Answer:
"42 m/s" is the appropriate answer.
Explanation:
Given:
Initial speed of Ball 1,
= 14 m/s
Now,
If the initial speed of ball 1 is doubled, then the speed of ball 2 will be:
⇒ [tex](V_{fx})_2=\frac{2m_1}{m_1+m_2}(v_1x_1)[/tex]
hence,
Doubling ([tex]v_1x_1[/tex]) will double [tex](V_{fx})_2[/tex],
⇒ [tex](V_{fx})_2=2\times 21[/tex]
[tex]=42 \ m/s[/tex]
An iron nail having threads along its cylindrical surface is
Answer:
A screw is a mechanism that converts rotational motion to linear motion, and a torque (rotational force) to a linear force.[1] It is one of the six classical simple machines. The most common form consists of a cylindrical shaft with helical grooves or ridges called threads around the outside.[2][3] The screw passes through a hole in another object or medium, with threads on the inside of the hole that mesh with the screw's threads. When the shaft of the screw is rotated relative to the stationary threads, the screw moves along its axis relative to the medium surrounding it; for example rotating a wood screw forces it into wood. In screw mechanisms, either the screw shaft can rotate through a threaded hole in a stationary object, or a threaded collar such as a nut can rotate around a stationary screw shaft.[4][5] Geometrically, a screw can be viewed as a narrow inclined plane wrapped around a cylinder.
Explanation:
A truck starts from rest with an acceleration of 0.3 m/ S^2 find its speed in km/h when it has moves through 150 m
Answer:
9.5 m/s
Explanation:
Distance, S = 150m
Acceleration, a = 0.3 m/s^2
Initial velocity, u = 0 m/s
Final velocity, v
Use kinematics equation
v^2 - u^2 = 2aS
v^2 - 0 = 2*0.3*150 = 90
v = sqrt(90) = 9.49 m/s
A table is moved using 60 N of force.
How far is the table moved if 900 J of work is done on the table?
960 m
840 m
15 m
0.06 m
Answer:
15m
Explanation:
W=f×s
[tex]s = \frac{w}{f} \\ s = \frac{900j}{60n} \\ s = 15m [/tex]
Given:
Force,
F = 60 NWork done,
W = 900 JWe know,
→ [tex]W = F\times s[/tex]
or,
→ [tex]s = \frac{W}{F}[/tex]
By putting the values, we get
[tex]= \frac{900}{60}[/tex]
[tex]= 15 \ m[/tex]
Thus the above response i.e., "Option c" is right.
Learn more about work done here:
https://brainly.com/question/8625856
If an electron (with a charge of 1.6 x10−19c) Experiences a force of 500 N at a certain point in an electric field, then find the strength of the electric field in that location
Answer:
3.125×10²¹ N/C
Explanation:
Electric Field: This can be defined as the force experienced per unit charge. The S.I unit of electric Field is N/C
Applying,
E = F/q.................. Equation 1
Where E = Electric Field, F = Force experienced, q = Charge of an electron.
