Answer:
Net force = 15 N
Explanation:
Given that,
Applied force on an object = 20 N
Frictional force = 5 N
We need to find the net force acting on the object.
Friction is an opposing force. It acts in the opposite direction.
Net force = Applied force - Frictional force
= 20 N - 5 N
= 15 N
Hence, the net force acting on the object is 15 N.
A 2 kg toy car moves at a speed of 5 m/s. If a child applies a 5N force for 2 m in the same direction the car is already moving, what is the change in kinetic energy of the car?
Answer:
[tex]10\: \mathrm{J}[/tex]
Explanation:
The kinetic energy of an object is [tex]KE=\frac{1}{2}mv^2[/tex], where [tex]m[/tex] is the mass of the object and [tex]v[/tex] is the velocity of the object.
The toy car's initial kinetic energy is [tex]KE_{i}=\frac{1}{2}\cdot 2\cdot 5^2=25\: \mathrm{J}[/tex].
After the child applies a 5N force on it in the same direction, its velocity will increase but its mass will stay the same.
To find the final velocity of the toy car, we can use kinematic equation [tex]v_f^2=v_i^2+2a\Delta x, \\ v_f=\sqrt{v_i^2+2a\Delta x}[/tex]
We are given [tex]v_i=5\: \mathrm{m/s}[/tex] and [tex]\Delta x = 2\: \mathrm{m}[/tex].
To find acceleration:
[tex]F=ma, a=\frac{F}{m}=\frac{5}{2}=2.5\: \mathrm{m/s^2}[/tex].
Now substitute [tex]v_i=5\: \mathrm{m/s}, \: a=2\: \mathrm{m/s^2}, \: \Delta x = 2\: \mathrm{m}[/tex] into [tex]v_f=\sqrt{v_i^2+2a\Delta x}[/tex] to get [tex]v_f\approx 5.92\: \mathrm{m/s}[/tex].
Using this, we can find the final kinetic energy of the toy car is [tex]KE_f=\frac{1}{2}\cdot 2\cdot 5.92^2[/tex].
Thus, the change in kinetic energy is [tex]KE_f-KE_i=\frac{1}{2}\cdot2\cdot 5.92^2-\frac{1}{2}\cdot 2\cdot 5^2=\fbox{$10\: \mathrm{J}$}[/tex] (one significant figure).
The change in the kinetic energy of the car is 10 J.
The given parameters;
mass of the car, m = 2 kginitial velocity of the car, u = 5 m/sforce applied by the child, F = 5 Ndistance traveled, s = 2 mThe acceleration of the car is calculated as follows;
[tex]F = ma\\\\a = \frac{F}{m} \\\\a = \frac{5}{2} \\\\a = 2.5 \ m/s^2[/tex]
The final velocity of the car is calculated as follows;
[tex]v^2 = u^2 + 2as\\\\v = \sqrt{u^2 + 2as} \\\\v = \sqrt{5^2 \ + \ 2(2.5)(2)} \\\\v = 5.92 \ m/s[/tex]
The change in the kinetic energy of the car is calculated as follows;
[tex]\Delta K.E = \frac{1}{2} m(v^2 - u^2)\\\\\Delta K.E = \frac{1}{2} \times 2 \times (5.92^2\ - \ 5^2)\\\\\Delta K.E = 10 \ J[/tex]
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describe measurement in our daily life
Which image shows the difference between the speed of molecules in hot and cold water? Explain your answer choice.
HELP ME,EVERYONE!!!!!!!! :(
Answer:
the answer is B
Explanation:
I think its B because on the top it shows the molecule speed and A looks like the water is cold, C shows that the hot
water is cooler, and D shows that both are cold
PLZ HELP WILL MARK BRAINLIEST!!
Amy has a mass of 50 kg, and she is riding a skateboard traveling 10 meters per second. What is her momentum?
5 kg·m/s
10 kg·m/s
50 kg·m/s
500 kg·m/s
Answer:
[tex]500 \: \mathrm{kg} \cdot \mathrm{m/s}[/tex]
Explanation:
The momentum of an object is given as [tex]p=mv[/tex]. Since Amy has a mass of 50 kg and is travelling 10 m/s, her momentum is [tex]p=mv=50\cdot 10 =\fbox{$500\: \mathrm{kg\cdot m/s}$}[/tex].
Answer:
500
Explanation:
Some giant ocean waves have a wavelength of 25 m and travel at 6.5 m/s with a frequency of 0.26 HZ. What is the period of such a wave ?
