Answer:
F = force
f = friction
u = coefficient of friction
R = normal reaction force
a = Acceleration
m = mass of block
g = gravity
f = uR
F = Ma
Say the block is moving to the right.
The 146N force thus acts to the right, and the friction force to the left, since it resists movement.
The 146N force acts to the right, but the horizontal component of it is 146 cos 50 = 93.84: So this is the force to the right.
Since F = uR and we're trying to find u, we need both F and R. R is easy to get since it is just m x g. This is in fact already given as the weight 350N. So R = 350.
The block is moving at a constant speed, so the force to the right must = the force to the left.
F = ma, so 93.84 - f = (350/g) x 0
This means f must be 93.84 also.
so we have f = uR,
93.84 = u x 350
so u = 0.268 or
0.27 to 2dp.
Hope you understand this.
Explanation:
The coefficient of friction is 0.26 if the 46N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N). Mathematically it is defined as the product of the coefficient of friction and normal reaction.
We know:
f = uR
Where f is the friction force and u is the coefficient of friction, R is the normal reaction force.
The horizontal component of 146N is:
F' = 146cos50°
F' = 93.84 N
Since F = mass×acceleration
Because the block is traveling at a constant speed, the right-hand force must equal the left-hand force.
F' = f = 93.84 N
93.84 = u x 350 (R = 350N)
u = 0.26
Thus, the coefficient of friction is 0.26 if the 46N force is needed to pull a 350 N block across a horizontal surface at a constant speed by a rope making an angle of 50 degrees with the floor.
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A 1250-kg car moves at 20.0 m/s. How much work must be done on the car to increase its speed to 30.0 m/s.
Answer:
312,497.5Joules
Explanation:
Work done = force × distance
W = FS
Get the force
F = ma
F = 1250×9.8
F = 12250N
Get the distances using the equation of motion
v² = u² +2gS
30² =20²+2(9.8)S
900 =400+19.6S
900-400 =19.6S
500 = 19.6S
S = 500/19.6
S = 25.51m
Get the work done
Work done = 12250×25.51
Workdone = 312,497.5Joules
Answer:
3.13 x 10^5 J
Explanation:
The work done increases the car's kinetic energy so that its new speed is 30 m/s. So, we set t, the net work required to increase the car's speed equal to the change in the car's kinetic energy.
Wnet = △K = Kf - Ki = 1/2mvf^2 - 1/2mvi^2 = 1/2m(vf^2-vi^2) = 1/2(1250kg)((30m/s)^2 - (20m/s)^2) = 3.13x10^5 J
15 points! Answer FAST!
When heat is transferred, it flows_____.
A. From the warmer object to the cooler object only if the cooler object is an insulator
B. From the cooler object to the warmer object regardless of the materials
C. From the warmer object to the cooler object regardless of the materials
D. From the warmer object to the cooler object only if the warmer object is a conductor.
Answer:heat brings it up then down
Explanation:
If force c is 30n and force d us 20 N what is the size of the horizontal force
What is calibration of thermometer?
Answer:
HACCP based food safety programs require accurate record keeping to be successful. Temperature is often the parameter of interest when monitoring a critical control point (CCP). ... To be considered accurate, a thermometer must be calibrated to measure within +/- 2° F (1.1° C) of the actual temperature.
Explanation:
What does the law of conservation of energy state? *
1. Total energy before transfer is equal to total energy after transfer
2. Total energy before transfer is less than total energy after transfer
3. Total energy before transfer is more than total energy after transfer
Answer:
total energy before transfer is equal to total energy after transfer
Answer:
1 is the correct one Answer
Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 0.2 in and a period of 0.1 s.
Answer:
The maximum velocity and maximum acceleration of the particle are approximately 12.566 inches per second and 789.568 inches per square second.
Explanation:
The equation of motion for the position of a particle experimentating a simple harmonic motion ([tex]x(t)[/tex]), measured in inches, is described by the following expression:
[tex]x(t) = A\cdot \cos \left(\frac{2\pi\cdot t}{T} +\phi\right)[/tex] (1)
Where:
[tex]A[/tex] - Amplitude, measured in inches.