From the question,
Given: F = 500 N, q = 1.6×10⁻¹⁹ C
Substitute these values into equation 1
E = 500/(1.6×10⁻¹⁹)
E = 312.5×10¹⁹
E = 3.125×10²¹ N/C
Suppose you have 10 helium-filled party balloons tied to a digital camera, so you can take photos of the core of a storm. With 10 balloons the camera will slowly but surely rise. As a storm approaches the atmospheric pressure drops. Assuming the air temperature and density are the same, the volume of each balloon will increase. When you let go, the speed of ascent will be
Answer:
a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]
Explanation:
Let's solve this problem in parts, with the initial data the camera rises slowly, so we can assume that at constant speed, we apply the equilibrium condition
B - W = 0
The thrust is given by Archimedes' law
B = rho_aire g V_body
The volume of the body can be found from the ideal gas ratio
P V = n R T
let's use the subscript "o" for the initial concisions
V₀ = (nR) T₀/P₀ 1
we substitute
10 ρ_air g (nR) T₀ /P₀ = W
when the storm approaches the pressure decreases, P
If we use the ideal gas equation
V = (nR) T₀ / P
we combine this equation with equation 1
V = V₀P₀ / P
if we write the pressure
P = P₀ -ΔP
we substitute
V = [tex]\frac{V_oP_o}{P_o - \Delta P} =V_o \ ( 1 - \frac{\Delta P}{P_o} )^{-1}[/tex]
we expand serie and eliminate higher order terms
V = V₀ ([tex]1+ \frac{\Delta P}{Po}[/tex])
with this expression we can write the thrust
B = B₀ + ΔB
Newton's second law for the new conditions is
B - W = m a
(B₀ + ΔB) - W = ma
ΔB = m a
a = ΔB / m
a = 10 ρ_air g [tex]\frac{\Delta P}{m P_o}[/tex]
This is the initial acceleration of the camera
A stunt performer falls off a wall that is 1.6 m high and then lands on a mat.
What is his impact velocity?
A. 5.6 m/s
B. 1.1 m/s
C. 4.7 m/s
o
D. 2.9 m/s
According to law of conservation of energy,
= ½mv² = mgh
= mv² = 2mgh
= v² = 2mgh/m
= v = √2gh
So, now just put the values of g & h, abd you are done;
= v = √2×9.8×1.6
= v = √31.36
= v = 5.6 m/s
A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same material and length stretch if its cross-sectional area is 8.00 mm2 and the same force is used to stretch it
Answer: [tex]0.05\ mm[/tex]
Explanation:
Given
Cross-sectional area of wire [tex]A_1=4\ mm^2[/tex]
Extension of wire [tex]\delta l=0.1\ mm[/tex]
Extension in a wire is given by
[tex]\Rightarrow \delta l=\dfrac{FL}{AE}[/tex]
where, [tex]E=\text{Youngs modulus}[/tex]
[tex]\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)[/tex]
for same force, length and material
[tex]\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)[/tex]
Divide (i) and (ii)
[tex]\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm[/tex]
A 0.2 kg hockey park is sliding along the eyes with an initial velocity of -10 m/s when a player strikes it with his stick, causing it to reverse its direction and giving it a velocity of +25 m/s the impulse the stick applies to the park is most nearly
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
[tex]I = m\cdot (\vec{v}_{2} - \vec{v_{1}})[/tex] (1)
Where:
[tex]I[/tex] - Impulse, in kilogram-meters per second.
[tex]m[/tex] - Mass, in kilograms.
[tex]\vec{v_{1}}[/tex] - Initial velocity of the hockey park, in meters per second.
[tex]\vec{v_{2}}[/tex] - Final velocity of the hockey park, in meters per second.
If we know that [tex]m = 0.2\,kg[/tex], [tex]\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right][/tex] and [tex]\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right][/tex], then the impulse applied by the stick to the park is approximately:
[tex]I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right][/tex]
[tex]I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
please help
a girl pulls a wheeled suitcase with a force of 3N. If the suitcase has a mass of 6 kg, what is the acceleration?
Explanation:
Start with what you know and list your knowns and unknowns
F = ma
F= 3N
m = 6kg
a =?
3N = 6kg x a
solve for a
3N / 6kg = a
plzzzz urgent
solve this
Answer:
[tex] \large{ \tt{☄ \: EXPLANATION}} : [/tex]
Before solving , You'll have to know - When an object starts from the state of rest , in this case , initial velocity ( u ) = 0Notice that we're provided the time ( t ) in minutes. So , first thing we have to do is convert the minutes into seconds. It would be - Time ( t ) = 5 minutes = 5 × 60 sec = 300 sec [ 1 min = 60 sec ]Here , We're provided - Initial velocity ( u ) = 0 , Final velocity ( v ) = 60 m / s , Time taken ( t ) = 300 seconds & We're asked to find out the acceleration ( a ) & distance covered by the jeep ( s ) .[tex] \large{ \tt{♨ \:LET'S \: START}} : [/tex]
Acceleration is defined as the rate of change of velocity. We know :[tex] \large{ \boxed{ \tt{❁ \: ACCELERATION \: (a) = \frac{FINAL \: VELOCITY(v) - INITIAL \: VELOCITY(u)}{TIME \: TAKEN \: ( \: t \: )}}}} [/tex]
- Plug the values & then simplify !