Answer:
3.85s
Explanation:
Given parameters:
Wavelength = 25m
Velocity = 6.5m/s
Frequency = 0.26Hz
Unknown:
Period of the wave = ?
Solution:
The period of a wave is the inverse of the frequency of the wave.
Period = [tex]\frac{1}{frequency}[/tex]
Period = [tex]\frac{1}{0.26}[/tex] = 3.85s
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated as particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/3 of the maximum possible value?
Answer:
Explanation:
Maximum value of force will be possible when both the sphere will have same charge . In that case charge on each sphere = Q / 2 =.5Q
F( max ) = k .5Q x .5Q / R²
=.25kQ² /R²
For the second case
F = k q ( Q-q)/ R²
F = .25kQ² /3R²
.25kQ² /3R² = k q ( Q-q)/ R²
.25 Q² = 3qQ - 3q²
3q² - 3qQ + .25 Q² = 0
q =
How does heat from the sun get to earth?
1. An engine absorbs 600 J of heat while doing 650 J of work. What is the change
in internal energy of the enginge? *
1250 J
-50 J
-1250 J
50 J
The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The absorb energy: Q = 600 Joule
Work done: W = 650 Joule.
Let, the change in internal energy of the engine= dU.
According to conservation of energy:
The absorb energy = change in internal energy + Work done
Q = dU + W
dU = Q - W
= 600 joule - 650 joule
= - 50 joule.
Hence, the change in internal energy of the engine is -50 joule.
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WILL MARK BRAINLIEST Why do we see sedimentary rocks more often than igneous and metamorphic rocks?
O Because they are extrusive
O Because they are intrusive
O because they are the rocks found in the ocean
O because they are the rocks that line the surface of our planet
Answer:
because they are the rocks that line the surface of our planet
Explanation:
We see sedimentary rocks more than other rock types because they are the rocks that line the surface of our planet.
Sedimentary rocks typically form the earth cover due to the way they are formed.
These rocks are produced by the weathering, transportation and deposition of sediments within a basin. In this basin, the sediment is lithified and converted to sedimentary rocks. These processes are driven by the external heat engineTherefore, it is confined to the surface. Igneous and metamorphic rock's processes are confined to the subsurface.A bottle rocket is fired off and has an acceleration of 14.5 m/s2 for the 2.25s until it burns out. If it starts at rest, what distance does it cover?
Answer:
S = 16.3125m
Explanation:
Given the following data;
Acceleration, a = 14.5m/s²
Time, t = 2.25secs
Since the bottle rocket starts from rest, its initial velocity is 0m/s.
To find the distance S, we would use the second equation of motion.
S = ut + ½at²
Substituting into the equation, we have
S = 0(2.25) + ½*14.5*2.25
S = 0 + 7.25*2.25
S = 16.3125m
Therefore, the bottle rocket covered a distance of 16.3125 meters.
Various amplifier and load combinations are measured as listed below using rms values. For each, find the voltage, current, and power gains ( A v , Ai , and Ap, respectively) both as ratios and in dB:
(a) vI= 100 mV, iI = 100 μA, vO = 10 V, RL = 100Ω
(b) vI = 10 μV, iI = 100 nA, vO =1 V, RL= 10 kΩ
(c) vI =1 V, iI = 1 mA, vO =5 V, RL = 10Ω
Answer:
The solution to this question can be defined as follows:
Explanation:
In point (a):
[tex]v_i= 100 \ mV\\\\ i_I = 100 \mu \ A\\\\ v_O = 10 \ V\\\\ R_L = 100 \ \Omega \\\\i_L = \frac{V_0}{R_L} = \frac{10}{100} = 100 \ MA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{100 \times 10^{-3}} =100 \\\\A_v(db) = 20 \lag (100) =40 \ db \\\\ A_i= \frac{i_L}{i_i} = \frac{100 \times 10^{-3}}{100 \times 10^{-6}} =1000 \\\\A_i(db) = 20 \lag (100) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_L}{v_i i_i} = \frac{ 10(100 \times 10^{-3})}{100 \times 10^{-6} \times 100 \times 10^{-3}} =100000\\\\ A_p(db) =10 \log (100000) =50 \ db \\\\[/tex]
In point (b):
