Answer:
4.933m/s
Explanation:
the wagon has a weight of 35.1kg*9.81m/s2 = 343.98N
of that weight 343.98N*sin(18.3)=108N are parallel to the hill and oposit tothe tension of the rope.
then, the force that is moving the wagon is 125N-108N=17N
F=m*a then 17N=35.1kg*a
a=0.4843 m/s2
we have two equations
[tex]v=a.t\\x=v.t+\frac{1}{2} . a . t^{2}[/tex]
then
[tex]x=a.t^{2} +\frac{1}{2}. a t^{2} \\x=\frac{3}{2}. a t^{2}[/tex]
[tex]t=\sqrt{\frac{2x}{3a} }[/tex]
t=10.188s
[tex]v=a.t[/tex]
v=4.933 m/s
A 500-kg object is accelerating to the right at 10 cm/s2. What is the magnitude of the rightward net force acting on it (in Newtons)?
Answer:
F = 50 N
Explanation:
As we know by Newton's II law
{tex]F = ma[/tex]
force is product of mass and acceleration
so we will have
[tex]m = 500 kg[/tex]
[tex]a = 10 cm/s^2[/tex]
[tex]a = 0.10 m/s^2[/tex]
now we have
[tex]F = 500 \times 0.10 [/tex]
[tex]F = 50 N[/tex]
The magnitude of the rightward net force acting on the object is 50 N.
What is force?Force can be defined as the product of mass and acceleration.
To calculate the magnitude of the rightward net force, we use the formula below.
Formula:
F = ma............. Equation 1Where:
F = Force on the objectm = mass of the objecta = acceleration due to gravityFrom the question,
Given:
m = 500 kga = 10 cm/s² = 0.1 m/s²Substitute these values into equation 1
F = 500(0.1)F = 50 NHence, The magnitude of the rightward net force acting on the object is 50 N.
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Your bedroom has a rectangular shape and you want to measure its size. You use a tape that is precise to 0.001 m and find that the shortest wall in the room is 3.547 m long. The tape, however is too short to measure the length of the second wall, so you use a second tape, which is longer but only precise to 0.01 m. You measure the second wall to be 4.79 m long. Which of the following numbers is the most precise estimate that you can obtain from your measurements for the area of your bedroom?
a. 17.0 m2
b. 16.990 m2
c. 16.99 m2
d. 16.9 m2
e. 16.8 m2
Answer:
a. 17.0 m²
Explanation:
The product of the two dimensions is 16.99013 m². The least precise contributor has 3 significant figures, so the most precise result available is one with 3 significant figures: 17.0 m².
__
Additional comment
Given that each measurement may be in error by 1/2 of a least-significant digit, their product can be as little as 16.9700025, or as great as 17.0102625. This amounts to 16.9901325 ± 0.0201300. The value 17.0 suggests a range from 16.95 to 17.05, which exceeds the actual range possible with the given measurements. On the other hand, a 4 significant-figure value (16.99) suggests a much smaller range in the product than there may actually be: (16.985, 16.995)
A car is stationary in front of a red stop light. As the light turns green, a truck goes past at a constant speed of 15 m/s. At the same moment the car begins to accelerate at 1.25 m/s2 ;when it reaches 25 m/s, the car continues at that speed. When (time) will the car pass the truck?
The car will pass the truck after 20 seconds.
Explanation:To find when the car will pass the truck, we need to determine the time it takes for the car to reach a speed of 25 m/s and the distance traveled by the truck in that time.
The car is accelerating at a rate of 1.25 m/s². Using the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration, and t is the time, we can solve for t: 25 m/s = 0 m/s + 1.25 m/s² * t.
Simplifying the equation, we get t = 20 seconds. Therefore, the car will pass the truck after 20 seconds.
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does air resistence decrease with speed
Answer:
yes
Explanation:
because when you slow down, the resistance slows with the speed.
Answer:
Yes, air resistance decrease with speed
Explanation:
Air resistance is a kind of fluid friction that acts when objects flow through fluids. It is affected by the velocity of moving objects and the area of the objects. When an object moves with a greater velocity the air resistance acting on them will be high, so the speed decreases.
Fluid friction acts in fluids namely liquids and gases. In liquids the friction is called buoyancy and in gases it is called air resistance or drag. When two objects having the same mass but different area move through air, the object with larger area will have less velocity compared with the other object.
Consider work being done on a positive test charge by an external force in moving the charge from one location to another. In this case, the potential energy _________ (increases, decreases) and the electric potential _________ (increases, decreases).
Answer:
the potential energy increases and the electric potential increases
Explanation:
we know here that Voltage i.e. (Electric potential) increases from the initial point to the final point
consider if 2 object of different charge 1 is twice the charge of other moveing in same distance in electric filed
than object of twice the charge require twice the force so twice amount of work
This work change the potential energy by equal amount of work done
so electric potential energy is depend on the amount of charge on object
so if work being done on a positive test charge by an external force in moving the charge from one location to another
the potential energy increases and the electric potential increases
When a positive test charge is moved from one location to another, the potential energy increases and the electric potential energy increases.
