A lumberjack is trying to drag a small tree that he cut down. If the static
coefficient of friction of the tree on the ground is 0.5 and the tree weighs 430
N, what is the minimum amount of horizontal force that he will need to apply
so that the tree will start moving?
A. 215 N
B. 430 N
C. 365 N
D. 500 N
Answer:
A
Explanation:
weight of the tree =normal force
Horizontal force =coefficient of friction x Fnormal
0.5×430=215
A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it moves. Which of the following must be true?
a) 9 joules of kinetic energy were produced
b) 9 joules of gravitational potential energy were produced
c) 9 joules of heat energy were produced
d) 9 joules of kinetic energy and heat were produced
Answer:
9 joules of heat energy was produced
Explanation: there is no acceleration therefore its not a kinetic energy
Energy= force × distance
= 3×3
=9
Using pascals principle, F1/A1=F2/A2, solve this like a proportion.
A force of 50 N is applied to an area of 200 sq feet, how much force will be applied of the area to be covered is 50 sq feet?
Answer:
NO CLUE
Explanation:
GOOD LUCK THOUGH
A 56 kg pole vaulter falls from rest from a height of 5.1 m onto a foam rubber pad. The pole vaulter comes to rest 0.29 s after landing on the pad.
Required:
a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision
a. The athlete's velocity just before reaching the pad is [tex]35.21m/s[/tex]
b. The constant force exerted on the pole vaulter is 6799.52 N
a. We use Newton's equation of motion,
[tex]v=u+at\\\\S=ut+\frac{1}{2}at^{2}[/tex]
Where u is initial velocity, v is final velocity, a is acceleration , t is time and S represent distance.
Given that, s = 5.1 m , t = 0.29s, u = 0
Substitute in above equation.
[tex]5.1=\frac{1}{2}*a*(0.29)^{2} \\\\a=\frac{5.1*2}{0.084}=121.42m/s^{2}[/tex]
the athlete's velocity, [tex]v=0+121.42*(0.29)=35.21m/s[/tex]
b. The constant force exerted on the pole vaulter due to the collision is given as, [tex]Force=mass*acceleration[/tex]
[tex]Force=56*121.42=6799.52N[/tex]
Learn more:
https://brainly.com/question/13532462
A gymnast of mass 70.0 kgkg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81m/s29.81m/s2 for the acceleration of gravity.
PART A Calculate the tension T in the rope if the gymnast climbs the rope at a constant rate.
PART B Calculate the tension TTT in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 1.00 m/s2
PART C Calculate the tension TTT in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 1.00 m/s2m/s2 .
Answer:
43994
Explanation:
Hope this helps!
two particles woth each charge magnitude 2.0×10^-7 c but opposite signs are held 15cm apart.what are the magnitude and direction of the electric field E at tge point midway between charges
Answer:
The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.
Explanation:
Given;
charge of each particle, Q = 2 x 10⁻⁷ C
distance between the two charges, r = 15 cm = 0.15 m
distance midway between the charges = 0.075 m
The magnitude of the electric field is calculated as;
[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]
The direction of the electric field is from positive particle to negative particle.
The food calorie, equal to 4186 J, is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit-and-cereal bar contains 160 food calories per bar.
Part A
If a 67.0 kg hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy?
Express your answer in meters.
Part B
If, as is typical, only 20.0 % of the food calories go into mechanical energy, what would be the answer to Part A? (Note: In this and all other problems, we are assuming that 100% of the food calories that are eaten are absorbed and used by the body. This is actually not true. A person's "metabolic efficiency" is the percentage of calories eaten that are actually used; the rest are eliminated by the body. Metabolic efficiency varies considerably from person to person.)
Express your answer in meters.
You throw a football straight up. Air resistance can be neglected. When the football is 4.00 mm above where it left your hand, it is moving upward at 0.500 m/sm/s. What was the speed of the football when it left your hand
Answer:
u=8.868 m/s
Explanation:
The displacement of the ball is 4 meters
The final speed of the ball is 0.5 m/s
The initial speed of the ball is to be calculated. Using the equation of the rectilinear motion,
[tex]v^{2} =u^{2} +2as[/tex]
Plugging the values in the above expression,
[tex]\\0.5^{2} =u^{2} +2*(-9.8)*4\\\\u^{2} =78.4+0.25\\\\u^{2} =78.65\\\\u=8.868 m/s[/tex]
A spring with a 10-kg mass and a damping constant 15 can be held stretched 2 meters beyond its natural length by a force of 6 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity.
Required:
Find the position of the mass at any time t.
