Answer:
The size will increase.
Explanation:
When you bend a plastic ruler, it's size will increase because it is elastic and will exhibit elastic deformation. When it is been bent, it will continue to stretch until it get to a point where it will not be able to regain it formal shape, it size wound of increase. Therefore when the ruler get to elastic limit and you have bend it to the point it cannot regain it's formal shape back, it will remain bent and if further force is apply on it,it will break.
the wire is 5.20 cm from the charge and carries a current of 68.5 A in a direction opposite to that of the moving charge. Calculate the magnitude of the force on the charge.
Complete Question
A 6.00-µc charge is moving with a speed of 7.50x10^4 m/s parallel to a very long, straight wire. the wire is 5.20 cm from the charge and carries a current of 68.5 A in a direction opposite to that of the moving charge. Calculate the magnitude of the force on the charge.
Answer:
The value is [tex]F =1.20 *10^{-4} \ N[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]Q = 6.00\muC = 6.00 *10^{-6} \ C[/tex]
The speed is [tex]v = 7.50 *10^{4} \ m/s[/tex]
The distance of the wire from the charge is [tex]d = 5.20 \ cm = 0.0520 \ m[/tex]
The current flowing through the wire in the opposite direction to the charge is [tex]I _2 = 68.5 \ A[/tex]
Gnerally the magnitude of force on that charge is mathematically represented as
[tex]F = qvB[/tex]
Here B is magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * I}{2\pi d}[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]F = qv[\frac{\mu_o * I}{2\pi d}][/tex]
=> [tex]F = 6.0*10^{-6} * (7.50 *10^{4})[\frac{4\pi * 10^{-7} * 68.5 }{2* 3.142 * 0.0520 }][/tex]
=> [tex]F =1.20 *10^{-4} \ N[/tex]
The lungs are large organs which contain smaller, expandable sacs.
These sacs greatly increase the surface area of the lungs.
Why is a large surface area important to the function of the lungs? A. Large amounts of liquid wastes and fatty tissue must be stored in the lungs. B. Oxygen in air taken in by the lungs must move quickly through the lung tissue and into the blood. C. The lungs must be very sturdy and rigid so they cannot move. D. The lungs must be able to filter oxygen from water while the organism is swimming.
Answer:
B. It is B because without this much space for oxygen and blood to flow and go through, we would not survive.
Explanation:
Centripetal acceleration must involve a change in
Answer:
Centripetal acceleration must involve a change in a. an object's tangential speed.
Explanation:
Centripetal acceleration must involve a change in direction of motion of the object.
What is centripetal acceleration?centripetal acceleration is the acceleration of a body traversing a circular path. Because velocity is a vector quantity (that is, it has both a magnitude, the speed, and a direction), when a body travels on a circular path, its direction constantly changes and thus its velocity changes, producing an acceleration. The acceleration is directed radially toward the center of the circle.
In your daily existence, you must have encountered numerous instances of centripetal acceleration. A centripetal acceleration occurs when you drive in a circle, and a centripetal acceleration also occurs when a satellite orbits the earth. Centripetal refers to being in the center.
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wavelength = 5.4 cm and frequency = 9.4 Hz
What is the speed
An unbalanced force of 23 N is applied to a 13 kg mass. What is the acceleration of the
mass?
Answer:
1.77 m/s^2
Explanation:
since it is unbalanced there must be net force(resultant force)
[tex]net \: force \: = accleration \: \times mass[/tex]
net force = 23N
mass = 13kg
a = x
x = 23/13
= 1.769...
approximately = 1.77 m/s^2
Which of the following choices is not part of the scientific method
Answer:
What are the choices?
Explanation:
A bus rolls to a stop along a horizontal road without the driver applying the brakes,
What answer choice BEST explains why this occurred?
A. The natural state of the bus is not to be in motion.
B. The bus ran out of momentum.
C. The force of gravity slowed the bus until it stopped.
D. The opposing force of friction stopped the train.
Answer:
should be d because friction allows things to go faster or slower
Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a 100 W incandescent bulb uses only 23.0 W of power. The compact bulb lasts 10000 hours, on the average, and costs $ 12.00, whereas the incandescent bulb costs only $ 0.76, but lasts just 750 hours. The study assumed that electricity cost $ 0.090 per kilowatt-hour and that the bulbs were on for 4.0 h per day.
