [(3.9m/s)^2/(5.0m)][45kg] = 140N
Answers go to two significant figures.
Find the height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground.
Answer:
s = 2.79 m
Explanation:
Given that,
A ball have a speed of 7.4 m/s just before it hits the ground.
Initial velocity of the ball was 0 (at rest)
We need to find the height from where the ball was dropped. It means we need to find the distance covered by it. Let it is h.
Using the third equation of motion to find it as follows :
[tex]v^2-u^2=2as\\\\s=\dfrac{v^2}{2g}\\\\s=\dfrac{(7.4)^2}{2\times 9.8}\\\\s=2.79\ m[/tex]
So, the ball is dropped from a height of 2.79 m.
The height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m
Equation of motionsAccording to the third equation of motion;
[tex]v^2=u^2+2as[/tex]
where;
v is the final velocity = 7.4m/s
Initial velocity = 0m/s
s is the distance
a = g= 9.8m/s²
Substitute the values into the formula to have:
[tex]7.4^2=0^2+2(9.8)s\\54.76 + 19.6s\\s=\frac{54.76}{19.6}\\ s=2.79m[/tex]
Hence the height from which you would have to drop a ball so that it would have a speed of 7.4 m/s just before it hits the ground is 2.79m
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A 47.5-g ball moves at 30.0 m/s. If its speed is measured to an accuracy of 0.20%, what is the minimum uncertainty in its position
Answer:
The minimum uncertainty in its position is 1.85 x 10⁻³² m.
Explanation:
Given;
mass of the ball, m = 47.5 g = 0.0475 kg
speed of the ball, v = 30 m/s
measuring accuracy of the speed, = 0.2% = 0.002
The uncertainty in measurement of momentum;
ΔP = mΔv
ΔP = (0.0475)(30 x 0.002)
ΔP = 2.85 x 10⁻³ kgm/s
The uncertainty in position is calculated as;
[tex]\delta x \geq \frac{h}{4\pi (\delta P)}[/tex]
where;
h is Planck's constant
[tex]\delta x \geq \frac{6.626 \ \times \ 10^{-34}}{4\pi (2.85 \ \times \ 10^{-3})} \\\\\delta x \geq 1.85 \ \times \ 10^{-32} \ m[/tex]
Thus, the minimum uncertainty in its position is 1.85 x 10⁻³² m.
Need help y’all ASAP please...physics
Answer:
t = 3/8 seconds
Explanation:
h=-16t^2 - 10t+6
h= 0 when it hits the ground
0=-16t^2 - 10t+6
factor out a -2
0= -2(8t^2 +5t -3)
divide by -2
0 = (8t^2 +5t -3)
factor
0=(8t-3) (t+1)
using the zero product property
8t-3 = 0 t+1 =0
8t = 3 t= -1
t = 3/8 t= -1
t cannot be negative ( no negative time)
t = 3/8 seconds
Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did you do
Answer:
40 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:
Workdone = force × distance
Wd = F × s
With the above formula, we can obtain the workdone as follow:
Force (F) = 10 N
Distance (s) = 4 m
Workdone (Wd) =?
Wd = F × s
Wd = 10 × 4
Wd = 40 J
Thus, 40 J of work was done.
14. After finishing her homework, Sue climbs up a 5.00 m high flight of stairs to her bedroom
Find the magnitude of Sue's weight
and how much
work Sue does in climbing the stairs if she
has a mass of 50.0 kg? (4.90 x 2 N, 2450J)
Explanation:
Given parameters:
Height = 5m
Mass of Sue = 50kg
Unknown:
Magnitude of Sue's weight = ?
Work done by Sue = ?
Solution:
Weight is the vertical force exerted by a body in the presence of gravity.
