Answer:
F = 118 N
Explanation:
Assume Ann and Bob lift at their respective ends of the table
Sum moments about Bob's position to zero.
Let F be Ann's upward force
F[2.25] - 18.5(9.80)[2.25 / 2] - 8.33(9.80)[0.750] = 0
F = 117.86133333...
Given:
Mass of table, [tex]m_t = 18.5 \ kg[/tex]Mass of box, [tex]m_b = 8.33 \ kg[/tex]Length of table, [tex]2.25 \ m[/tex]Length of box, [tex]0.75 \ m[/tex]The weight of table will be:
→ [tex]W_t = m_t g[/tex]
[tex]= 18.5\times 9.8[/tex]
[tex]= 181.3 \ N[/tex]
Now,
→ [tex]\sum M_A_{nn} = -81.634\times 0.750-181.3\times 1.125+R_{bob}\times 1.125[/tex]
or,
→ [tex]R_{bob} = \frac{61.2255+203.9625}{2.25}[/tex]
[tex]= \frac{265.188}{2.25}[/tex]
[tex]= 117.9 \ N[/tex]
Thus the above answer is right.
Learn more about force here:
https://brainly.com/question/19427453
gAn optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be
Answer:
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explanation:
The complete question is as follows:
An optical engineer needs to ensure that the bright fringes from a double-slit are 15.7 mm apart on a detector that is 1.70m from the slits. If the slits are illuminated with coherent light of wavelength 633 nm, how far apart should the slits be?
The answer can be given by using the formula derived from Young's Double Slit Experiment:
[tex]y = \frac{\lambda L}{d}\\\\d =\frac{\lambda L}{y}\\\\[/tex]
where,
d = slit separation = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = distance from screen (detector) = 1.7 m
y = distance between bright fringes = 15.7 mm = 0.0157 m
Therefore,
[tex]d = \frac{(6.33\ x\ 10^{-7}\ m)(1.7\ m)}{0.0157\ m}\\\\[/tex]
d = 68.5 x 10⁻⁶ m = 68.5 μm
Explain why it takes much more effort to stop a freight train compared with a car?
Answer:
Train wheels and rails are both made of steel, and the steel-steel friction coefficient is around 0.25. As a result, the stopping time and distance will be three to four times that of a car.
what's impulse of
force
Answer:
The impulse experienced by the object equals the change in momentum of the object. In equation form, F.t = m. Δv. In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum.
The upward normal force exerted by the floor is 710 N on an elevator passenger who weighs 720 N . You may want to review (Pages 107 - 110) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Weighing yourself in an elevator. Part A What is the reaction force to the upward normal force exerted by the floor
Answer:
If the person is to remain the floor the reaction force will be equal to the normal force exerted by the floor.
F(normal) - F(reaction) = 0
That means the person is not moving with respect to the elevator.
Expanding the applied forces we have:
Fw - Fn = 720 - 710 = 10 N where the positive direction is chosen as down
Fw is the weight of the person and Fn the force exerted on the person by the elevator,
The acceleration of the person the becomes F = m a = m * 10 N and will be downward agreeing with our choice of coordinate axes.
Two people, who have the same mass, throw two different objects at the same velocity. If the first object is heavier than the second, compare the velocities gained by the two people as a result of recoil.
a. The first person will gain more velocity as a result of recoll.
b. The second person will gain more velocity as a result of recoll.
c. Both people will gain the same velocity as a result of recoll.
d. The velocity of both people will be zero as a result of recoil
Answer:
The first person will gain more velocity as a result of recoil.
Explanation:
Let us recall that from Newton's third law of motion, action and reaction are equation and opposite. A consequence of this law is the proposition that ''momentum can neither be created nor destroyed.''
Hence, when two people who have the same mass, throw two different objects at the same velocity but the first object is heavier than the second, the first object possesses greater momentum than the second object hence the first person will gain more velocity as a result of recoil.