Potential energyThe work done in moving a unit positive test charge from infinity to a certain point in the electric field is known as potential energy.
V = E x d
V = (F/q) x d
where;
V is the potential energyE is electric fieldd is the distanceF is the electric forceThus, when a positive test charge is moved from one location to another, the potential energy increases and the electric potential energy increases.
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When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time of the impact about 10 times as great as for a stiff-legged landing. In this way the average force your body experiences is
A) less than 1/10 as great.
B) more than 1/10 as great.
C) about 1/10 as great.
D) about 10 times as great.
Answer:
C) about 1/10 as great.
Explanation:
We use the relation between Impulse, I, and momentum, p:
[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ \\[/tex]
[tex]F=m(v_{f}-v{o})/t=-mv{o}/t\\ \\[/tex] the final speed is zero
We can see that the average Force is inversely proportional to the time, so if the time is 10 times bigger, the average Force is 1/10 as great
Average force body experience is about 1/10 as great as when the legs are kept stiff
Impact forcefrom Newtons second law, force. f is the product of mass, m and acceleration, a
f = m * a
a = ( v - u ) / t
f = m * ( v- u ) / t
so force is inverrsely related to time, making the time of the impact 10 times as great as stiff-legged landing impiles mathematically as follows:
for stiff-legged
f1 = m * ( v - u ) / t
for bent knees
f2 = m * ( v - u ) / 10t
f2 = f1 / 10
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What is the mass of a dog that weighs 382 N?(unit=kg)
Answer:
The answer to your question is: mass = 38.93 kg
Explanation:
Data
mass = ?
Weight = 382 N
gravity = 9.81 m/s2
Formula
Weight = mass x gravity
mass = weight / gravity
mass = 382 / 9.81 substitution
mass = 38.93 kg result
The mass of a dog that weighs 382 N can be calculated using the weight formula w=mg, rearranged as m=w/g. Substituting the values, we find that the mass of the dog is approximately 39 kg.
Explanation:To calculate the mass of a dog that weighs 382 N, we use the formula w = mg, where w is weight, m is mass, and g is the acceleration due to gravity, which on Earth is approximately 9.8 m/s². So, to find the mass, we rearrange the formula to m = w/g.
Substituting the given values into the equation, m = 382 N / 9.8 m/s² = 39.0 kg. So, the mass of the dog is approximately 39 kg when rounded to the nearest whole number.
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A rock is dropped from the top of a diving platform into the swimming pool below. Will the distance traveled by the rock in a 0.1-second interval near the top of its flight be the same as the distance covered in a 0.1-second interval just before it hits the water? Explain.
Answer:
No the distance traveled in last 0.1 s is not same as that the distance traveled in first 0.1 s
so it will cover more distance in last 0.1 s then the distance in first 0.1 s
Explanation:
As we know that when stone is dropped from the diving board then its velocity at the time of drop is taken to be ZERO
so here we can say that its displacement from the top position in next 0.1 s is given as
[tex]d_1 = v_y t + \frac{1}{2}at^2[/tex]
[tex]d_1 = 0 + \frac{1}{2}(9.81)(0.1)^2[/tex]
[tex]d_1 = 0.05 m[/tex]
Now during last 0.1 s of its motion the stone will attain certain speed
so we will have
[tex]d_2 = v_y(0.1) + \frac{1}{2}(9.81)(0.1)^2[/tex]
[tex]d_2 = 0.1 v_y + 0.05 m[/tex]
so it will cover more distance in last 0.1 s then the distance in first 0.1 s
Final answer:
The distance a rock travels during a 0.1-second interval while falling will vary because of acceleration due to gravity. Early in its fall, it will travel less distance, and just before impact, it will cover more distance due to increased speed.
Explanation:
The distance traveled by a rock in free fall will not be the same in two intervals of time if these intervals occur at different stages of the fall, because the rock is accelerating due to gravity.
Near the top of its flight, it will have just started to accelerate, so it will cover a smaller distance in the first 0.1-second interval. However, just before the rock hits the water, it will have been accelerating for the entire duration of the fall, meaning it will be traveling much faster and will cover a greater distance in the last 0.1-second interval before impact.
Jack observed his coworker Jane crying when she was called into the boss's office. Jack thinks that crying at work is a sign of weakness that makes women unfit for employment in his industry. Jack is displaying _____.
... prejudice based on misogyny.
Also a good bit of stupidity.