[tex]v_i = 10 \mu V\\\\ i_i = 100 \ nA \\\\ v_O =1 \ V \\\\ R_L= 10 \ k \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{1}{10 \ K} = 100 \ \muA \\\\A_v = \frac{V_0}{V_i} = \frac{10}{10 \times 10^{-6}} =100000 \\\\A_v(db) = 20 \lag (100000) =100 \ db \\\\ A_i= \frac{i_0}{i_i} = \frac{100 \times 10^{-6}}{100 \times 10^{-9}} =1000 \\\\A_i(db) = 20 \lag (1000) =60 \ db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 1 \times 100 \times 10^{-6})}{10 \times 10^{-6} \times 100 \times 10^{-9}} =100000000\\\\A_p(db) =10 \log (100000000) =80 \ db \\\\[/tex]
In point (C):
[tex]v_i =1\ V\\\\ i_I = 1 \ mA\\\\ v_O =5\ V \\\\ R_L = 10 \ \Omega \\\\i_0 = \frac{V_0}{R_L} = \frac{5}{10 } = 0.5 \ A \\\\A_v = \frac{V_0}{V_i} = \frac{5}{1} =5 \\\\A_v(db) = 20 \log 5 =13.97 \ db = 14 \db \\\\ A_i= \frac{i_0}{i_i} = \frac{0.5}{1\times 10^{-3}} =500 \\\\A_i(db) = 20 \log (500) =53.97 \ db = 54 \db \\\\[/tex]
[tex]A_p= \frac{P_0}{p_i} =\frac{v_0 i_0}{v_i i_i} = \frac{ 5 \times 0.5 }{1 \times 1 \times 10^{-3}} =2500\\\\A_p(db) =10 \log (2500) = 33.97 \ db = 34 \db\\\\[/tex]
help me help me help me
Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?
Answer:
Explanation:
Expression for fundamental frequency of tone produced in a wire under tension of T and length L is given as follows
[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
m is mass per unit length .
We shall apply this formula for given wires .
For shorter wire
[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
For longer wire for second harmonic
length of wire is 2L , tension is 4T ,
[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]
[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]
f = 2 x 60 = 120 Hz .
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
Particle A with charge q and mass ma and particle B with charge 2q and mass
mb, are accelerated from rest by a potential difference AV and subsequently
deflected by a uniform magnetic field into semicircular paths. The radii of the
trajectories by particle A and B are R and 3R, respectively. The direction of
the magnetic field is perpendicular to the velocity of the particle. Determine
their mass ratio?
what are ribosomes?
I'm tired. But I have insomnia. Big ugh moment. <.<.
Answer:
Ribosomes are organelles the make protein for the cell.
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
Suppose it takes a constant force a time of 6.0 seconds to slow a 2500 kg truck
from 26.0 m/sec to 18.0 m/sec. What is the magnitude of the force? Give
your answer in scientific notation rounded correctly.
Answer:
[tex]3.3\cdot 10^3\:\mathrm{N}[/tex]
Explanation:
Impulse on an object is given by [tex]\mathrm{[impulse]}=F\Delta t[/tex].
However, it's also given as change in momentum (impulse-momentum theorem).
Therefore, we can set the change in momentum equal to the former formula for impulse:
[tex]\Delta p=F\Delta t[/tex].
Momentum is given by [tex]p=mv[/tex]. Because the truck's mass is maintained, only it's velocity is changing. Since the truck is being slowed from 26.0 m/s to 18.0 m/s, it's change in velocity is 8.0 m/s. Therefore, it's change in momentum is:
[tex]p=2500\cdot 8.0=20,000\:\mathrm{kg\cdot m/s}[/tex].
Now we plug in our values and solve:
[tex]\Delta p=F\Delta t,\\F=\frac{\Delta p}{\Delta t},\\F=\frac{20,000}{6}=\fbox{$3.3\cdot 10^3\:\mathrm{N}$}[/tex](two significant figures).
b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?
Answer:
Twice.
Explanation:
The momentum of an object is given by :
p = mv
Where
m is mass and v is the velocity
If the mass of the ball were doubled, m'=2m and v'=v=3 m/s
New momentum,
p'=m'v'
p'=2m × v
p'=2mv
or
p'=2p
So, the new momentum becomes twice the initial momentum.
Sultan walks for 15 km at 35° south of east.
Which of the following journeys would result in the same displacement?
Answer:
☝
Explanation:
I will give brainly
Defend Democritus' work on the atom and its contribution to the modern atomic model.
If two objects have different temperatures when they come in contact, heat will flow from the warmer
object to the cooler one UNTIL
Answer:
They both have an equal temperature.
Explanation:
Any material or object that allow the conduction (transfer) of electric charge or thermal energy is generally referred to as a conductor. Some examples of a conductor are metals, copper, aluminum, graphite, etc.