Catching a wave, a 77-kg surfer starts with a speed of 1.3 m>s, drops through a height of 1.65 m, and ends with a speed of 8.2 m>s. How much nonconservative work was done on the surfer
Answer:
W = 2523.67 J
Explanation:
Given that,
The mass of surfer, m = 77 kg
He starts with a speed of 1.3 m/s
It drops through a height of 1.65 m and ends with a speed of 8.2 m/s.
We know that, the work done is equal to the change in kinetic energy. So,
[tex]W=\dfrac{1}{2}m(v_2^2-v_1^2)\\\\=\dfrac{1}{2}\times 77\times (8.2^2-1.3^2)\\W=2523.67\ J[/tex]
So, the required work done is equal to 2523.67 J.
describe the movement of the man when the resultant horizontal force is 0 N
can anyone help in both questions please
Answer:
Force A newton Law first law
F = M.A which Force in 0 N as you Questions Above
Force B
Newton Law 3
Action = -Reaction
Hope you can explain this formula as you want to scribe to explaining
An astronaut weighs 202 lb. What is his weight in newtons?
A vehicle is used to transport material down a straight aisle. The max acceleration of the vehicle is 1 m/s/s and the max speed of the vehicle is 5m/s. The vehicle starts at the beginning of the aisle. How long does it take to move down the aisle and come to a stop at the other end if:
a) the aisle is 100 meters long?
b) the aisle is 9 meters long?
Answer:
(a) 14.14 s
(b) 4.24 s
Explanation:
maximum acceleration, a = 1 m/s^2
maximum speed, v = 5 m/s
initial speed, u = 0 m/s
(a) distance, s = 100 m
Let the time is t.
Use second equation of motion
[tex]s = u t 0.5 at^2\\\\100 = 0 + 0.5 \times 1 \times t^2\\\\t = 14.14 s[/tex]
(b) distance, s = 9 m
Let the time is t'.
Use second equation of motion
[tex]s = u t + 0.5 at^2\\\\9= 0 + 0.5 \times 1 \times t'^2\\\\t' = 4.24 s[/tex]
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the x-axis does the third piece move
Answer:
Explanation:
There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.
momentum of first piece = .320 kg x 2 m/s
= 0.64 kg m/s along x -axis.
momentum of second piece = .355 kg x 1.5 m/s
= 0.5325 kg m/s along y- axis .
Let the velocity of third piece be v and it is making angle of θ with x -axis .
Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ
vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ
For making total momentum in the plane zero
.1 v cosθ = 0.64 kg m/s
.1 v sinθ = 0.5325 kg m/s
Dividing
Tanθ = .5325 / .64 = .83
θ = 40⁰.
The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.
Answer:
8.3 m/s, 2196 degree from + X axis
Explanation:
m = 320 g , u = 2 m/s along X axis
m' = 355 g, u' = 1.5 m/s along Y axis
m'' = 100 g, u'' = v
Let the speed of the third piece is v makes an angle A from the X axis.