An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0 m from time t = 1.00 s to time t = 2.00 s. What is the acceleration of the object?a) 5.20 m/s^2 b) 10.4 m/s^2 c) 8.67 m/s^2 d) 6.93 m/s^2 e) 12.1 m/s^2
Answer:
The acceleration of the object is 9.3 m/s²
Explanation:
For a straight movement with constant acceleration, this equation for the position applies:
x = x0 + v0 t + 1/2 a t²
where
x = position at time t
x0 = initial position
v0 = initial velocity
a = acceleration
t = time
we have two positions: one at time t = 1 s and one at time t = 2 s. We know that the difference between these positions is 14.0 m. These are the equations we can use to obtain the acceleration:
x₁ = x0 + v0 t + 1/2 a (1 s)²
x₂ = x0 + v0 t + 1/2 a (2 s)²
x₂ - x₁ = 14 m
we know that the object starts from rest, so v0 = 0
substracting both equations of position we will get:
x₂ - x₁ = 14
x0 + v0 t + 1/2 a (2 s)² - (x0 + v0 t + 1/2 a (1 s)²) = 14 m
x0 + v0 t + 2 a s² - x0 -v0 t - 1/2 a s² = 14 m
2 a s² - 1/2 a s² = 14 m
3/2 a s² = 14 m
a = 14 m / (3/2 s²) = 9.3 m/s²
The constant acceleration of the object is 9.33 m/s².
The given parameters;
distance traveled by the object, s = 14 minitial time of motion of the object, t = 1 sfinal time of motion of the object, t = 2.0sThe acceleration of the object is calculated by applying the second kinematic equation as follows;
s = ut + ¹/₂at²
where;
u is the initial velocity of the object = 0at t = 1.0 s
x₁ = 0 + ¹/₂(1²)a
x₁ = ¹/₂(a)
at t = 2.0 s
x₂ = 0 + ¹/₂(2²)a
x₂ = 2a
The change in position;
Δx = x₂ - x₁ = 14
14 = 2a - ¹/₂(a)
[tex]14 = \frac{3a}{2} \\\\3a = 2(14)\\\\3a = 28\\\\a = \frac{28}{3} = 9.33 \ m/s^2[/tex]
Thus, the constant acceleration of the object is 9.33 m/s².
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An operation has a 20 percent scrap rate. As a result, 80 pieces per hour are produced. What is the potential labor productivity that could be achieved by eliminating the scrap?
Answer:
25%
Explanation:
If there is 20% scrap, then
80% of (production + scrap) = 72 pcs/hr
Now,
[tex]\frac{80}{100} \times (72+scrap)= 72 pcs/hr\\0.80\times (72+Scrap) = 72 \\ 72+Scrap = \frac{72}{0.80}\\72 + Scrap = 90\\Scrap = 18 pcs /hr[/tex]
The percent increase in labor productivity is, [tex]\frac{18}{72} =0.25[/tex].
Or it can be written as 25% of which would give 80 pieces per hour.
The magnitude of the electrostatic force between two identical ions that are separated by a distance of 5.0A is 3.7×10^-9N.a) what is the charge of each ion?b) how many electrons are missing from each ion( thus giving the ion its charge imbalance)?
Explanation:
Given that,
Electrostatic force, [tex]F=3.7\times 10^{-9}\ N[/tex]
Distance, [tex]r=5\ A=5\times 10^{-10}\ m[/tex]
(a) [tex]F=\dfrac{kq^2}{r^2}[/tex], q is the charge on the ion
[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]
[tex]q=\sqrt{\dfrac{3.7\times 10^{-9}\times (5\times 10^{-10})^2}{9\times 10^9}}[/tex]
[tex]q=3.2\times 10^{-19}\ C[/tex]
(b) Let n is the number of electrons are missing from each ion. It can be calculated as :
[tex]n=\dfrac{q}{e}[/tex]
[tex]n=\dfrac{3.2\times 10^{-19}}{1.6\times 10^{-19}}[/tex]
n = 2
Hence, this is the required solution.
1. A man stands on top of a 25 m tall cliff. The man throws a rock upwards at a 25-degree angle above the horizontal with an initial velocity of 7.2 m/s. What will be the vertical position of the rock at t = 2 seconds? What will be the horizontal position of the rock at that time?
2. A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how far would the cannonball travel before it lands on the ground?
3. Look at the following picture. What would be the resultant vector of A+B?
4. An airplane undergoes the following displacements: First, it flies 72 km in a direction 30° east of north. Next, it flies 48 km due south. Finally, it flies 100 km 30° north of west. Using analytical methods, determine how far the airplane ends up from its starting point.
5.The following picture shows a golf ball being hit and given an initial velocity of v0. The ball is hit at an unknown angle above the ground. What are TWO values that are known, even in the absence of all other numbers?
Answer:
These are the answers for 1, 2 and 3
Explanation:
Sorry I couldn't help you with 4 and 5
1.
Answer:
y = 11.48 m
x = 13.0 m
Explanation:
Components of initial velocity is given as
[tex]v_x = 7.2 cos25 = 6.52 m/s[/tex]
[tex]v_y = 7.2 sin25 = 3.04 m/s[/tex]
Now after t = 2 s the vertical position is given as
[tex]y = y_0 + v_y t + \frac{1}{2}a_y t^2[/tex]
[tex]y = 25 + 3.04(2) - \frac{1}{2}(9.8)(2^2)[/tex]
[tex]y = 11.48 m[/tex]
Now horizontal position is given as
[tex]x = v_x t[/tex]
[tex]x = 6.52 \times 2[/tex]
[tex]x = 13.04 m[/tex]
2.