[tex] \large{ \bf{↬a = \frac{60 - 0}{300} = \frac{60}{300} = \boxed{ \bold{ \bf{0.2 \: m {s}^{ - 2} }}} }}[/tex]
The acceleration of the jeep is 0.2 m/s²[tex] \large{ \tt{۵ \: AGAIN, \: USING\: SECOND \: EQUATION \: OF \: MOTION}} : [/tex]
[tex] \boxed{ \large{ \bf{✾ \: s = \frac{u + v}{2} \times t}}}[/tex]
- Plug the values & then simplify !
[tex] \large{ \bf{↦s = \frac{0 + 60}{2} \times 300 = \boxed{ \bold{ \bf{9000 \: m}}}}}[/tex]
The distance covered by the jeep is 9000 m .❃ The days that break you are the days that make you ! ♪
♡ Hope I helped! ツ
☃ Have a wonderful day / evening ! ☼
# StayInAndExplore ! ☂
▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁
Answer:
What she said is right I know it is
A tank has the shape of an inverted circular cone with height 16m and base radius 3m. The tank is filled with water to a height of 9m. Find the work required to empty the tank by pumping all of the water over the top of the tank. Use the fact that acceleration due to gravity is 9.8 m/sec2 and the density of water is 1000kg/m3. Round your answer to the nearest kilojoule.
Answer:
[tex]W=17085KJ[/tex]
Explanation:
From the question we are told that:
Height [tex]H=16m[/tex]
Radius [tex]R=3[/tex]
Height of water [tex]H_w=9m[/tex]
Gravity [tex]g=9.8m/s[/tex]
Density of water [tex]\rho=1000kg/m^3[/tex]
Generally the equation for Volume of water is mathematically given by
[tex]dv=\pi*r^2dy[/tex]
[tex]dv=\frac{\piR^2}{H^2}(H-y)^2dy[/tex]
Where
y is a random height taken to define dv
Generally the equation for Work done to pump water is mathematically given by
[tex]dw=(pdv)g (H-y)[/tex]
Substituting dv
[tex]dw=(p(=\frac{\piR^2}{H^2}(H-y)^2dy))g (H-y)[/tex]
[tex]dw=\frac{\rho*g*R^2}{H^2}(H-y)^3dy[/tex]
Therefore
[tex]W=\int dw[/tex]
[tex]W=\int(\frac{\rho*g*R^2}{H^2}(H-y)^3)dy[/tex]
[tex]W=\rho*g*R^2}{H^2}\int((H-y)^3)dy)[/tex]
[tex]W=\frac{1000*9.8*3.142*3^2}{9^2}[((9-y)^3)}^9_0[/tex]
[tex]W=3420.84*0.25[2401-65536][/tex]
[tex]W=17084965.5J[/tex]
[tex]W=17085KJ[/tex]
'
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A 15-watt bulb is connected to a circuit that has a total of 60. Ω of resistance. How many electrons are passing through that bulb every second?
Answer:
3.2075*10^16
Explanation:
Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.