[tex]t[/tex] - Time, measured in seconds.
[tex]T[/tex] - Period, measured in seconds.
[tex]\phi[/tex] - Phase, measured in radians.
Then, we obtain the formulas for the velocity and acceleration of the particle by differentiating (1):
[tex]v(t) = -\frac{2\pi\cdot A}{T}\cdot \sin \left(\frac{2\pi\cdot t}{T}+\phi \right)[/tex] (2)
[tex]a(t) = -\left(\frac{2\pi}{T} \right)^{2}\cdot A\cdot \cos \left(\frac{2\pi\cdot t}{T}+\phi \right)[/tex] (3)
From (2) and (3) we find that maximum velocity ([tex]v_{max}[/tex]), measured in inches per second, and maximum acceleration ([tex]a_{max}[/tex]), measured in inches per square second, are defined by the following formulas:
[tex]v_{max} = \frac{2\pi\cdot A}{T}[/tex] (4)
[tex]a_{max} = \left(\frac{2\pi}{T} \right)^{2}\cdot A[/tex] (5)
If we know that [tex]A = 0.2\,in[/tex] and [tex]T = 0.1\,s[/tex], then the maximum velocity and maximum acceleration of the particle are, respectively:
[tex]v_{max} = \frac{2\pi\cdot (0.2\,in)}{0.1\,s}[/tex]
[tex]v_{max} \approx 12.566\,\frac{in}{s}[/tex]
[tex]a_{max} = \left(\frac{2\pi}{0.1\,s} \right)^{2}\cdot (0.2\,in)[/tex]
[tex]a_{max} \approx 789.568\,\frac{in}{s^{2}}[/tex]
The maximum velocity and maximum acceleration of the particle are approximately 12.566 inches per second and 789.568 inches per square second.
When an object executes a to and fro motion is said to be a simple harmonic motion. The maximum velocity and maximum acceleration of the particle will be 12.566 in/s and 789.568 in/s².
What is simple harmonic motion?When an object executes a to and fro motion in the definite plane when it is tied with the string. The type of motion will be the simple harmonic motion.
Simple harmonic motion is a form of periodic motion in mechanics and physics in which the restoring force on the moving item is directly proportional to the size of the object.
The value of maximum velocity is given by the formula in the SHM motion is
[tex]\rm V_{max = \frac{2\pi A}{T}[/tex]
[tex]\rm V_{max} = \frac{2\times 3.14 \times 0.2 }{0.1}[/tex]
[tex]\rm V_{max = 12.566 \;in/sec[/tex]
The value of maximum acceleration is given by the formula in the SHM motion is
[tex]\rm a_{max}=(\frac{2\pi}{T} )^2.A\\\\ \rm a_{max}=(\frac{2\times 3.14 }{0.1} )^2 \times 0.2 \\\\ \rm a_{max}=789.568\; in/sec^2[/tex]
Hence the maximum velocity and maximum acceleration of the particle will be 12.566 in/s and 789.568 in/s²
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A 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.03 kg shooter marble moving to the left at 0.18 m/s. After the collision, the smaller marble moves to the left at 0.315 m/s. What is the velocity of the 0.03 kg marble after the collision
Answer:
0.135m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = m1v1 +m2v2
m1 and m2 are the masses
u1 and u2 are the initial velocities
v1 and v2 are thee final velocities
Substitute the given parameters and get v2
0.015(0.225)+0.03(0.18) = 0.015(0.315)+0.03v2
0.003375+0.0054 = 0.004725 + 0.03v2
0.008775 - 0.004725 = 0.03v2
0.00405 = 0.03v2
v2 = 0.00405/0.03
v2 = 0.135m/s
Hence the final velocity of the 0.03kg marble after collision is 0.135m/s
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
what kind of mathematical relationship do you predict wll exist between velocity and number of filters
Answer:
[tex]PK_T = constant[/tex]
Explanation:
[tex]PK_T[/tex] = constant
terminal velocity is the maximum const. velocity to achieved by the partical when particle is free fall with the effect of gravity
so that here
eq of motion is
Fnet = mg - kv ................1
here a = 0
v = terminal velocity
0 = mg - kV
v = [tex]\frac{mg}{k}[/tex] m/s
A helicopter flies with an air speed of 175 km/h, heading south. The wind is blowing at 85 km/h to the east relative to the ground. Calculate the speed and direction of the helicopter.
Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
[tex]x = arctan (\frac{85}{175} )[/tex] = 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
A low rider truck accelerates from 0 m/s to 30.0 m/s with a displacement of 50m. What is the vehicle's acceleration?
a. 2.00m/s
b. 5.00m/s
c. 9.0m/s
d. 10.0m/s
Answer:
a = 9 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}^{2}=v_{o}^{2} +2*a*x[/tex]
where:
Vf = final velocity = 30 [m/s]
Vo = initial velocity = 0
a = acceleration [m/s²]
x = displacement = 50 [m]
Now replacing:
[tex]30^{2}=0 +2*a*50\\100*a=900\\a=9[m/s^{2} ][/tex]
What is the distance of separation between objects of masses 5.6 x 10 5 kg and 8.8 x 10 6 kg if the force of gravity between them is 440 N
Answer:
2.87m
Explanation:
Using the law of gravitation to solve this question
F = GMm/r²
G is the gravitational constant
M and m are the masses
r is the distance between the masses
Substitute the given values
G = 6.67×10^-11 m³/kgs²
M =8.8 x 10^6 kg
m = 5.6 x 10^5 kg
F =440N
400 = 6.67×10^-11×8.8 x 10^6 ×5.6 x 10^5/r²
400r² = 328.698×10
400r² = 3286.98
r² = 3286.98/400
r² = 8.21745
r = √8.21745
r = 2.87m
Hence the distance of separation is 2.87m
The distance of separation will be "0.86 m".
Given:
Mass,
[tex]m_1 = 5.6\times 10^5 \ kg[/tex][tex]m_2 = 8.8\times 10^6 \ kg[/tex]Force,
[tex]F = 440 \ N[/tex]As we know,
→ [tex]F = \frac{Gm_1 m_2}{r^2}[/tex]
By putting the values, we get
→ [tex]440=\frac{6.67\times 10^{-11}\times 5.6\times 10^5\times 8.8\times 10^6}{r^2}[/tex]
→ [tex]r^2 = \frac{6.67\times 5.6\times 8.8}{440}[/tex]
→ [tex]r^2 = 0.74704[/tex]
→ [tex]r = \sqrt{0.74704}[/tex]
→ [tex]= 0.86 \ m[/tex]
Thus above approach is correct.
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Which term is the product of force and distance? power work net force acceleration please help
Answer: WORK
Explanation:
Answer:
work
Explanation:
got it right on edge:p
Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are tripled and their separation is also tripled, what happens to the magnitude of the electrostatic force that each charge exerts on the other
Answer:
The force remains the same.
Explanation:
Let the magnitude of the forces in each case be F1 and F2 respectively.
The charges are Q1 and Q2.
The distance of separation between them is R.
Hence, for F1;
F1 = KQ1Q2/R^2
For F2:
F2 = K * 3Q1 * 3Q2/(3R)^2
F2 = 9KQ1Q2/9R^2
F2 =KQ1Q2/R^2
Hence F1=F2
The force is the same in both cases!
A 105 kg football player runs at 8.5 m/s and plows into the back of an 85 kg referee running at 3.5 m/s on the field causing the referee to fly forward at 6.0 m/s.
What must the momentum of the football player be after the collision?
Answer:
680 Kg.m/s
Explanation:
Mass of player; m_p = 105 kg.
Speed of player before Collision; v_pi = 8.5 m/s
Mass of referee; m_r = 85 kg
Speed of referee before collision; v_ri = 3.5 m/s
Speed of referee after collision; v_rf = 6 m/s
From conservation of momentum,
Initial momentum = final momentum
Thus;
(m_p × v_pi) + (m_r × v_ri) = (m_p × v_pf) + (m_r × v_rf)
Where (m_p × v_pf) is the momentum of the player after collision.