Answer:
d = 26.6 m
Explanation:
Initial position on y axis is given as
[tex]y = 1.5 m[/tex]
velocity of ball in y direction
[tex]v_y = 0[/tex]
now we have
[tex]\Delta y = v_y t + \frac{1}{2}gt^2[/tex]
[tex]1.5 = \frac{1}{2}(9.8) t^2[/tex]
[tex]t = 0.55 s[/tex]
now the distance moved by the ball in horizontal direction is given as
[tex]d = v_x t[/tex]
[tex]d = 48.1 \times 0.55[/tex]
[tex]d = 26.6 m[/tex]
3
Answer:
[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]
Explanation:
Here we can see the two vectors inclined at different angles
so two components of vector A is given as
[tex]A = 11.3 cos21\hat i - 11.3 sin21\hat j[/tex]
[tex]A = 10.55 \hat i - 4.05 \hat j[/tex]
Similarly for other vector B we have
[tex]B = 4.78cos67 \hat i + 4.78 sin67\hat j[/tex]
[tex]B = 1.87\hat i + 4.4 \hat j[/tex]
now we need to find A + B
so we have
[tex]A + B = (10.55\hat i - 4.05\hat j) + (1.87\hat i + 4.4 \hat j)[/tex]
[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]
4.
Answer:
d = 81.86 m
Explanation:
Displacement of airplane is given as
[tex]d_1[/tex] = 72 km in direction 30 degree East of North
[tex]d_1 = 72sin30\hat i + 72cos30\hat j[/tex]
[tex]d_1 = 36\hat i + 62.35\hat j[/tex]
[tex]d_2[/tex] = 48 km South
[tex]d_2 = -48\hat j[/tex]
[tex]d_3[/tex] = 100 km in direction 30 degree North of West
[tex]d_3 = -100 cos30\hat i + 100 sin30\hat j[/tex]
[tex]d_3 = -86.6\hat i + 50\hat j[/tex]
so net displacement is given as
[tex]d = d_1 + d_2 + d_3[/tex]
[tex]d = 36\hat i + 62.35\hat j - 48\hat j - 86.6\hat i + 50 \hat j[/tex]
[tex]d = -50.6\hat i + 64.35\hat j[/tex]
now magnitude of displacement is given as
[tex]d = \sqrt{50.6^2 + 64.35^2}[/tex]
[tex]d = 81.86 m[/tex]
5.
Answer:
1) final speed at which it will hit the ground again
2) acceleration during the motion of the ball
Explanation:
As we know that the speed at which the ball is thrown is always same to the speed by which it will hit back on the ground
so we know that final speed will be same as initial speed
Also we know that during the motion the acceleration of ball is due to gravity so it will be
[tex]a = - g[/tex]
If fe represents the electrostatic force in N that point charge q1 in C exerts on point charge q2 in C, and r represents the distance between the point charges in m, what is the unit for the electrostatic constant k in the formula F = kq1q2/r^2?
The electrostatic constant is also known as Coulomb's constant or the electric constant. Its unit is [tex]\rm \frac{Nm^2}{C^2}[/tex].
It is a fundamental physical constant that appears in Coulomb's law, which describes the interaction between electric charges. It's denoted by the symbol K.
Given information:
Unit of force (F) = Newton
Unit of charge (Q) = Coloumb
Unit of distance (r) = Meters
Now,
On substituting the values of given units:
[tex]\rm F = \frac{Kq_1q_2}{r^2}\\N =\frac{K C\times C}{m^2}\\N = \frac{K C^2}{m^2}\\K = \frac{Nm^2}{C^2}[/tex]
The unit of electrostatic constant (K) is [tex]\rm \frac{Nm^2}{C^2}[/tex]. This constant (K) plays a crucial role in understanding and calculating the forces and interactions between charged particles in the realm of electrostatics.
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Final answer:
The unit for the electrostatic constant k is Newton meters squared per Coulomb squared (N·m²/C²), and k is also known as Coulomb's constant with a value of approximately 8.99 × 10⁹ N·m²/C².
Explanation:
The unit for the electrostatic constant k in the formula F = k[tex]q_{1}[/tex][tex]q_{2}[/tex] /r² can be determined by rearranging the formula to solve for k, which gives us k = Fr² / [tex]q_{1}[/tex][tex]q_{2}[/tex] . Since the unit for the electrostatic force F is Newtons (N), the distance r is in meters (m), and the charge [tex]q_{1}[/tex]and [tex]q_{2}[/tex] are in Coulombs (C), the units for k would be Newton meters squared per Coulomb squared (N·m²/C²). The value of the electrostatic constant k is approximately 8.99 × 10⁹ N·m²/C², also known as Coulomb's constant.
A 5.0-kg clay putty ball and a 10.0-kg medicine ball are headed towards each other. Both have the same speed of 20 m/s. If they collide perfectly inelastically, what approximately is the speed of the blob of clay and ball immediately after the collision?