use conservation of momentum along X axis
0 = 320 x 2 + 100 x v cos A
v cos A = - 6.4 ..... (1)
Use conservation of momentum along Y axis
0 = 355 x 1.5 + 100 x v sin A
v sinA = - 5.3 ... (2)
Squaring and adding
[tex]v^2 = (-6.4)^2 +(-5.3)^2\\\\v= 8.3 m/s[/tex]
The angle is given by
[tex]tan A = \frac{-5.3}{-6.4}\\\\A = 219.6 degree[/tex] from + X axis
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. A
Complete Question
The barometer of a mountain hiker reads 980 mbars at the beginning of a hiking trip and 790 mbars at the end. Neglecting the effect of altitude on local gravitational acceleration, determine the vertical distance climbed. Assume an average air density of 1.20kg/m^2
Answer:
[tex]h=1614m[/tex]
Explanation:
From the question we are told that:
Initial Pressure [tex]P_1=980mbar=>98000Pa[/tex]
Final Pressure [tex]P_2=790mbar=>79000Pa[/tex]
Density [tex]\rho=1.20kg/m^2[/tex]
Generally the equation for Height climbed is mathematically given by
[tex]h=\frac{P_1-P_2}{\rho*g}[/tex]
[tex]h=\frac{P_1-P_2}{1.20*9.81}[/tex]
[tex]h=1614m[/tex]
A police car travels towards a stationary observer at a speed of 15m/s. the siren on the car emits a sound of frequency 250Hz. Calculate the observer frequency. the speed of sound is 340m/s
Observer Frequency = sound frequency x ( speed of sound / speed of sound - speed of car)
= 250 x (340/( 340-15))
= 261.54 Hz
question 1+1677-789909
Answer:
your answer is -788231
Explanation:
1+1677=1678
1678-789909=-788231
A CD is spinning on a CD player. In 12 radians, the cd has reached an angular speed of 17 r a d s by accelerating with a constant acceleration of 3 r a d s 2 . What was the initial angular speed of the CD
Answer:
The initial angular speed of the CD is equal to 14.73 rad/s.
Explanation:
Given that,
Angular displacement, [tex]\theta=12\ rad[/tex]
Final angular speed, [tex]\omega_f=17\ rad/s[/tex]
The acceleration of the CD,[tex]\alpha =3\ rad/s^2[/tex]
We need to find the initial angular speed of the CD. Using third equation of kinematics to find it such that,
[tex]\omega_f^2=\omega_i^2+2\alpha \theta\\\\\omega_i^2=\omega_f^2-2\alpha \theta[/tex]
Put all the values,
[tex]\omega_i^2=(17)^2-2\times 3\times 12\\\\\omega_i=\sqrt{217}\\\\\omega_i=14.73\ rad/s[/tex]
So, the initial angular speed of the CD is equal to 14.73 rad/s.
a fixed mass of gas occupies a volume of 1000 CM3 at 0 degree celsius if it is heated at constant pressure of 100 degree celsius calculate the new volume
Answer:
P V = N R T ideal gas equation
V1 = k * T1 if P is constant and also N and R will be constant
V2 = k * T2 where k is some constant
Or V2 = (T2 / T1) * V1 also known as "Charles Law" for expansion at
constant pressure
V2 = (373 / 273) * 1000 cm^3 = 1366 cm^3 where T is absolute temperature
A car of mass 2100 kg collides with a motorcycle of mass 290 kg. After the collision, the car and motorcycle stick and slide together. The car's velocity just before the collision was<30,-10>m/s , and that of the motorcycle was <10,10>m/s. Determine the velocity of the stuck-together car and motorcycle just after the collision.
Answer:
V = 29.49 m/s
Explanation:
Given that,
The mass of a car,[tex]m_c=2100\ kg[/tex]
The mass of a motorcycle, [tex]m_m=290\ kg[/tex]
The initial velocity of the car,[tex]v_c=30i-10j[/tex]
[tex]|v_c|=\sqrt{30^2+(-10)^2} =31.62\ m/s[/tex]
The initial velocity of the motorcycle,[tex]v_m=10i+10j[/tex]
[tex]|v_m|=\sqrt{10^2+10^2} =14.14\ m/s[/tex]
As they stick together. Let V is the speed. So, using the conservation of momentum,
[tex]m_cv_c+m_mv_m=(m_c+m_m)V\\\\V=\dfrac{m_cv_c+m_mv_m}{(m_c+m_m)}\\\\V=\dfrac{2100\times 31.62+290\times 14.14}{(2100+290)}\\\\V=29.49\ m/s[/tex]
So, the velocity of the stuck together car and the motorcycle after the collision is 29.49 m/s.
What process provides the sun with its energy
Answer:
nuclear fusion
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom. During the fusion process, radiant energy is released.Answer:
nuclear fusion
Explanation:
The sun generates energy from a process called nuclear fusion. During nuclear fusion, the high pressure and temperature in the sun's core cause nuclei to separate from their electrons. Hydrogen nuclei fuse to form one helium atom.