Answer:
Explanation:
Given that:
mass = 10kg
damping constant C = 15 kg/s
length = 2 m
Force F = 6N
Using the Hooke's law:
F = kx
6 = 15x
k = 6 N /2 m
spring constant k = 3 N/m
For the critical damping
C² - 4k*m= 0
m = C²/4k
m = (15)²/4(3) kg
m = 225/12 kg
m = 18.75 kg
a 2100-kg car drives with a speed of 18 m/s onb a flat road around a curve that has a radius of curvature of 83m. The coefficient of static friction between the car and the road is 0.78. What is the magnitude of the force of static friction acting on the car
Answer:
The magnitude of the friction force is 8197.60 N
Explanation:
Using the definition of the centripetal force we have:
[tex]\Sigma F=ma_{c}=m\frac{v^{2}}{R}[/tex]
Where:
m is the mass of the carv is the speed R is the radius of the curvatureNow, the force acting in the motion is just the friction force, so we have:
[tex]F_{f}=m\frac{v^{2}}{R}[/tex]
[tex]F_{f}=2100\frac{18^{2}}{83}[/tex]
[tex]F_{f}=8197.60 \: N[/tex]
Therefore the magnitude of the friction force is 8197.60 N
I hope it helps you!
Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the beam, and moves at 314 m/s. What is the maximum magnetic force it can feel
Answer:
The maximum magnetic force is 2.637 x 10⁻¹² N
Explanation:
Given;
Power, P = 8.25 m W = 8.25 x 10⁻³ W
charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C
speed of the charge, v = 314 m/s
area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²
The intensity of the radiation is calculated as;
[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]
The maximum magnetic field is calculated using the following intensity formula;
[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]
The maximum magnetic force is calculated as;
F₀ = qvB₀
F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)
F₀ = 2.637 x 10⁻¹² N
The drag force Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
Fd =C dA 2 pV2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and \rhorho is the air density.
Determine the power, in hp, required to overcome aerodynamic drag for an automobile moving at
(a) 25 miles per hour,
(b) 70 miles per hour.
Assume Cd=0.28,
A= 25ft2
and p=0.075Ib/ft2
Answer:
Explanation:
a)
Given that:
V = 25 mi/hr
To ft/sec, we have:
[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{110}{3} ft/s[/tex]
[tex]\rho = 0.075 \ lb/ft^3[/tex]
[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]
[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]
[tex]C_d = 0.28[/tex]
A = 25ft²
Recall that:
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]
[tex]F_d =10.967 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]
For 70 miles per hour, we have:
[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]
[tex]V = \dfrac{308}{3} ft/s[/tex]
The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]
[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]
[tex]F_d =85.99 \ lbf[/tex]
[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]
At 20 ◦C a copper wire has a resistance of 4×10−3 Ω and a temperature coefficient of resistivity of 3.9×10−3 (C◦)−1, its resistance at 100 ◦C is
A.
52.5 × 10-3 Ω
B.
5.25 × 10-3 Ω
C.
5.25 × 10-4 Ω
D.
5.25 × 10-2 Ω
E.
25.5 × 10-3 Ω
Answer:
[tex]R _{t} = R _{0}( \alpha t + 1) \\ = 4 \times {10}^{ - 3} (3.9 \times {10}^{ - 3} \times 20 + 1) \\ = 4 \times {10}^{ - 3} (1.078) \\ = 4.312 \times {10}^{ - 3} \: Ω[/tex]
A pressure sensor was used to measure the unsteady pressure in a cylinder. The sensor output was acquired for 15 seconds at a rate of 202 Hz and spectral analysis was performed using FFT. What is the maximum frequency that will be displayed on the power spectrum plot
Answer:
Maximum frequency on power spectrum plot = 101 Hz
Explanation:
Given:
Time taken for output = 15 seconds
Frequency rate = 202 Hz
Find:
Maximum frequency on power spectrum plot
Computation:
Maximum frequency = Given frequency rate / 2
Maximum frequency on power spectrum plot = Frequency rate / 2
Maximum frequency on power spectrum plot = 202 / 2
Maximum frequency on power spectrum plot = 101 Hz
how to calculate sound of an echo
by an echo meter
please flw me and thank my answers
#Genius kudi
Find the ratio of speeds of a proton and an alpha particle accelerated through the same voltage, assuming nonrelativistic final speeds. Take the mass of the alpha particle to be 6.64 ✕ 10−27 kg.
Answer:
The required ratio is 1.99.
Explanation:
We need to find the atio of speeds of a proton and an alpha particle accelerated through the same voltage.
We know that,
[tex]eV=\dfrac{1}{2}mv^2[/tex]
The LHS for both proton and an alpha particle is the same.
So,
[tex]\dfrac{v_p}{v_a}=\sqrt{\dfrac{m_a}{m_p}} \\\\\dfrac{v_p}{v_a}=\sqrt{\dfrac{6.64\times 10^{-27}}{1.67\times 10^{-27}}} \\\\=1.99[/tex]
So, the ratio of the speeds of a proton and an alpha particle is equal to 1.99.