Mathematically;
W = mg
m is the mass
g is the acceleration due to gravity = 9.8m/s²
Weight = 50 x 9.8 = 490N
Work done = Force x distance = weight x height
Work done = 490 x 5 = 2450J
What are the units for measuring specific heat?
a. degrees Celsius per gram
b. joules per degrees Celsius
c. joules per gram degree Celsius
d. degrees Celsius per joule gram
Answer:
c. joules per gram degree Celsius
Explanation:
edg 2021
Q3) Salman walks to the mosque with speed 2.4 m/s. If it takes him 3 min to
reach the mosque. Find the distance.
Answer: 432m
Explanation:
Convert 3 min to seconds
1 min = 60 sec
3 min = 180 sec
Multiply the speed times time to get distance.
2.4 x 180 = 432m
The device shows the relative humidity at 22°C. What’s the water vapor density if the maximum water vapor in air at this temperature is 20 grams/cubic meter?
A device showing that at 22 degrees Celsius the relative humidity is 58%.
Answer:
The answer would be 11.6 g/m^3
Explanation:
Knowing that vapor's density at 100% humidity is 20g/m^3 we can calculate the answer using the following math:
0.58 * 20 = 11.6 grams per cubic meter
the units can also be written as g/m^3
The density of water vapor is 11.6 grams/cubic meter.
What is relative humidity?The quantity of atmospheric moisture that is present compared to the amount that would be there if the air were saturated is known as relative humidity, and it is stated as a percentage. Relative humidity depends on both moisture content and temperature because the latter quantity is temperature-dependent. The related temperature and dew point for the specified hour are used to calculate relative humidity. it is expresses as %.
Now given that the maximum water vapor in air at this temperature is 20 grams/cubic meter at 22°C, that is, density of vapor at 100% humidity is 20g/[tex]m^3[/tex] at 22°C.
And, The device showing that at 22 degrees Celsius the relative humidity is 58%
So, water vapor density at that moment is = 20g/[tex]m^3[/tex] × 58%
= 11.6 g/[tex]m^3[/tex] .
= 11.6 grams/cubic meter.
Hence, according to the measurement of device, the water vapor density is 11.6 grams/cubic meter.
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Economics these two PLEASE
Answer:
7. d. England during the Age of Enlightenment
8. a. State governments do not usually act together
Explanation:
The Age of Enlightenment occurred in the 18th century and at this time England was a Constitutional monarchy with a monarch and parliament ruling the country with different powers.
The third and fourth methods of amending the Constitution will be much harder to use because state governments simply do not usually act together as they most times have ideological differences. The sheer number of states it would take to unite for these methods to be used makes this impractical.
Calculate the heat energy needed to change the temperature of 2 kg of copper from 10°C to 110°C.
If you could show your process and equations used, that would be very helpful! Thanks!
Answer:
77000 J
Explanation:
Formula for the heat energy is;
Q = m•c•Δt
We are given;
mass; m = 2 kg
Change in temperature; Δt = 110 - 10 = 100 °C
From online values, specific heat capacity of copper is; c = 385 J/kg.°C
Thus;
Q = 2 × 100 × 385
Q = 77000 J
Answer:
heat = 20 Kcal
Explanation:
write the difference between convert and
Concave minores
What type of telescope did Galileo use to observe Jupiter?
reflector telescope
refractor telescope
radio telescope
latter telescope
Answer:
a refractor telescope
Explanation:
Answer:
a. refractor telescope
short essay your teacher have provided you
A boat is drifting to the right with a speed of 5.0 m/s when the driver turns on the motor. The motor runs for 6.0 seconds causing a constant leftward acceleration of magnitude 4.0 m/s squared. What is the displacement of the boat over the 6.0 second time interval?
Answer:
[tex]D= -0.42km[/tex]
Explanation:
From the question we are told that
Drifting right with speed 5.0m/s
The motor runs for 6.0 seconds
Leftward acceleration of magnitude 4.0 m/s squared
Generally the equation [tex]V=ut+1/2at^2[/tex] can be used here
[tex]V=ut+1/2at^2[/tex]
Mathematically solving with the newton equation above we have that
[tex]D=5*6 + \frac{1}{2} (-4)*6^2[/tex]
[tex]D=30-72[/tex]
[tex]D=-42m[/tex] [tex]or -0.42km[/tex]
Therefore having this the Displacement is [tex]D= -0.42km[/tex] leftward
1. A 75.0 kg man pushes on a 500,000kg wall for 250s but it does not move. How
much work does he do on the wall?