Required:
a. What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years?
b. How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb?
c. What is the resistance of a "100-W" fluorescent bulb?
Answer:
a
[tex]The \ price \ for \ Compact\ fluorescent\ bulbs \ is \ C_T = \$ 21.067\\The \ price \ for \ incandescent\ bulb \ is \ C_t = \$ 43.98[/tex]
b
[tex]The \ amount \ saved \ is \ C_S = \$ 22.913[/tex]
c
[tex]The \ resistance \ is \ R = 626.1 \Omega[/tex]
Explanation:
From the question we are told that
The power used by incandescent bulb [tex]P_r = 23.0 W[/tex]
The power used by Compact fluorescent bulb [tex]P_p = 100W[/tex]
The life span of Compact fluorescent bulb is [tex]t = 1000 \ hours[/tex]
The prices of Compact fluorescent bulb is [tex]C_c = \$ 12 [/tex]
The prices of incandescent bulb is [tex]C_i = \$ 0.76 [/tex]
The life span of incandescent bulb is [tex]t_k = 750 \ hours[/tex]
The cost of electricity is [tex]C_e = \$ 0.090/ kilowatt-hour = \frac{0.090}{1000} = \$0.00009 / W / h[/tex]
The duration of daily bulb usage is [tex]t_d = 4 \ hours[/tex]
Generally 3 years of bulb usage in hours is mathematically evaluated as
[tex]t_3 = 3 * 365 * t_d[/tex]
=> [tex]t_3 = 3 * 365 * 4[/tex]
=> [tex]t_3 = 4380 \ h [/tex]
Generally the cost to power a 23W bulb for the above duration is mathematically represented as
[tex]Z = 23 * t_3 * C_e[/tex]
=> [tex]Z = 23 * 4380 * 0.00009[/tex]
=> [tex]Z = \$ 9.0667[/tex]
The number of Compact fluorescent bulbs required to achieve this about of time is mathematically represented as
[tex]N = \frac{t_3}{ t}[/tex]
=> [tex]N = \frac{4380}{1000}[/tex]
=> [tex]N = 0.438 [/tex]
=> [tex]N \approx 1 [/tex]
The total cost of usage for the Compact fluorescent bulb is mathematically represented as
[tex]C_T = Z +C_c[/tex]
=> [tex]C_T = 9.0667 +12[/tex]
=> [tex]C_T = \$ 21.067 [/tex]
The number of incandescent bulbs required to achieve this about of time is mathematically represented as
[tex]M = \frac{t_3}{ t_k}[/tex]
=> [tex]M = \frac{4380}{750}[/tex]
=> [tex]M = 6 [/tex]
Generally the cost to power a 100W bulb for the above duration is mathematically represented as
[tex]Z = 100 * t_3 * C_e[/tex]
=> [tex]Z = 100 * 4380 * 0.00009[/tex]
=> [tex]Z = \$ 39.42[/tex]
The total cost of usage for the incandescent bulbs is mathematically represented as
[tex]C_t = C_e + [6 *C_i][/tex]
=> [tex]C_t = 39.42 + [0.76 * 6][/tex]
=> [tex]C_t = \$ 43.98[/tex]
Generally the total amount saved is mathematically represented as
[tex]C_S = C_t - C_T[/tex]
=> [tex]C_S = 43.98 - 21.067[/tex]
=> [tex]C_S = \$ 22.913[/tex]
Generally the resistance of the Compact fluorescent bulb is
[tex]R = \frac{V^2}{P}[/tex]
[tex]R = \frac{120^2}{23}[/tex]
[tex]R = 626.1 \Omega[/tex]
A 10 [kg] object is dropped from rest. a. How far will it drop in 2 [s]?
Answer:
y=0.5 g t^2
=0.5*10*2^2
=20 m
A ball is thrown horizontally from a 12 m
building with a speed of 2.0 m/s.
-high
How far from the base of the building does the ball hit the ground ?