A mass of 5 kg stretches a spring 20 cm. The mass is acted on by an external force of 10 sint4 N (newtons) and moves in a medium that imparts a viscous force of 4 N when the speed of the mass is 2 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 7 cm/s.
Required:
Formulate the initial value problem describing the motion of the mass. Assume that g = 9.8 m/s^2.
Answer:Answer:
Initial value problem is:
u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0
u'(0) = 0.03m/s
Explanation:
The directions of Fd(t*) and U'(t*) are not specified in the question, so we'll take Fd(t*) to be negative and U'(t*) to be positive. This is due to the fact that the damping factor acts in the direction opposite the direction of the motion of the mass.
M = 5kg; L= 10cm or 0.1m;
F(t) = 10 sin(t/2) N ; Fd(t*) = - 2N
U'(t*) = 4cm/s or 0.04m/s
u(0) = 0
u'(0) = 3cm/s or 0.03m/s
Now, we know that W = KL.
Where K is the spring constant.
And L is the length of extension.
So, k = W/L
W= mg = 5 x 9.81 = 49.05N
So,k = 49.05/0.1 = 490.5kg/s^(2)
Now from spring damping, we know that; Fd(t*) = - γu'(t*)
Where,γ = damping coefficient
So, γ = - Fd(t*)/u'(t*)
So, γ = 2/0.04 = 50 Ns/m
Therefore, the initial value problem which describes the motion of the mass is;
5u'' + 50u' + 490u = (10 sin(t/2) N
Divide each term by 5 to give;
u'' + 10u' + 98u = (2 sin(t/2) N for u(0) = 0
u'(0) = 0.03m/s
Explanation:
your neighbour is throttling his recent bought motorbike to show off. the sound intensity measured at your window 16m away is 0.25W/m^2. what is the sound intensity level in dB at your friend's house, a distance of 28m away from the noisy bike?
Answer:
hi your pinterest I'd or Twitter I'd pls.
Hannah wants to create a record keeping system to track the inventory needed to efficiently run her lawn and landscape business, such as spare parts, gas cans, string trimmers, etc. Her crew manager will also be using the system. Hannah is considering whether to use Excel or Access. Which one of the following is NOT a benefit of using Access?
a. More data storage
b. Multiuser capability
c. Easier setup
d. Additional reporting features
Answer:
c). Easier setup
Explanation:
As per the question, 'easier setup' cannot be characterized as the advantage of using Access because it comprises of plenty of steps that must be followed in the sequential order to establishing a database or carrying transactions based on time. However, there are plenty of advantages of using Microsoft access like 'enhanced and increased storage of data,' 'hassle free database systems,' 'easy importing of data,' 'highly economical,' 'capability to allow multiple users,' 'extra features for reporting,' and much more. Hence, option c is the correct answer.
Physics question on picture
Answer:
B. according to Newton's Third Law of Motion, the force of the Moon on the Earth and the force of the Earth on the Moon are equal in magnitude and opposite in direction
How does an airpump work?
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 22 ft/s2. What is the distance covered before the car comes to a stop
Answer:
The correct solution is "122.2211".