Answer:
0J
Explanation:
No work is being done on the wall by the man pushing on it.
Given parameters:
Mass of man = 75kg
Mass of wall = 500000kg
Time = 250s
Unknown:
Work done = ?
Solution:
Work done is the force applied on a body that moves it along a particular path.
For work to be done, distance must be move or displacement must occur.
Since the wall is not moving the distance is 0;
Work done = Force x distance
Since distance is 0m, work done is 0J
The work done on the wall by the man is 0 J.
To calculate the amount of work done by the man, we use the formula below.
Formula:
W = (ma)d............. Equation 1Where:
W = Work done on the wall by the manm = mass of the walla = acceleration of the walld = distance.from the question,
Given:
m = 500000 kga = 0 m/s² (not moving)d = 0 m.Substitute these values into equation 1
W = 500000(0)(0)W = 0 J.Hence, the work done on the wall by the man is 0 J.
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Find the applied voltage of a telephone circuit that draws 0.017A through a resistance Of 5,000 ohms
A horizontal force of 90.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.08 m/s, what is the magnitude of the force of kinetic friction (in N) acting on the crate?
Why is 1/2kx^2=gym not a linear equation
What is the momentum of a 12 kg ball with velocity of 24 m/s?
Answer:
288 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 12 × 24
We have the final answer as
288 kg.m/sHope this helps you
a suspension bridge cable is connected to its anchor at a 20 degree angle. calculate the vertical component force on the anchor by the cable.
The question is missing some parts. Here is the complete question.
A suspension bridge cable is connected to its anchor at a 20° angle. Find the vertical and horizontal component of the force on the anchor by the cable.
Answer: [tex]F_{x}=[/tex] 14095.4 N
[tex]F_{y}=[/tex] 5130.3 N
Explanation: The force applied to the anchor is not perpendicular to the horizontal plane. So, it can be decomposed into 2 components: a vertical component, which is on the y-axis, and a horizontal component, which is on the x-axis.
The force and its components forms a right triangle, so we can calculate the components by using trigonometric relations:
Horizontal
[tex]cos(20)=\frac{F_{x}}{F}[/tex]
[tex]F_{x}=F.cos(20)[/tex]
[tex]F_{x}=15,000(0.9397)[/tex]
[tex]F_{x}=[/tex] 14,095.4 N
Vertical
[tex]sin(20)=\frac{F_{y}}{F}[/tex]
[tex]F_{y}=Fsin(20)[/tex]
[tex]F_{y}=[/tex] 15,000(0.3420)
[tex]F_{y}=[/tex] 5,130.3 N
The vertical and horizontal components of force on the anchor by the cable are 5130.3 N and 14095.4 N, respectively.
A certain planet has a radius of 4990 km. If, on the surface of that planet, a 95.0 kg object has a weight of 591 N, then what is the mass of the planet?

Answer:
3743.489 kg
Explanation:
F_g = 591 N
G = 6.674x10^-11 constant of gravity
m_1 = 95 kg
m_2 = unknown
r = 4990*1000 =
F_g = G[(m_1*m_2)/r^2]
591 N = 6.674x10^-11[(95*m_2)/4990^2]
8.855 = [(95*m_2)/4990^2]
355631.472 = 95*m_2
m_2 = 3743.489 kg
The mass of the planet, which has a radius of 4990 km, is 1.81×10²³kg.
What is Newton's law of universal gravitation ?Newton's law of universal gravitation states that
The force of attraction between any two bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their masses.
Given parameters:
Mass of the object: m = 95 kg
Radius of the planet: r = 4990 km = 4990 × 1000 m.
Weight f the object: F= 591 N
We know that: universal gravitational constant: G = 6.674x10^-11 SI unit.
We have to find: mass of the planet: M = ?