Answer:
24
Explanation:
you multiply 2.0x12
Which of the following are examples of
molecules? Check all that apply.
two carbon atoms bonded to each other
a carbon atom
a chain of one hundred carbon atoms bonded
to one another
a carbon atom bonded to an oxygen atom
Answer:
two carbon atoms bonded to each other
a chain of one hundred carbon atoms bonded to one another
a carbon atom bonded to an oxygen atom
two carbon atoms bonded to each other a carbon atom, a chain of one hundred carbon atoms bonded to one another, a carbon atom bonded to an oxygen atom all these are examples of molecules, therefore all the given options are the correct answer
What is a molecule?The smallest component of a substance that still has its content and characteristics is called a molecule. It is constructed of two or more atoms connected by chemical bonds. Moles are the foundation of chemistry. Molecules are designated by their elemental symbol and a subscript indicating their atom count.
Atoms are the fundamental unit of an element. They consist of a nucleus and surrounding electrons. When an atom has an incomplete electron shell, it is said to have valence electrons. When two or more atoms come together to share outer shell valence electrons, a covalent bond is formed,
Since molecules include two carbon atoms bonded to each other, one carbon atom, a chain of hundred carbon atoms bonded to one another, and a carbon atom bonded to an oxygen atom, all of the suggested answers are correct.
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2) A boat travels 12.0 m while it reduces its velocity from 9.5 m/s to 5.5 m/s. What is the
boat's acceleration while it travels that distance?
Answer:
2.5m/s^2
Explanation:
Note that the boat is reducing its speed. It is having negative acceleration or deceleration.
V^2 = u^2 -2as ( minus sign is used because of speed is reduced)
Given that,
s = 12 m
v = 5.5 m/s
u = 9.5 m/s
a = (v^2 - u^2) ÷ (-2s)
a= ( 5.5^2 - 9.5^2) ÷ ( -2× 12)
a = 2.5 m/s^2
The acceleration of the boat is -2.5 m/s²
The given paramters;
distance traveled by the boat, d = 12 m
initial velocity of the boat, u = 9.5 m/s
final velocity of the boat, v = 5.5 m/s
The acceleration of the boat is calculated as;
[tex]v^2 = u^2 + 2as\\\\2as = v^2 - u^2\\\\a = \frac{v^2 - u^2}{2s} \\\\a = \frac{(5.5)^2 -(9.5)^2 }{2(12)} \\\\a = -2.5 \ m/s^2[/tex]
Thus, the deceleration of the boat is 2.5 m/s²
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Add the following vectors:
(a) (12, 5) + (6, 3)
(b) (−3, 8) + (6, −2)
(c) (3, 8, −7) + (7, 2, 17)
(d) (a, b, c) + (d, e, f)
Answer:
a. (18,8)
b. (3, 6)
c. (10, 10, 10)
d. (a+d, b+e, c+f)
Give me brainliest for this... Thanks
Like chargers attract one another
nope. opposite charges attract. like charges repel.
I need help pls !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
figure d
Explanation:
What kind of workplace is a mine?
When a 600 g mass is suspended from a spring, the spring stretches 1.2 cm. What is the spring constant of the spring?
Well first you need to know the formula for the spring force. It’s -1/2k*d
K being the constant
D being the displacement
We can actually find the force on the spring by calculating the weight of the block. It would be mass multiplied by gravity, convert the grams to kilograms first tho.
600/1000 = 0.6kg
0.6kg * 9.8 m/s^2 = 5.88N
Now plug everything into the equation, also convert the centimeters to meters:
1.2/100 = 0.012 meters
5.88 = -1/2k*(0.012)
5.88 = k*(0.006)
We could drop the negative sign because we really just want the magnitude of the spring constant.
K = 980N
a single frictionless roller coaster car of mass m=750 kg tops the first hill with speed v=15 m/s ar height h =40m as shown
1) Find the speed of the car at B and C
2) if mass m were doubled, would the speed at B increase, decrease, or remain the same?
1. The speed of the roller coaster car at point B and C: are 19.8 m/s and 0 m/s respectively.
2. If mass (m) of the roller coaster car were doubled, the speed at B would increase because energy is dependent on the mass of an object.
Given the following data:
Mass of roller coaster car = 750 kgSpeed of roller coaster car = 15 m/sHeight = 40 meters1. To find the speed of the roller coaster car at point B and C:
First of all, we would determine the potential energy of the roller coaster car by using the formula:
[tex]P.E = mgh[/tex]
Where:
m is the mass of object.g is the acceleration due to gravity ([tex]9.8\;m/s^2[/tex]).h is the height of an object.From the diagram, height at point B = [tex]\frac{h}{2} = \frac{40}{2} = 20[/tex] meters
[tex]P.E = 750(9.8)(20)[/tex]
P.E = 147000 Joules.