Answer:
- 3.33 m /s
Explanation:
m1 = 5 kg
m2 = 10 kg
u1 = 10 m/s
u2 = - 10 m/s
Let the velocity of the combination of the blob is v.
Use the conservation of momentum
[tex]m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v[/tex]
5 x 10 - 10 x 10 = (5 + 10)v
50 - 100 = 15 v
v = - 3.33 m /s
Thus, the velocity of the blob after sticking together is - 3.33 m/s.
On a boat ride, the skipper gives you an extra-large life preserver filled with lead pellets. When he sees the skeptical look on your face, he says that you'll experience a greater buoyant force if you fall overboard than your friends who wear regular-sized Styrofoam-filled life preservers. True or False
Answer:
True.
Explanation:
He is correct but what he does not tell is that you will be drown. Your life preserver will submerge and displace more water than those of your friends who float at the surface. Although buoyant force on you will be greater , the net downward force on you will still be greater. Hence you will be drown inside the water.
A car traveling at a speed of v can brake to an emergency stop in a distance x , Assuming all other driving conditions are all similar , if the traveling speed of the car double, the stopping distance will be (1) √2x,(2)2x, or(3) 4x:(b) A driver traveling at 40.0km/h in school zone can brake to an emergency stop in 3.00m. What would be braking distance if the car were traveling at 60.0 km/h?
Answer:
D = 6.74 m
Explanation:
Let the f newtons be the braking force
distance to stop = x meters
speed = v m/s
we know that work done is given as
work done W = fx joules
[tex]fx = \frac{1}{2} mv^2[/tex]
If the speed is doubled ,
[tex]fx' = \frac{1}{2} m(2v)^2[/tex]
[tex] = 4[\frac{1}{2}mv^2] = 4fx[/tex]
stooping distance is D = 4x
[tex]40km/h = \frac{40*1000}{60*60} = 11.11 m/s[/tex]
60 km/hr = 16.66 m/s
braking force = f
[tex]f*3 = [\frac{1}{2}mv^2] = [\frac{1}{2}m*11.11^2][/tex]
[tex]f*D = [\frac{1}{2}m*16.66^2][/tex]
[tex]\frac{D}{3} = \frac{16.66^2}{11.11^2}[/tex]
[tex]\frac{D}{3} = 2.2489[/tex]
D = 6.74 m
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.90 m/s at an angle of 23.0° below the horizontal. It strikes the ground 5.00 s later.(a) How far horizontally from the base of the building does the ball strike the ground?1 m(b) Find the height from which the ball was thrown.2Your response differs from the correct answer by more than 10%. Double check your calculations. m(c) How long does it take the ball to reach a point 10.0 m below the level of launching?3 s
Answer:
Part a)
[tex]\Delta x = 36.35 m[/tex]
Part b)
height will be 107.2 m
Part c)
[tex]t = 1.78 s[/tex]
Explanation:
Initial velocity of the ball is given as
v = 7.90 m/s
angle of projection of the ball
[tex]\theta = 23^o[/tex]
now the two components of velocity of ball is given as
[tex]v_x = 7.90 cos23 = 7.27 m/s[/tex]
[tex]v_y = 7.90 sin23 = 3.09 m/s[/tex]
Part a)
Since ball strike the ground after t = 5 s
so the distance moved by the ball in horizontal direction is given as
[tex]\Delta x = v_x \times t[/tex]
[tex]\Delta x = 7.27 (5)[/tex]
[tex]\Delta x = 36.35 m[/tex]
Part b)
Now in order to find the height of the ball we can find the vertical displacement of the ball
[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]
[tex]\Delta y = (3.09)(5) - \frac{1}{2}(9.81)(5^2)[/tex]
[tex]\Delta y = -107.2 m[/tex]
So height will be 107.2 m
Part c)
when ball reaches a point 10 m below the level of launching then the displacement of the ball is given as
[tex]\Delta y = -10 m[/tex]
so we will have
[tex]\Delta y = v_y t + \frac{1}{2}at^2[/tex]
[tex]-10 = 3.09 t - \frac{1}{2}(9.81)t^2[/tex]
so we will have
[tex]t = 1.78 s[/tex]
For an object in free-fall (ignoring air resistance), which of the following statements is false: You are free to choose the origin and direction of the positive axis. The acceleration due to gravity is constant. The acceleration may point in the opposite direct as the velocity. The acceleration always points in the same direct as the velocity.
Answer:hh
The acceleration may point in the opposite direct as the velocity
Explanation:
By definition free-fall is an uniform accelerate movement so speed always is increasing in fact acelertion and velocity always point in the same direction
Answer:
T h e a c c e l . e r a t i o n m a y p o i n t i n t h e o p p o s i t e d i r e c t a s t h e v e l o c i t y .
Explanation:
PSYCHOLOGY! A _________ is a graphical representation of association between variables.
A. scatterplot
B. regression plot
C. graphical plot
D. variable plot
Answer:
A. Scatterplot
Explanation:
because it is
For questions 18-21, . In each space, classify the reaction type.
