Answer:
(A - B).A = -2.91
Explanation:
First, let's define the sum and dot product of vectors.
For two vectors V = (x₁, y₁) and W = (x₂, y₂) we have:
sum (or subtraction):
V + W = (x₁, y₁) + (x₂, y₂) = (x₁ + x₂, y₁ + y₂)
dot product:
V.W = (x₁, y₁).(x₂, y₂) = x₁*x₂ + y₁*y₂
Here remember the notation:
V = x₁*i + y₁*j = (x₁, y₁)
Now let's solve our problem, we have:
A = (3.7, 1.0)
B = (3.0, 6.5)
Then:
(A - B).A = ( (3.7, 1.0) - (3.0, 6.5) ).(3.7, 1.0)
= (3.7 - 3.0, 1.0 - 6.5).(3.7, 1.0)
= (0.7, -5.5).(3.7, 1.0) = (0.7*3.7) + (-5.5)*(1.0) = -2.91
The colors that make up white light are called what?
Answer:
The ROYGBIV
Explanation:
R - red
O - orange
Y - yellow
G - green
B - blue
I - indigo
V - violet
A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the bucket is gradually increased by the addition of sand. At some point, the bucket will accumulate enough sand to set the block in motion. The coefficients of static and kinetic friction are 0.60 and 0.50 respectively.
Required:
a. Determine the mass of sand in [kg], including the bucket, needed to start the block moving.
b. Find the blocks acceleration, in [m/s^2] up the plane?
Answer:
a). M = 20.392 kg
b). am = 0.56 [tex]m/s^2[/tex] (block), aM = 0.28 [tex]m/s^2[/tex] (bucket)
Explanation:
a). We got N = mg cos θ,
f = [tex]$\mu_s N$[/tex]
= [tex]$\mu_s mg \cos \theta$[/tex]
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + [tex]$\mu_s mg \cos \theta$[/tex] .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
[tex]$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$[/tex]
[tex]$M=2(m \sin \theta + \mu_s mg \cos \theta)$[/tex]
[tex]$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$[/tex]
M = 20.392 kg
b). [tex]$(h-x_m)+(h-x_M)+(h'+x_M)=l$[/tex] .............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
[tex]$-\ddot{x}-2\ddot x_M=0$[/tex]
[tex]$\ddot x_M=\frac{\ddot x_m}{2}$[/tex]
[tex]$a_M=\frac{a_m}{2}$[/tex] .....................(iv)
We got, N = mg cos θ
[tex]$f_K=\mu_K mg \cos \theta$[/tex]
∴ [tex]$T-(mg \sin \theta + f_K) = ma_m$[/tex]
[tex]$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$[/tex] ................(v)
Mg - 2T = M[tex]a_M[/tex]
[tex]$Mg-Ma_M=2T$[/tex]
[tex]$Mg-\frac{Ma_M}{2} = 2T$[/tex] (from equation (iv))
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}=T$[/tex] .....................(vi)
Putting (vi) in equation (v),
[tex]$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$[/tex]
[tex]$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$[/tex]
[tex]$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$[/tex]
[tex]$a_m= 0.56 \ m/s^2$[/tex]
Using equation (iv), we get,
[tex]a_M= 0.28 \ m/s^2[/tex]
an object is 70 um long and 47.66um wide. how long and wide is the object in km?
Answer:
length = 7*10^(-8)km
width = 4.666*10^(-8) km
Explanation:
We know that:
1 μm = 1*10^(-6) m
and
1km = 1*10^3 m
or
1m = 1*10^(-3) km
if we replace the meter in the first equation, we get:
1 μm = 1*10^(-6)*1*10^(-3) km
1 μm = 1*10^(-6 - 3)km
1 μm = 1*10^(-9)km
Now with this relationship we can transform our measures:
Length: 70 μm is 70 times 1*10^(-9)km, or:
L = 70*1*10^(-9)km = 7*10^(-8)km
And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:
W = 46.66*1*10^(-9)km = 4.666*10^(-8) km
A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.
Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?
Answer:
a) μ = 0.0136, b) F = 22.8 N
Explanation:
This exercise must be solved in parts. Let's start by using conservation of moment.
a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved
initial instant. Before the crash
p₀ = m v₀
final instant. After inelastic shock
p_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m + M} \ v_o[/tex]
We look for the speed of the block with the bullet inside
v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]
v = 0.448 m / s
Now we use the relationship between work and kinetic energy for the block with the bullet
in this journey the force that acts is the friction
W = ΔK
W = ½ (m + M) [tex]v_f^2[/tex] - ½ (m + M) v₀²
the final speed of the block is zero
the work between the friction force and the displacement is negative, because the friction always opposes the displacement
W = - fr x
we substitute
- fr x = 0 - ½ (m + M) vo²
fr = ½ (m + M) v₀² / x
the friction force is
fr = μ N
μ = fr / N
equilibrium condition
N - W = 0
N = W
N = (m + M) g
we substitute
μ = ½ v₀² / x g
we calculate
μ = ½ 0.448 ^ 2 / 0.75 9.8
μ = 0.0136
b) Let's use the relationship between work and the variation of the kinetic energy of the block
W = ΔK
initial block velocity is zero vo = 0
F x₁ = ½ M v² - 0
F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]
F = ½ 2.50 0.448² / 0.0110
F = 22.8 N
according to the law of conservation of vhange , what must always be true in a nuclear reaction?