Explanation:
Given:
deceleration,
a = 22 ft/sec²
Initial velocity,
[tex]V_i=50 \ m/h[/tex]
Now,
[tex]V_i=50 \ m/h\times 5280 \ ft/m\times hr/3600 \ s[/tex]
[tex]=73.333 \ ft/sec[/tex]
Now,
Final velocity,
[tex]V_f=0[/tex]
Initial velocity,
[tex]V_{initial} = 73.333 \ ft/sec[/tex]
hence,
⇒ [tex]V_f^2=V_i^2+2aD[/tex]
By putting the values, we get
[tex]0=(73.333)^2+2\times( -22) D[/tex]
[tex]44D=(73.333)^2[/tex]
[tex]D=\frac{(73.333)^2}{44}[/tex]
[tex]=122.2211[/tex]
An oscillating dipole antenna 1.73 m long with a maximum 36.0 mV potential creates a 500 Hz electromagnetic wave. (a) What is the maximum electric field strength created
Answer:
[tex]E_0=0.021v/m[/tex]
Explanation:
From the question we are told that:
Length [tex]l=1.73m[/tex]
Voltage [tex]V=36.0mV[/tex]
Frequency [tex]F=500Hz[/tex]
Generally the equation for maximum Electric Field is mathematically given by
[tex]E_0=\frac{\triangle}{r}[/tex]
[tex]E_0=\frac{36*10^{-3}}{1.73}[/tex]
[tex]E_0=0.021v/m[/tex]
Which car has the greatest potential energy? What type of potential energy does the
car have? What characteristics or properties does potential energy depend on? Which
cars have the least potential energy? Explain.
Explanation:
*car A has the greatest PE
*gravitational potential energy
*mass and height
*car D has the least PE
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.
Answer:
C
Explanation:
Similarity
What is the electric field strength between two parallel conducting plates separated by 10 cm and having a potential difference between them of 2000 V?
a.
2000 V/m
b.
200 V/m
c.
20 kV/m
d.
200000 V/m
Answer:
• Potential Difference (V) = 2000 V
• Distance b/w the two parallel plates (d) = 10 cm = 10/100 = 1/10 = 0.1 m
• Electric field (E) = ?
[tex]\implies V = E.d[/tex]
[tex]\implies E = \dfrac{V}{d} [/tex]
[tex]\implies E = \dfrac{2000}{0.1} [/tex]
[tex]\implies E = \dfrac{2000}{ {10}^{ - 1} } [/tex]
[tex]\implies E = 2000 \times {10}^{1} [/tex]
[tex]\implies\bf E = 20000 \:V/m[/tex]
[tex]\implies\bf E = 20\:kV/m[/tex]
Hence, option C) the correct answer.
An inductive circuit contains resistance of 20 ohm and an inductance of 20 H. If an ac voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be
A 0.0159
A 0.017
A 0.02
A 0.16
Answer:
answer : option (b) 0.016 amp
explanation : resistance of resistor , R = 10 Ω
inductance of inductor , X_LX
L
= 20H
voltage of AC circuit , V = 120volts
frequency, ff =60Hz
so, angular frequency, \omega=2\pi fω=2πf = 2 × π × 60 = 120π rad/s
now, current , i=\frac{V}{\sqrt{R^2+\omega^2L^2}}i=
R
2
+ω
2
L
2
V
= 120/√{10² + (120π)² × 20²}
= 120/√{100 + 14400π² × 400}
after solving this we get, i = 0.016 amp
15. Calculate the resistive forces acting on a sports car if it is travelling at a steady speed of 25 m/s when the engine is providing 200 kW. gnt V 10X50 - soot Grade 9
Answer:
8000 N
Explanation:
Applying
P = F×V.............. Equation 1
Where P = Power of the sport car, F = resistive force acting on the sport car, V = Speed of the sport car.
make F the subject of the equation
F = P/V............ Equation 2
From the question,
Given: P = 200 kW = 200000 W, V = 25 m/s
Substitute these values into equation 2
F = 200000/25
F = 8000 N
During the Moment of Inertia experiment, a group of students decided to mount a solid sphere on the top of the horizontal disk. The sphere and disk have the same mass, M, and radius, R. They recorded the values for the initial and final positions for the falling mass, and the height above the floor when falling mass is at its lowest position. These values are respectively 90cm, 68.5 cm, and 23.5cm. The radius of the pulley, r, was 1/10 of the radius of the sphere. Furthermore, the mass of the disk, M, was 1.4kg, what was the mass, m, for the falling mass
Answer:
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