18. ____________________ 2LiBr + Pb(NO3) - PbBr2 + 2LiNO3
19. ____________________ Fe + 2HCl - FeCl2 + H2
20. ____________________ CaO + H2O - Ca(OH)2
21. ____________________ NiCl2 - Ni + Cl2
Answer:
[tex]2LiBr+Pb(NO_3) = PbBr_2+ 2LiNO_3[/tex] is a double displacement reaction
[tex]Fe+2HCl = FeCl_2+2H_2[/tex] is a redox reaction [tex]CaO+H_2 O = Ca(OH)_ 2[/tex]is a combination reaction [tex]NiCl_2 = Ni+Cl_2[/tex] is a decomposition reactionExplanation:
[tex]2LiBr+PbNO_3 = PbBr_2+2LiNO_3[/tex]
It is of the form [tex]AX+BY = AY+BX[/tex]
[tex]Fe+2HCl = FeCl_2+H_2[/tex]
Both oxidation and reduction occur in this reaction. Fe gets oxidized and H gets reduced.
[tex]Fe = Fe^(2+)+2e^-[/tex]
[tex]2H^++2e^- = H_2[/tex]
[tex]CaO+H_2 O = Ca(OH)_2[/tex]
It is an exothermic combination reaction. Calcium reacts with water to produce calcium hydroxide and heat is released in this process.
[tex]NiCl_2 = Ni+Cl_2[/tex]It is a decomposition reaction where nickel chloride decomposes to nickel and chlorine.
Answer:
18. Double Displacement Reaction
19. Single Displacement Reaction
20. Synthesis Reaction
21. Decomposition Reaction
Explanation:
Double Displacement formula:
AB + CD --> AD + CB
Single Displacement formula:
A + BC --> AC + B
Synthesis formula:
A + B --> AB
Decomposition formula:
AB --> A + B
The front brakes on a vehicle do more work than do the rear brakes. Technician A says that this is because the front wheel cylinders are closer to the master cylinder than are the rear wheel cylinders. Technician B says that this is because the weight of a vehicle shifts to the front during a stop. Which technician is correct?
Answer:
Technician A says that this is because the front wheel cylinders are closer to the master cylinder than are the rear wheel cylinders.
Explanation:
Since the position of cylinder is near the front wheel so the normal force of the front wheel is more than the normal force on the rear wheel
as we know that the center of mass of the wheel is shifted towards the front wheel as the balancing is done with reference to its center of mass
so we will have
[tex]N_1 d_1 = N_2 d_2[/tex]
also we have
[tex]N_1 + N_2 = W[/tex]
so here we can say
friction force on front wheel will be more
[tex]F_f = \mu N_1[/tex]
so here front wheel has to do more work to stop the vehicle
Technician B is correct; the front brakes do more work because the vehicle's weight shifts to the front during braking due to dynamic load transfer. This is a physics principle related to force and motion, rather than the hydraulic pressure distribution which is equal for all brakes due to Pascal's principle.
Technician B suggests that the front brakes do more work than the rear brakes because the weight of a vehicle shifts to the front during a stop. This is correct, as deceleration causes the vehicle's weight to transfer to the front due to inertia, resulting in greater pressure on the front brakes. This phenomenon is known as weight transfer or dynamic load transfer. The claim by Technician A that proximity to the master cylinder affects braking force is not accurate, as hydraulic systems utilize Pascal's principle to ensure equal pressure distribution throughout the brake fluid regardless of the distance from the master cylinder.
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium when the string makes a 30 degrees angle with the vertical, what is the net charge on the ball?
Q = _______ C
Answer:
[tex]Q = \frac{0.068}{E}[/tex]
where E = electric field intensity
Explanation:
As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical
So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field
while weight of the ball is vertically downwards
so here we have
[tex]QE = F_x[/tex]
[tex]mg = F_y[/tex]
since string makes 30 degree angle with the vertical so we will have
[tex]tan\theta = \frac{F_x}{F_y}[/tex]
[tex]tan30 = \frac{QE}{mg}[/tex]
[tex]Q = \frac{mg}{E}tan30[/tex]
[tex]Q = \frac{0.012\times 9.81}{E} tan30[/tex]
[tex]Q = \frac{0.068}{E}[/tex]
where E = electric field intensity
The net charge on the small plastic ball which is suspended by a string in a uniform is 0.068/E C.
What is electric field?The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
A point charge on a string in equilibrium with an electric field can be given as,
[tex]E=\dfrac{mg \tan\theta}{Q}[/tex]
Here, (m) is the mass, (g) is acceleration due to gravity, and (Q) is the charge.
A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field.
The ball is in equilibrium when the string makes a 30 degrees angle with the vertical. Thus, by the above formula,
[tex]E=\dfrac{(0.012)(9.81)\tan(30)}{Q}\\Q=\dfrac{0.068}{E}\rm\; C[/tex]
Thus, the net charge on the small plastic ball which is suspended by a string in a uniform is 0.068/E C.