Answer:
The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same
Explanation:
Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.
Two resistors, A and B, are connected in parallel across a 6.0-V battery. The current through B is found to be 2.0 A. When the two resistors are connected in series to the 6.0- V battery, a voltmeter connected across resistor A measures a voltage of 4.0 V. Find the resistances of A and B
Answer:
The resistance of A is 6 ohms and the resistance of B is 3 ohms
Explanation:
Step 1: For the first connection (parallel connection), the resistance of B will be calculated.
Note: in a parallel connection, the voltage through each resistor is the same.
[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]
Step 2: The resistance of A will be calculated from the second connection (series connection)
Note: in series connection, the current flowing in each resistor is the same
[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
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identify the word being referred to choose your answer from the words below
Answer:
1:Rotation
2:Axis
3:Aphelion
4:orbit
What is Velocity in physics
Answer:
hii
Explanation:
i hope this helps you
Answer:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it
Explanation:
hope it helps
pls maek me as brainliest thanks❤
Electricity is the result of moving electrons, so it's classified as
A. Kinetic Energy
B. Gravitational Energy
C. Potential Energy
D. Elastic Energy
Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed in m/s of a satellite in an orbit 980 km above the Earth's surface.
Answer:
564
Explanation:
Two identical conductors have charge -1.8 C and 5.5 C on them, respectively. They are connected by a conducting wire for a short period of time and then disconnected. What is the net charge on each of the conductors after the interaction? g
Answer: 1.85 C
Explanation:
Given
charges on the conductors are [tex]-1.8\ C[/tex] and [tex]5.5\ C[/tex]
They are connected by a conducting wire for a short period of time and then disconnected. During this time charge flow from the wire and net charge becomes [tex]5.5-1.8=3.7\ C[/tex]
This charge will be equally distribute among the two conductors i.e. 1.85 C on each conductor.
Which pair of magnets has the strongest attraction between them?
Equal masses of two different liquids are put into identical beakers.
Liquid 1 is heated for 100s and liquid 2 is heated for 200s by heaters of the same power.
The temperature of both liquids increases by the same amount.
Which statement is correct?
A Both liquids receive the same amount of energy.
B. Liquid 1 receives more energy than liquid 2.
C. Both liquids have equal thermal capacity.
D. The thermal capacity of liquid 1 is less than the thermal capacity of liquid 2.
Answer:
C
Explanation:
Because they both received the same temperature
Kilometer is a unit of length where as kilogram is a unit of mass
By George, you've nailed it, Stacy !
That's a fact, uh huh.
Truer words were never written.
Your statement is one of unquestionable veracity.
The pure truthiness of it cannot be denied.
Was there a question you wanted to ask ?
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they stick together. What is the total momentum of the system after the collision? What is the total momentum of the system before the collision? What is the velocity of the cars after the collision?
Answer:
The total momentum of the cars before the collision is 61,000 kg.m/s
The total momentum of the cars after the collision is 61,000 kg.m/s
The velocity of the cars after the collision is 27.727 m/s
Explanation:
Given;
mass of the first car, m₁ = 1000 kg
initial velocity of the car, u₁ = 25 m/s
mass of the second car, m₂ = 1200 kg
initial velocity of the second car, u₂ = 30 m/s
The common velocity of the cars after collision = v
The total momentum of the cars before collision is calculated as;
P₁ = m₁u₁ + m₂u₂
P₁ = (1000 x 25) + (1200 x 30)
P₁ = 61,000 kg.m/s
The total momentum of the cars after collision is calculated as;
P₂ = m₁v + m₂v
where;
v is the common velocities of the cars after collision since they stick together.
P₂ = v(m₁ + m₂)
To determine "v" apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
(1000 x 25) + (1200 x 30) = v(1000 + 1200)
61,000 = 2,200v
v = 61,000/2,200
v = 27.727 m/s
The total momentum after collsion = v(m₁ + m₂)
= 27.727(1000 + 1200)
= 61,000 kg.m/s
Thus, momentum before and after collsion are equal.
A ratio that compares the width and length of a garden is what type of model?
Answer:
physical
PLEASE MARK ME AS A BRAINLIEST
Answer: Mathematical
Explanation: I took the quiz
A small block, with a mass of 0.05 kg compresses a spring with spring constant 350 N/m a distance of 4 cm. It is released from rest, then slides around the loop and up the incline before momentarily comes to rest at point A. The radius of the loop is 0.1 m.
Required:
Find the elastic potential energy.