In the process of heat conduction, thermal energy is usually transferred from fast moving particles to slow moving particles during the collision of these particles. Also, thermal energy is typically transferred between objects that has different degrees of temperature and materials (particles) that are directly in contact with each other but differ in their ability to accept or give up electrons.
Hence, if two objects have different temperatures when they come in contact, heat will flow from the warmer object to the cooler one until they both have an equal temperature.
Why is damage from sound waves is an issue on the launchpad but not in the air
(I would have done more points for answering but I'm almost out sry. just pls answer and help.)
The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.
When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.
Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.
Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
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What is the instantaneous velocity of a freely falling object 11 s after it is released from a position of rest
Answer:
v= -107.8 m/s
Explanation:
Since the object is in free fall, this means that is moving at an accceleration equal to the one due to gravity.Since it starts at rest, we can apply the definition of acceleration, rearranging terms as follows:[tex]v_{f} = v_{o} + a*t = a*t = -g*t = 9.8m/s2*11s = -107.8 m/s (1)[/tex]
(Assuming as positive the upward direction)A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
David is driving a steady 28.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90 m/s^2 at the instant when David passes.
Required:
a. How far does Tina drive before passing David?
b. What is her speed as she passes him?
Answer:
Explanation:
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion S = ut + 1/2at^2..... 1
For David u = 28.0 m/s where 'a' is set to nought
S = ut
S = 28t.......2
For Tina consider equation 1
Where acceleration = 2.90m/s^2 and u is set at nought
S = 1/2×2.90 m/s×t^2.......3
Equate 2 and 3
28t = 1.45t^2
Divide through by t
28 = 1.45t
t = 28/1.45
t = 19.31seconds
Now put the value of t into equation 3
S = 1/2×2.90 m/s×t^2.......3
= 1.45×20×20
= 580m
Tina must have driven 580meters before passing David
Considering the equation of linear motion : V^2 = U^2+2as
Where u is set at nought
V^2 = 2as
V^2 = 2×2.9×580
V^2 = 3364
V = √3364
V = 58m/s
Her speed will be 58m/s
(a) Tina should drive for 580 m, before passing the David.
(b) The speed of Tina during her passage through the David is 58 m/s.
Given data:
The initial velocity of the David is, u = 28.0 m/s.
The magnitude of acceleration is, [tex]a = 2.90 \;\rm m/s^{2}[/tex].
(a)
We can use the second kinematic equations of motion to obtain the distance covered by Tina, before passing the David. As per the second kinematic equation of motion,
[tex]s= u't + \dfrac{1}{2}at^{2}[/tex]
Here, u' is the initial speed of Tina and t is the time interval. Then,
Let t represent the time for Tina to catch David.
Hence, considering the equation of linear motion as,
S = ut + 1/2at²...............................................................(1)
Also,
S = ut
S = 28t ...........................................................................(2)
For Tina consider equation 1
S = 1/2×2.90t²................................................................(3)
Equate 2 and 3
28t = 1.45t²
28 = 1.45t
t = 28/1.45
t = 19.31 seconds
Now put the value of t into equation (3)
S = 1/2×2.90 t².
= 1.45×20×20
= 580m
Thus, we can conclude that Tina should drive for 580 m, before passing the David.
(b)
Now, using the third kinematic equation of motion to obtain the speed of Tina during her passage through David as,
v² = u²+2as
Solving as,
v² = 28.0² + 2(2.90)(580)
v = √3364
v = 58m/s
Thus, we can conclude that the speed of Tina during her passage through the David is 58 m/s.
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Solve each of the following problems to 3 sig figs and correct Sl units, showing all work.
1. A cart with a mass of 45.0 kg is being pulled to the right with a force of 250 N giving it an
acceleration of 1.30 m/s2. The wheels of the cart are locked and the cart must be dragged.
a) Draw a free body diagram of the cart.
b) Calculate the net force acting on the cart.
c) Create a force table and fill it in.
d) Find the coefficient of kinetic friction.
Answer:
every number to 3 sf = 1) 45.0 2) 250 3) 1.30
Explanation:
your welcome :)
PLEASEEEEEE HELPPPPPPP
Define resistance and discuss how it affects current.
Answer:
Resistance is the opposing of the flow of current through a conductor.
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
How long ago did most Middle Eastern countries gain their independence?
A.
10-20 years ago
B.
50-100 years ago
C.
200-300 years ago
D.
400-500 years ago
Please select the best answer from the choices provided
A
B
C
D
Answer:
the correct answer is B
Explanation:
Answer:
.B
Explanation:
This is the answer on edge 2021
Have a good day!