Thus, Plugging in the relevant values, we have;
(105 × 8.5) + (85 × 3.5) = (m_p × v_pf) + (85 × 6)
(m_p × v_pf) = (105 × 8.5) + (85 × 3.5) - (85 × 6)
(m_p × v_pf) = 680 Kg.m/s
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
h. If the mass of an object is 390 g and its volume is 50 cm?, then its density is ....... g/cm' or ....... kg/m”.
- The relative density of the object in part "h" is
Edit: nvm I'm an idiot I remembered this but thanks anyway
Answer: See explanation
Explanation:
The density would like be calculated by dividing the mass by volume. This will be:
Density = Mass / Volume
Density = 390 / 50
Density = 7.8g/cm
A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h; The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s2. What is the shortest time in which the driver of the car can complete the passing operation? (assuming that the driver of the car does not pay any attention to the speed limit
Answer:
The shortest time in which the driver can complete the operation is approximately 6.32456 seconds
Explanation:
The given parameters are;
The speed of the truck and the car = 35 mi/h ≈ 51.33 ft./s
Let "t" represent the time it takes the car to pass the truck by 40 ft., we have;
The distance covered by the truck = 35 mi/h × t
The distance covered by the car = 35 mi/h × t + 80 ft = 51.33
Let the distance over which the car accelerates = d
We have;
d = 51.33×t₁ + 1/2×5×t₁²
51.33·t + 80 - d = (51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)²
51.33 = 51.33 + 5·t₁ - 20·(t - t₁)
∴ 5·t₁ = 20·(t - t₁)
5·t₁ = 20·t - 20·t₁
25·t₁ = 20·t
t₁ = 4·t/5
(51.33 + 5·t₁)·(t - t₁) - 1/2·20·(t - t₁)² + 51.33×t₁ + 1/2×5×t₁² = 51.33·t + 80
We get;
(51.33 + 5·(4·t/5))·(t - (4·t/5)) - 1/2·20·(t - (4·t/5))² + 51.33×(4·t/5) + 1/2×5×(4·t/5)² = 51.33·t + 80
(51.33 + 4·t)·(t/5) - 2·t²/5 + 51.33 × (4·t/5) + 8·t²/5 = 51.33·t + 80
51.33×t/5 + 51.33×4·t/5 + 4·t²/5 - 2·t²/5 + 8·t²/5 = 51.33·t + 80
51.33·t + 10·t²/5 = 51.33·t + 80
2·t² = 51.33·t + 80 - 51.33·t = 80
t² = 80/2 = 40
t = √40 = 2·√10 ≈ 6.32456
The shortest time in which the driver can complete the operation is "t" ≈ 6.32456 seconds
A speedboat moving at 29.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.4 m/s2 by reducing the throttle. (a) How long does it take the boat to reach the buoy
Answer:
[tex]t =4.8sec[/tex]
Explanation:
From the question we are told that
Velocity of speed boat [tex]V=29.0m/s[/tex]
Distance to Marker [tex]d=100[/tex]
Acceleration of [tex]a=-3.4m/s^2[/tex]
Generally the Newtons 3rd motion equation is given as
[tex]v^2 = u^2 + 2 * a* s[/tex]
[tex]v^2 = 29^2 + 2 * -3.4* 100[/tex]
[tex]v = \sqrt{161}[/tex]
[tex]v=12.68m/s[/tex]
Generally the Newton's first equation of motion is given as
[tex]v = u + a*t[/tex]
[tex]12.68 = 29 -3.4*t[/tex]
[tex]12.68-29 = -3.4t[/tex]
[tex]-16.32 = -3.4t[/tex]
[tex]t =\frac{-16.32}{-3.4}[/tex]
[tex]t =4.8sec[/tex] .
prove principle of conversation of linear momentum in accordance to principle of classical relativity
Answer:
Cuz im batman hahahhahah
What do you think causes the changes in sunrise and sunset times over the course of the year?