Next, we would determine the speed by applying the law of conservation of energy.
[tex]Kinetic\;energy = Potential\;energy\\\\\frac{1}{2} mv^2 = mgh[/tex]
Substituting the values, we have;
[tex]147000 = \frac{1}{2} (750)v^2\\\\147000 = 375v^2\\\\v^2 = \frac{147000}{375} \\\\v^2 = 392\\\\v = \sqrt{392}[/tex]
Speed, v = 19.8 m/s
From the diagram, at point C, the speed is equal to zero (0) meters per seconds.
2. If mass (m) of the roller coaster car were doubled, the speed at B would increase because energy is dependent on the mass of an object.
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Select the correct answer.
Which scientist discovered the relationship between electricity and magnetism?
OA.
Nikola Tesla
OB
Alessandro Volta
OC.
Hans Christian Oersted
OD
Isaac Newton
Answer:
Hans Christian Oersted
Explanation:
A storm is moving east toward your house at an average speed of 35 km an hour if the storm is currently 80 km from your house and how much time do you expect it to arrive
Answer: 2 hours 17 minutes
Explanation: 80/35h = 16/7 = 2hrs 17 minutes
A fundraising company agrees to donate an extra $75 for every $100 the school raises through selling cookies. Part A: What is the constant of proportionality
Answer:
3/4 0r 0.75
Explanation:
Let E represent the extra cash.
Let R represent the amount raised.
From the question given above, we can say that:
The extra cash (E) is directly proportional to the amount raised (R) i.e
Extra cash (E) ∝ Amount raised (R)
E ∝ R
E = KR
NOTE: K is the constant of proportionality.
With the above formula, we can obtain the value of K as follow:
Extra cash (E) = $ 75
Amount raised (R) = $ 100
Constant of proportionality (K) =?
E = KR
75 = K × 100
Divide both side by 100
K = 75/100
K = 3/4 0r 0.75
Thus, the constant of proportionality is 3/4 0r 0.75
This is about acceleration, velocity and all that mumbo jumbo
Which best depicts the data?
1, From 0 to 2 seconds, Anna experienced negative acceleration.
2, From 3 to 6 seconds, Elan accelerated while Anna did not.
3, From 6 to 8 seconds, Elan accelerated faster than Anna.
4, From 8 to 10 seconds, Elan experienced negative acceleration.
Answer:
[tex]3. \: From \: 6 \: to \: 8 \: seconds, \: Elan \: \\ accelerated \: \: faster \: than \: Anna. [/tex]
Tell which is the answer
help please
Answer: a constant? im not sure that just sounds like itd be right.
Explanation:
Answer:
b
Explanation:
because he isn't moving
What is the average velocity of a boy moving from initial position of 4 m to a final position of 8 min 20 s. Show steps of solving:
please please heeeeelp
9514 1404 393
Answer:
0.2 m/s
Explanation:
The average velocity is the change in position divided by the change in time:
(8 m -4 m)/(20 s) = (4/20) m/s = 1/5 m/s = 0.2 m/s
The boy's average velocity is 0.2 m/s.
Imagine that you throw a rock into the air. When you release the rock, it is moving at 19.4 m/s. The rock flies straight up into the air and returns back to your hand. For how long was the rock in the air? Note: acceleration due to gravity is -9.81 m/s/s.
Answer:
4 s
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 19.4 m/s.
Acceleration due to gravity (g) = –9.81 m/s²
Total time (T) =.?
Next, we shall determine the time take by the rock to get to the maximum height. This can be obtained as follow:
NOTE: At maximum height, the final velocity is zero.
Initial velocity (u) = 19.4 m/s.
Acceleration due to gravity (g) = –9.81 m/s²
Final velocity (v) = 0 m/s
Time taken to reach the maximum height (t) =.?
v = u + gt
0 = 19.4 + (–9.81 × t)
0 = 19.4 – 9.81t
Collect like terms
0 – 19.4 = –9.81t
–19.4 = –9.81t
Divide both side by –9.81
t = –19.4 / –9.81
t ≈ 2 s
Thus, it took approximately 2 s for the rock to reach its maximum height.