Now, F= GMm/r^2
591 N = 6.674x10^-11[(95×M)/(4990×1000)^2]
⇒ M =1.81×10²³kg
Hence, mass of the planet is 1.81×10²³kg.
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List two examples of how the land can have a dramatic change in temperature throughout the day.
Answer:
one could freeze and the second would thaw
Explanation:
sorry if its wrong
The two examples of how land can have a dramatic change in temperature
is during freezing and thawing.
Cold temperatures which is common during the winter period is
characterized by the formation of snow and freezing of smaller water
bodies.
There may be a short phase in which there is relative sunlight which melts
the frozen substances thereby forming liquids . This is usually as result of
the temperature being on the increase in the atmosphere.
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Use the graph and the table above to complete the table by estimating the amount of energy transferred near location B.
A.
1600 kilojoules
B.
300 kilojoules
C.
800 kilojoules
D.
400 kilojoules
E.
3200 kilojoules
Answer:
A. 1600 kilojoules
Explanation:
Looking at the graph, at 4 meters, we can guess the energy transferred would be around 1500 kilojoules. 1600 kilojoules is the closest to that value, so A would be the best choice.
A 1.5kg object moving with a speed of 2.5m/s strikes a wall and the ball rebounds with a speed of 1.5m/s. The ball is in contact with the wall for 0.045s. What is the magnitude of the average force exerted on the ball by the wall?
Answer:
F = 133.33[N]
Explanation:
This problem can be solved by the principle of momentum conservation, which tells us that momentum is preserved before and after the bounce of the ball on the wall.
In such a way that the movement towards the wall we will take it with a positive sign, and the force of the rebound to the left as negative. The movement to the left will be taken as a negative sign.
[tex]m_{1}*v_{1}-F*t=-m_{1}*v_{2}[/tex]
where:
m₁ = mass of the object = 1.5 [kg]
v₁ = velocity of the ball before hitting the wall = 2.5 [m/s]
F = average force [N]
t = time contact = 0.045 [s]
v₂ = velocity of the ball after hitting the wall = 1.5 [m/s]
Now replacing:
[tex](1.5*2.5)-F*0.045=-(1.5*1.5)\\3.75+2.25=F*0.045\\F=6/0.045\\F=133.33[N][/tex]
Help Help!
You can experience loss of coordination with a BAC as low as 0.02 -0.03.
A. True
B. False
Suppose that when spring was wound, 100J of work was done but 15J escaped to the surrounding as heat. The change in internal energy of the spring is?
Answer: 85J
Explanation:
From the question, we are informed that when spring was wound, 100J of work was done but 15J escaped to the surrounding as heat.
Therefore, the change in internal energy of the spring will be calculated as:
ΔU = q + w
where, q = -15J
w = 100J
ΔU = -15J + 100J
= 85J
An automobile which set the world record for acceleration increase speed from rest to 96 km/h in 3.07 seconds what distance traveled by the time the final speed was achieved
Answer:
41.02m
Explanation:
Given parameters:
Initial velocity = 0m/s
Final velocity = 96km/hr
Time taken = 3.07s
Unknown:
Distance traveled by the time the final speed was achieved = ?
Solution:
To solve this problem, we first find the acceleration of the car;
Acceleration = [tex]\frac{v - u }{t}[/tex]
v is the final velocity
u is the initial velocity
t is the time taken
Now convert the the final velocity to m/s;
96km/hr to m/s;
1 km/hr = 0.278m/s
96km/hr = 96 x 0.278 = 26.7m/s
Now;
Acceleration = [tex]\frac{26.7 - 0}{3.07}[/tex] = 8.69m/s²
So;
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
26.7² = 0² + 2 x 8.69 x s
712.89 = 17.38s
s = 41.02m
A 1248 kg car is pulled by a tow truck that has an acceleration of 2.0 m/s2. What is the
net force on the car?
Answer: = 2,496 N
Explanation:
Force is the push or pull on an object and it is calculated as:
= Mass * Acceleration
= 1,248 * 2.0
= 2,496 N