Learn more about electric field here;
https://brainly.com/question/14372859
A wide river flows from North to South at a steady rate of 2 m/s. The motor boat has been tested on a calm pond and it was found that it goes through the water at 5 m/s. Make the downstream direction positive. www During a 5 second period how much distance does the boat move over the water as seen by the person in the inner tube.
Answer:
35 meters
Explanation:
The river flows at 2 m/s from North to South and motor can propel the boat at 5m/s. As the downstream direction is positive, we are considering the river flow also propels the boat adding its speed to the boat. It means the boat and the person in the inner tube are in fact moving at 7 m/s. Distance can be calculated as follows:
[tex]v = d/t[/tex] ⇒[tex]d = vt[/tex] ⇒[tex]d = 7\frac{m}{s}x5s[/tex]
[tex]d = 35m[/tex]
The motorboat will cover a distance of 35 meters downstream in 5 seconds. This is calculated by adding the boat's velocity relative to the water (5 m/s) to the velocity of the river current (2 m/s), giving a resultant velocity of 7 m/s and multiplying it by the time interval.
We need to calculate the distance the motorboat will cover downstream in a 5-second interval. The speed of the river current is 2 m/s and the speed of the motorboat relative to the water is 5 m/s. Since we are considering the downstream direction as positive, the boat's velocity relative to an observer on the shore would be the sum of these two speeds.
The boat's speed relative to the shore (resultant velocity) is:
Velocity of the boat relative to the shore = Velocity of the boat relative to the water + Velocity of the river = 5 m/s + 2 m/s = 7 m/s
The distance covered by the boat over a period of 5 seconds would be:
Distance = Velocity × Time = 7 m/s × 5 s = 35 m
Therefore, the motorboat will move 35 meters downstream as observed by someone outside of the water, like the person in the inner tube.
a 0.0818 kg salt shaker on a rotating table feels an inward frictional force of 0.108 N when it is moving 0.333m/s. what is the radius of its motion?(unit=m)
Answer:
The required radius of its motion is [tex]0.084m[/tex].
Explanation:
The formula for calculating the required radius of its motion is given by
[tex]F = (mv^2)/r[/tex]
Where m= mass
V= moving velocity
F=frictional force
r = radius of its motion
Then the required radius of its motion is given by
[tex]r =(mv^2)/F[/tex]
Given that
mas =0.0818 kg
Frictional force= 0.108 N
Moving with Velocity of = 0.333 m/s
radius of its motion = [tex]\frac{[0.0818 kg \times (0.333 m/s)^2]}{0.108 N}[/tex]
Hence the required radius of its motion is r = [tex]8.4 cm=0.084m[/tex]
Answer:
0.084
Explanation:
Acellus
A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration of 10 cm/s2 for another 5.0 s. The particle moves in the same direction along a straight line. The average speed over the whole time interval is?
Answer:
V=20cm/s
Explanation:
The average speed is the distance total divided the time total:
[tex]V=X/T[/tex]
First stage:
T1=5s
[tex]v_{f} =v_{o} - at[/tex]
But, [tex]v_{f} =0[/tex] (decelerates to rest)
then: [tex]a =v_{o} /t=0.3/5=0.06m/s^{2}[/tex]
on the other hand:
[tex]x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m[/tex]
X1=75cm
Second stage:
T2=5s
[tex]x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m[/tex]
X2=125cm
Finally:
X=X1+X2=200cm
T=T1+T2=10s
V=X/T=20cm/s
The average speed of the particle is 20 centimetre per second.
Average speed is given as total distance covered by the particle divided by the total time of the journey.
The particle decelerates from 30 cm/s to rest (0 cm/s) in 5.0 s. Hence, the deceleration can be given as,
a = (0 - 30 cm/s) / 5 s = - 6 cm/s²
From the kinematic equation:
s = ut + [tex]\frac{1}{2}at^2[/tex], where s = displacement, a = acceleration, t = time, and u = initial velocity of the particle, we can determine the displacement.
so, s = (30 cm/s) (5 s) - [tex]\frac{1}{2} (6 cm/s^2)(5 s)^2[/tex] = 150 cm - 75 cm = 75 cm
The particle then accelerates uniformly at 10 cm/s² for another 5.0 s. The final velocity after this period is given as:
v = u + at
Here, u = 0 cm/s, a = 10 cm/s². t = 5s. Hence,
v = (10 × 5) m/s = 50 cm/s
Using the kinematic equation:
v² = u² + 2as, we get:
(50 cm/s)² = 2 (10 cm/s²) s
or, s = 125 cm
Total displacement of the particle = 75 cm + 125 cm = 200 cm
total time of journey = 5s + 5s = 10s
so, average velocity = 200 cm / 10 s = 20 cm/s
An electron with speed of 104 m/s enters a ""forbidden"" region where an electric force tries to push it back along its path with a constant acceleration of 107 m/s2 . How far will the electron go into the ""forbidden"" region? How long will it be in that region?