A train accelerates from 30 km/h to 45 km/h in 15.0 second. Find its acceleration and the distance it travels during this time
Answer:
a. Acceleration, a = 0.28 m/s²
b. Distance, S = 156 meters
Explanation:
Given the following data;
Initial velocity = 30 km/h
Final velocity = 45 km/h
Time = 15 seconds
a. To find the acceleration;
Conversion:
30 km/h to m/s = 30*1000/3600 = 8.33 m/s
45 km/h to m/s = 45*1000/3600 = 12.5 m/s
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation;
[tex]a = \frac{12.5 - 8.3}{15}[/tex]
[tex]a = \frac{4.2}{15}[/tex]
Acceleration, a = 0.28 m/s²
b. To find the distance travelled, we would use the second equation of motion given by the formula;
[tex] S = ut + \frac {1}{2}at^{2}[/tex]
Where;
S represents the displacement or height measured in meters.
u represents the initial velocity measured in meters per seconds.
t represents the time measured in seconds.
a represents acceleration measured in meters per seconds square.
Substituting into the equation, we have;
[tex] S = 8.3*15 + \frac {1}{2}*(0.28)*15^{2}[/tex]
[tex] S = 124.5 + 0.14*225[/tex]
[tex] S = 124.5 + 31.5 [/tex]
S = 156 meters
Please Help with thia question
Answer:
the types of force shown in the picture is
balanced force.
Science question, can someone please help ? I’ll give brainliest !!
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10 nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10-5 C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell wall is filled with air. What is the magnitude of the electric field between the membranes
Answer:
E = 1.1 10⁶ N / C
Explanation:
In this case they indicate that we can approximate the membrane as a parallel plate capacitor, we can use
E = [tex]\frac{\sigma}{\epsilon_o }[/tex]
note that in this case the electric field created by each plate goes in the same direction, they are added
let's calculate
E = [tex]\frac{10^{-5}}{8.85 \ 10^{-12}}[/tex]
E = 1.1 10⁶ N / C
When a wave steepens until it collapses it becomes a ________. wave of oscillation forced wave breaker wave of translation swell
Answer:
Breaker
Explanation: A wave is a motion or disturbance that transfers energy from one location to another without any permanent displacement of the particles involved in the wave motion.
Wave motion can occur in various media such as water, air etc a wave is described as a breaker when it steepens and before it finally stops or losses its energy/collapses.
A 5 kg mass compresses a horizontal spring by .06 meters. The spring has a spring constant of 2 N/m. If the surface is frictionless, find the velocity of the mass when the spring is released.
Answer:
Explanation:
The frictionless surface implies that the speed of the spring is at a max. When the speed of the spring is at its max, the potential energy in the spring is 0. Use the equation for the Total Energy in a Spring/Mass System:
KE + PE = [tex]\frac{1}{2}kA^2[/tex] where KE is the Kinetic Energy available to the spring, PE is the potential energy available to the spring, and the sum of those is equal to one-half times the spring constant, k, times the amplitude of the spring's movement away from the equilibrium position. Sometimes this amplitude is the same as the displacement of the spring. This can be tricky. But since we are only given one value for the distance, we are going to use it as an amplitude. Keeping in mind that the PE is 0 when KE is at its max, then the equation becomes
KE + 0 = [tex]\frac{1}{2}kA^2[/tex] or to put it simpler terms:
KE = [tex]\frac{1}{2}kA^2[/tex] We need to find the value for KE before we can fully solve the problem we are being tasked with.
Filling in using the info given:
[tex]KE=\frac{1}{2}(2.0)(.06)^2[/tex] Notice I added another place of significance to the 2 because 1 simply isn't enough and the physics teacher in me can't handle that. Simplifying a bit:
[tex]KE=(.06)^2[/tex] because the k = 2 cancels out the 2 in the denominator of the 1/2. So
KE = 3.6 × [tex]10^{-3[/tex]
Now plug that in for KE and solve for v:
KE = [tex]\frac{1}{2}mv^2[/tex]:
[tex]3.6*10^{-3}=\frac{1}{2}(5.0)v^2[/tex] and
[tex]v=\sqrt{\frac{2(3.6*10^{-3})}{5.0} }[/tex] gives us a velocity of
v= [tex]3.8*10^{-2[/tex]
car moves a distance of 420 m. Each tire on the car has a diameter of 42 cm. Which shows how many revolutions each
tire makes as they move that distance?