Finally, we shall determine the total time spent by the rock in the air as follow:
Time taken to reach the maximum height (t) = 2 s
Total time spent in air (T) =?
The total time (T) spent by the rock in the air will be two times the time taken (t) to reach the maximum height i.e
T = 2t
t = 2 s
T = 2 × 2
T = 4 s
Therefore, the rock spent approximately 4 s in the air.
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3. A cat is moving at 18 m/s when it accelerates at 4 m/s? for 2 seconds. What is his new
velocity?
Answer:
26 m/s
Explanation:
v = u + at
Given that,
u(initial speed) = 18 m/sa (acceleration) = 4m/s^2t ( time) = 2sand v ( final velocity) = ?
v = 18 + ( 4 × 2)
v = 18 + 8
v = 26 m/s
One end of a horizontal string is attached to a vibrating blade, and the other end passes over a pulley as in Figure (a). A sphere of mass 1.00 kg hangs on the end of the string. The string is vibrating in its second harmonic. A container of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates in its second harmonic (or the third normal mode of oscillation) as shown in Figure (b). What is the radius of the sphere?
Answer:
when it weighs 2kg it is 7.38
Explanation:
hope this helps
A 100 kg disc with radius 1.6 m is spinning horizontally at 25 rad/s. You place a 20 kg brick quickly and gently on the disc so that it sticks to the edge of the disc. Determine the final angular speed of the disc-brick system.
Answer:
20.83rad/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the mass of the disc and Brick respectively
u1 and u2 are the linear velocities respectively
v is their final velocity of the block-disc system
Given
m1 = 100kg
m2 = 20kg
u1 = wr =25×1.6
u1 = 40m/s
u2 = 0m/s (brick is stationary)
Substitute and the final linear velocity v
100(40)+20(0) = (100+20)v
4000 = 120v
v = 4000/120
v = 33.3m/s
From the formula;
v = wr
w = v/r
w= 33.3/1.6
w = 20.83rad/s.
Hence the final angular speed of the disc-brick system is 20.83rad/s
A net force of 275 N accelerates a 18.0 kg mass. What is the resulting acceleration? (5 points)
Answer:
The answer is 15.28 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
m is the mass
f is the force
We have
[tex]a = \frac{275}{18} \\ = 15.277777...[/tex]
We have the final answer as
15.28 m/s²Hope this helps you
As part of a carnival game, a 5.00 kg target is freely hanging from a very long and very light wire. Contestants can use one of two 1.5 kg balls to try to hit the target and deflect it high enough to win a prize. Ball A will have an elastic collision and bounce back toward you while ball B will have a nearly perfectly inelastic collision, but rather than sticking to the target, the ball will just drop straight downward to the ground after the collision. You can throw each ball with a velocity of 12 m/s. You are the first to try the game and which ball should you throw? Calculate the expected height the target will reach after each is thrown.
Answer:
Height ball A will deflect the target is 2.65 m
Height ball A will deflectthe target is 2.2 m
Ball A, will deflect the target to greater height, thus i will throw ball A
Explanation:
Given;
mass of the target, m₁ = 5.00 kg
mass of the ball, m₂ = 1.5 kg
initial velocity of each throw, u = 12 m/s
Throwing ball A; apply the principle of conservation linear momentum for elastic collision;
m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂
The initial velocity of the target, u₁ = 0
The ball bounced back at the same speed, v₂ = -12 m/s
the velocity of the target after collision, = v₁
0 + 1.5 x 12 = v₁(5) + (-12 x 1.5)
18 = 5v₁ - 18
18 + 18 = 5v₁
36 = 5v₁
v₁ = 36/5
v₁ = 7.2 m/s
The vertical height reached by the target is given by;
v₁² = u₁² + 2gh
v₁² = 0 + 2gh
v₁² = 2gh
h = v₁² / 2g
h = (7.2)² / (2 x 9.8)
h = 2.65 m
Throwing ball B; apply the principle of conservation energy for the inelastic collision;
the kinetic energy of the ball will be converted to the potential energy of the target.
¹/₂m₁u₂² = mgh
¹/₂(1.5)(12)² = (5 x 9.8)h
108 = 49h
h = 108 / 49
h = 2.2 m
Ball A will deflect the target to greater height, thus i will throw ball A.