Answer:
The elastic potential energy of the spring is 0.28 J
Explanation:
Given;
mass of the block, m = 0.05 kg
spring constant, k = 350 N/m
extension of the spring, x = 4 cm = 0.04 m
The elastic potential energy of the spring is calculated as;
[tex]U_x = \frac{1}{2}kx^2\\\\U_x = \frac{1}{2} \times 350 \times (0.04)^2\\\\U_x = 0.28 \ J[/tex]
Therefore, the elastic potential energy of the spring is 0.28 J
An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?
a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min
Answer:
T = 188.5 s, correct is C
Explanation:
This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved
initial instant. Before the crash
L₀ = r m v₀ + I₀ w₀
the angular speed of the fan is zero w₀ = 0
final instant. After the crash
L_f = I₀ w + m r v
L₀ = L_f
m r v₀ = I₀ w + m r v
angular and linear velocity are related
v = r w
w = v / r
m r v₀ = I₀ v / r + m r v
m r v₀ = (I₀ / r + mr) v
v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]
let's calculate
v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]
v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]
v = 0.02 m / s
To calculate the time of a complete revolution we can use the kinematics relations of uniform motion
v = x / T
T = x / v
the distance of a circle with radius r = 0.6 m
x = 2π r
we substitute
T = 2π r / v
let's calculate
T = 2π 0.6/0.02
T = 188.5 s
reduce
t = 188.5 s ( 1 min/60 s) = 3.13 min
correct is C
An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth
Answer:
0 N
Explanation:
This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N
The power in an electrical circuit is given by the equation P= RR, where /is the current flowing through the circuit and Ris the resistance of the circuit. What is the current in a circuit that has a resistance of 100 ohms and a power of 15 watts?
[pleas ee helpppp)
I= 0.39 A
OPTION B is the correct answer.
calculate the voltage that is being applied across a 10W bulb if a current of 0.2A flows through it
Answer:
below
Explanation:
from P= I * V
v = p/I
v = 10/0.2
v = 50 volts
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?
a) 8.03 x 10^16 nuclei
b) 4.01 x 10^16 nuclei
c) 2.02 x 10^16 nuclei
d) 1.61 x 10^17 nuclei
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
c) 2.02 x 10^16 nucleiEffects of global warming is
A-decrease in temperature
B-melting of polar ice caps
C-breathing problems
Answer:
B- the melting of polar ice caps
Explanation:
As the world's temperature increases, polar ice caps will no longer be able to remain solid.
List and briefly explain the incidents leading to the occurrence of any five nuclear accidents that have taken place in different parts of the world.
Answer:
Chernobyl Nuclear Disaster Nuclear Disaster. Japan 2011 Kyshtym Nuclear Disaster. Russia 1957 Windscale Fire Nuclear Disaster. Sellafield, UK 1957 Three Mile Island Nuclear Accident. Pennsylvania, USA 1979
Explanation:
Hope this helps... pls vote as brainliest
PLEASE HELP ME WITH THIS ONE QUESTION
What is the rest energy of a proton? (c = 2.9979 x 10^9 m/s, mp = 1.6726 x 10^-27)
A) 8.18 x 10^-14 J
B) 2.73 x 10^-22 J
C) 1.5053 x 10^-10 J
D) 1.5032 x 10^-10 J
Answer:
djfjci3jsjdjdjdjdjddndn
ds
You send a traveling wave along a particular string by oscillating one end. If you increase the frequency of oscillations, does the speed of the wave increase, decrease, or remain the same?
Answer:
The speed of the wave remains the same
Explanation:
Since the speed of the wave v = √(T/μ) where T is the tension in the string and μ is the linear density of the string.
We observed that the speed, v is independent of the frequency of the wave in the string. So, increasing the frequency of the wave has no effect on the speed of the wave in the string, since the speed of the wave in the string is only dependent on the properties of the string.
So, If you increase the frequency of oscillations, the speed of the wave remains the same.
1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
following:
blue chip
b. yellow chip
c. yellow chip
Answer:
Explanation:
Blue: 10/30
Red: 5/30
Yellow: 15/30
The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
What is the probability?
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
Given that
Blue chips =10
Red chips = 5
Yellow chips = 15
Number of the total samples =10+15+5=30
Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]
Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]
Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]
Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
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A ball is thrown with an initial velocity of 30.0 m/s and makes an angle of
30.0° with the ground. Find the
A.Horizontal Distance
B.Maximum Height
C.Total Time The Ball is Traveling
Statements imply it is thrown with velocity 30cos30° horizontally and 30sin30° vertically.
Vertically:
Total time taken = 2 x time to go up
= 2(v - u)/a
= 0 - 30sin30°)/(-g)
= 30/g
Therefore, it would travel 30/g sec in horizontal direction as well.
Horizontally :
Distance = horizontal speed x time
= 30cos30° (30/g)
= 450√3 /g
If g = 10, distance is 45√3 m.
Vertically,
Distance = vert. speed x (time of flight/2)
= 30sin30° x (30/g)/2
= 90 m.
Time taken = 30/g = 3 sec
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
A) 1.67 x 10^-4 s^-1
B) 5.43 x 10^-4 s^-1
C) 1.40 x 10^-4 s^-1
D) 2.22 x 10^-4 s^-1
OPTION C is the correct answer.