Answer:
The combination of Earth's elliptical orbit and the tilt of its axis results in the Sun taking different paths across the sky at slightly different speeds each day. This gives us different sunrise and sunset times each day.
Explanation:
(Are these statements about the spring constant true or
false?
a. The spring constant is a measure of the stiffness of
the spring.
b. The spring constant tells you how many newtons of
force it takes to stretch the spring one meter.
c. If a spring stretches easily, it will have a high spring
constant.
d. The spring constant of a spring varies with the
amount of stretch or compression of the spring.
Answer:
true
true
false
true
pls follow me and thank you
How much extra water does a 147-lb concrete canoe displace compared to an ultra lightweight 36-lb Kevlar canoe of the same size carrying the same load
Answer:
1.7791 ft^3
Explanation:
The computation of the extra water is as follows;
Here we use the following formula
Volume of the water = ÷ Weight of concrete canoe ÷ value of the weight of the specific water
= 147 lb ÷ 62.4 lb /ft^3
= 2.356 ft^3
Now the volume of water occupied in ultra lightweight
= 36 lb ÷ 62.4 lb /ft^3
= 0.5769 ft^3
Now the water volume displaced is
= 2.356 - 0.5769
= 1.7791 ft^3
A 10 kg weight is suspended in the air by a strong cable. How much work is done, per unit time, in suspending the weight
Answer:
There is no work done.
Explanation:
Given the following data;
Mass = 10 kg
To find the work done?
In Physics, work done can be defined as the amount of energy transfered when an object or body is moved over a distance due to the action of an external force.
Mathematically, work done is given by the formula;
Work done = force * distance
[tex] W = F * d[/tex]
Where,
W is the work doneF represents the force acting on a body. d represents the distance covered by the body.However, the weight suspended in the air by a strong cable does not move or experience any form of displacement. Therefore, there is no work done.
A constant force of 20. newtons applied to a box causes it to move at a constant speed of 4.0 meters per second. How much work is done in the boxin 6.0 seconds?
1) 240 joules
2) 120 joules
3) 480 joules
4) 80. joules
The work done is 480 Joules.
Option 3 is correct.
The work done is computed as multiplication of force and displacement.
Given that, Force = 20 N
Displacement = 4 meter per second
For 6 second,
Displacement [tex]=4*6=24m[/tex]
[tex]Workdone=Force*displacement\\\\Workdone=20*24=480J[/tex]
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Sam is driving along the highway towards Savannah. He travels 150 m in 3 seconds. What is his speed
Answer:
50m/s
Explanation:
Speed = distance/ time
Distance = 150m
Time = 3 secs
Therefore Speed =150/3
Speed = 50 m/s
Convert 41.3 kilocalories into joules.
Explanation:
41.3 kilocalories = 172 799.2 joules
[tex]hope \: it \: will \: help[/tex]
2- A car on a straight highway goes in the positive direction for 8 km and then backs up for 3.6 km. What are the distance and displacement covered by the train?
Answer:
11.6km
4.4km in the negative direction
Explanation:
Distance is the total length of path covered and traveled by a body.
So, for this car on a straight line;
Total distance = 8km + 3.6km = 11.6km
Displacement is the distance traveled along a path and the direction it takes.
It is a vector quantity with magnitude and directional attributes.
For this journey;
Displacement = 8km - 3.6km = 4.4km in the negative direction.
Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is released
Answer:
171.5m
Explanation:
The velocity of sound in water = 343m/s
Time taken = 1.00secs
using the formula to calculate the distance
2x = vt
x is the distance
v is the speed of sound
t is the time
x = vt/2
x = 343(1)/2
x = 171.5m
hence their separation 1.00 s after the second object is released is 171.5m
In a chemical reaction, the mass of the reactants is 13,3 grams. Which of the
following could represent the mass of the products?
Answer:
my assuytyyhyyyyyy gftdrrdtiifyb tvhyvth rv b yy