Answer:
The distance travelled is 151.22m and it took 0.97s
Explanation:
Well, this is an ARM problem, so we will need the following formulas
[tex]x(t)=x_{0} +v_{0} *(t-t_{0} )+0.5*a*(t-t_{0} )^{2}[/tex]
[tex]v(t)=v_{0} +a*(t-t_{0} )[/tex]
where [tex]x_{0}[/tex] is the initial position (we can assume is zero), [tex]v_{0}[/tex] is the initial speed of 104 m/s, [tex]t_{0}[/tex] is the initial time (we also assume is zero), a is the acceleration of 107 m/s2, v is speed, x is position and t is time.
Now that we have the formulas, we know that when the electron stops it has no speed. Then we calculate how much time it takes to stop.
[tex]0=104m/s-107m/s^{2} *t\\t=0.97s[/tex]
Finally, we calculate the distance travelled in this time
[tex]x(0.97s)=104m/s*0.97s+0.5*107m/s^{2}*(0.97s)^{2}=151.22m[/tex]
Answer:
Part a)
[tex]d = 5 m[/tex]
Part b)
[tex]T = 2\times 10^{-3} s[/tex]
Explanation:
Part a)
The electron will move in this forbidden region till its speed will become zero
So here we will have
[tex]v_f = 0[/tex]
[tex]v_i = 10^4 m/s[/tex]
also its deceleration is given as
[tex]a = - 10^7 m/s^2[/tex]
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - (10^4)^2 = 2(-10^7) d[/tex]
[tex]d = 5 m[/tex]
Part b)
Now the time till its speed is zero
[tex]v_f - v_i = at[/tex]
[tex]0 - 10^4 = -10^7 t[/tex]
[tex]t = 10^{-3} s[/tex]
so total time that it will be in the region is given as
[tex]T = 2 t[/tex]
[tex]T = 2\times 10^{-3} s[/tex]
Two objects are thrown vertically upward, first one, and then, a bit later, the other. Is it (a) possible or (b) impossible that both objects reach the same maximum height at the same instant of time?
Answer:
No, it is impossible
Explanation:
Kinematics equation:
[tex]Vf^{2} =Vo^{2} -2gy[/tex]
if height is maximum:
y=H and Vf=0
so:
[tex]Vo^{2} =2gH[/tex][tex]H=Vo^{2} /2g[/tex]Analysis: From the last equation we see that the maximum height depends ONLY on the initial speed. This means that if both objects reach the same maximum height, then they necessarily need to have the SAME initial velocity. If they have the same initial velocity and in order to reach the maximum height at the SAME time the only way is that they are released at the SAME TIME.
A standing wave is set up in a 2.0 m length string fixed at both ends. The string is then made to vibrate in 5 distinct segments when driven by a 120 Hz source. What is the natural, fundamental frequency of this string?
Answer:
The natural, fundamental frequency of this string is 24 hertz.
Explanation:
Length of the string, l = 2 m
Number of segments, n = 5
Frequency, f = 120 Hz
Let f' is the natural fundamental frequency of this string. The frequency for both side ended string is given by :
[tex]f=\dfrac{nv}{2l}[/tex]
[tex]v=\dfrac{2fl}{n}[/tex]
[tex]v=\dfrac{2\times 120\times 2}{5}[/tex]
v = 96 m/s
For fundamental frequency, n = 1
[tex]f'=\dfrac{v}{2l}[/tex]
[tex]f'=\dfrac{96}{2\times 2}[/tex]
f' = 24 Hertz
So, the natural, fundamental frequency of this string is 24 hertz. Hence, this is the required solution.
The natural, fundamental frequency of this standing wave is equal to 24 Hertz.
Given the following data:
Length of the string = 2.0 meters.
Number of segments = 5.
Frequency = 120 Hertz.
How to calculate natural, fundamental frequency.First of all, we would determine the velocity of the standing wave by using this formula:
[tex]F=\frac{nV}{2L} \\\\V=\frac{2FL}{n} \\\\V=\frac{2 \times 120 \times 2.0}{5}\\\\V=\frac{480}{5}[/tex]
V = 96 m/s.
Now, we can calculate the natural, fundamental frequency by using this formula:
[tex]F'=\frac{V}{2L} \\\\F'=\frac{96}{2 \times 2.0} \\\\F'=\frac{96}{4.0}[/tex]
F' = 24 Hertz.
Read more on frequency here: brainly.com/question/3841958
A plane starting from rest (vo = 0 m/s) when t0 = 0s. The plane accelerates down the runway, and at 29 seconds, its velocity is +72.2 m/s. (+ indicates to the right). Determine the average acceleration for the plane.
Answer:
Acceleration, [tex]a=2.48\ m/s^2[/tex]
Explanation:
Given that,
The plane is at rest initially, u = 0
Final speed of the plane, v = 72.2 m/s
Time, t = 29 s
We need to find the average acceleration for the plane. It can be calculated as :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{72.2}{29}[/tex]
[tex]a=2.48\ m/s^2[/tex]
So, the average acceleration for the plane is [tex]2.48\ m/s^2[/tex]. Hence, this is the required solution.