B. Complete the lists:
Things that I must do for my family
Things I must never do to my family
1.
2.
2.
3.
3.
4.
5.
5.
Answer:
Things you should do for your family
help your parentstreat them kindlylisten and obey themappreciate them for anything they do for you talk softlythings you shouldn't
backanswering them Disobey And anything that's harsh or make it parents sadWhat is the main way in which heat transfer occurs in liquids and gases?
Which statement describes an action-reaction pair?
O A. You push on a car, and the car pushes back on you.
B. A book pushes down on a table, and the table pushes down on the
Earth.
C. The Moon pulls on Earth, and Earth pulls on the Sun.
D. You push down on your shoe, and Earth's gravity pulls down on the
shoe.
Answer:
A
Explanation:
a pex
How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?
Answer:
Power = 70 W
Explanation:
Given that,
Force, F = 70 N
Height, h = 5 m
Time, t = 5 s
We need to find the power of the object. We know that,
Power = work done/time
Put all the values,
[tex]P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W[/tex]
So, the required power is 70 W.
According to Newton’s law of universal gravitation, which statements are true?
PLEASE HELP ME WITH THIS ONE QUESTION
Given the atomic mass of Boron-9 is 9.0133288 u, what is the nuclear binding energy of Boron-9? (Mproton = 1.0078251, Mneutron = 1.0086649, c^2 = 931.5 eV/u)
A) 59 eV
B) 58 eV
C) 57 eV
D) 56 eV
Answer:
a. 59 ev. helpful answer
A ball is thrown vertically upward at 24.0 ms can reach a height of 28.8m ( neglecting air resistance).The speed,in m/s,when it is halfway to its highest point is (using g= 10 ms ^2)
Answer:
The answer is "[tex]16.79\ \frac{m}{s}[/tex]"
Explanation:
In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]
Using formula:
[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]
[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]
why does a spherometer have three legs?
spherometer is a device used to measure curved in surface
it have 3 legs which form equivalent triangle.
geometry says that 3 point determine a plane that's why it have 3 legs
Calculate the current flowing when the voltage across is 35V and the resistance is 7ohms.
Explanation:
V= IR
35=I×7
I=35/7
I=5amperes
pls give brainliest
A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute
Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
what is the difference between VELOCITY and SPEED?
Answer:
Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object's movement. Put another way, speed is a scalar value, while velocity is a vector. ... In its simplest form, average velocity is calculated by dividing change in position (Δr) by change in time (Δt).
Explanation:
A tennis ball of mass of 0.06 kg is initially traveling at an angle of 47o to the horizontal at a speed of 45 m/s. It then was shot by the tennis player and return horizontally at a speed of 35 m/s. Find the impulse delivered to the ball.
Answer:
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
Explanation:
By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:
[tex]Imp = m\cdot (\vec v_{f} - \vec v_{o})[/tex] (1)
Where:
[tex]m[/tex] - Mass of the ball, in kilograms.
[tex]\vec v_{o}[/tex] - Vector of the initial velocity, in meters per second.
[tex]\vec v_{f}[/tex] - Vector of the final velocity, in meters per second.
[tex]Imp[/tex] - Impulse, in meters per second.
If we know that [tex]m = 0.06\,kg[/tex], [tex]\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})[/tex] and [tex]\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)[/tex], then the impulse delivered to the ball is:
[tex]Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right][/tex]
[tex]Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
A 35 kg child slides down a playground slide at a constant speed. The slide has a height of 3.8 m and is 8.0 m long. Find the magnitude of the kinetic friction force acting on the child.
Answer:
The magnitude of the kinetic frictional force acting on the child is 162.93 N
Explanation:
Given;
mass of the child, m = 35 kg
height of the slide, h = 3.8 m
length of the slide, d = 8.0 m
The change in thermal energy associated with the kinetic frictional force is calculated as follows;
[tex]\Delta E_{th} + \Delta K.E + \Delta U = 0\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i) =0\\\\since \ the \ speed \ is \ constant, \ v_f = v_i \ and \ \Delta K.E = 0\\\\Also, \ final \ height \ , h _f= 0\\\\\Delta E_{th} - mgh_i = 0\\\\\Delta E_{th} = mgh_i\\\\\Delta E_{th} = 35 \times9.8 \times 3.8\\\\\Delta E_{th} = 1303.4 \ J[/tex]
The magnitude of the kinetic frictional force that produced this thermal energy is calculated from the work done by frictional force;
[tex]\Delta E_{th} = F \times d\\\\F = \frac{\Delta E_{th} }{d} \\\\F = \frac{1303.4}{8} \\\\F = 162.93 \ N[/tex]
Therefore, the magnitude of the kinetic frictional force acting on the child is 162.93 N
With neat circuit diagrams where possible show any two (2) ways of direct current motor
excitations,