Plzzzz help asap
Answer:
10 is the correct answer
Answer:
Total Distance: 420 meters
Diameter: 42 cm
Notice the units meters vs cm
420÷ 42 = 10 total revolutions
In a pinball machine, the launching spring with a spring constant of 30 N/m is compressed 0.15 m. How fast will it launch a 0.20 kg pinball?
Answer:
1.84 m/s
Explanation:
Applying,
The kinetic energy of the pinball = Elastic energy of the spring
mv²/2 = ke²/2
mv² = ke².................. Equation 1
Where m = mass of the pin ball, v = velocity of the pin ball, k = force constant of the spring, e = extension of the spring.
make v the subject of the equation
v = √(ke²/m)................ Equation 2
From the question,
Given: e = 0.15 m, k = 30 N/m, m = 0.20 kg
Substitute these values into equation 2
v = √[(30×0.15²)/0.2]
v = 1.84 m/s
30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pushing the box with a horizontal force of 35N. What is the magnitude of the force of static friction between the box and floor
Answer:
84.05
Explanation:
F=mg×0.25F=20x9.81×0.25f=49.05NF=35NF=f+F
F=49.05+35
=84.05
A bike, a truck, and a train—all without passengers, motors, or engines—roll down the same hill. Put the vehicles in order from the least amount of motion energy to the most.
Answer:
Train Bike Truck
Explanation:
what is the dimension formula of power and pressure
Answer:
Power = ML²T⁻³
Pressure = ML⁻¹T⁻²
Explanation:
Applying,
Power(P) = Work(W)/Time(t)
P = W/t..................... Equation 1
But
W = Fd............. Equation 2
Where F = force, d = distance
Also,
F = ma.............. Equation 3
Where m = mass, a = acceleration.
Also,
a = v/t................ Equation 4
Where v = velocity
Also,
v = d/t............... Equation 5
Where d = distance
Substitute equation 5 into equation 4
a = d/t²................. Equation 6
Substitute equation 6 into equation 3
F = m(d/t²)........... Equation 7
Susbtitute equation 7 into equation 2
W = m(d/t²)×d
W = md²/t²........... Equation 8
Substitute equation 8 into equation 1
P = (md²/t²)/t
P = md²/t³............ Equation 9
In dimension,
mass(m) = M, distance(d) = L, time(t) = T
Substitute into equation 9
P = ML²/T³
P = ML²T⁻³
And
Pressure(R) = Force(F)/Area(A)
R = F/A................ Equation 10
F = md/t²,
A = d²
Susbtitute into equation 10
R = (md/t²)/d²
R = m/t²d
Therefore,
R = ML⁻¹T⁻²
Which of these describes a real image?
As a ball is released from 10ft above the ground, it falls freely (without friction)
toward the ground, at what point does the ball have the maximum gravitational
potential energy?
Can't be determined
At the top, at the release point.
Half
way
down the ground
At the floor
Answer:
Explanation:
The formula to find Potential Energy is PE = mgh, where m is the mass of the object, g is gravity, and h is the height from which the object can potentially fall. Because this is linear, then PE will increase as either the mass or the height increase (gravity is constant at 9.8 m/s/s). If the mass of the ball being dropped doesn't change, then the only thing that determines this ball's max PE is the height from which it is dropped; max PE ALWAYS occurs at the highest point from which an object can potentially fall. So your answer is "At the top".
what do you mean by work?
Work is transfer of energy in an object when it travel some distance by external force,
Work= force × displacement
what is the direction of acceleration due to gravity ?
The direction of acceleration due to gravity is always towards earth, going downwards.
mark me brainliesttt :))
A football player kicks a football in a field goal attempt. When the football reaches its maximum height, what is the relationship between the direction of the velocity and acceleration vectors
Answer:
The correct answer is - At the maximum height, the velocity and acceleration vectors are perpendicular to each other.
Explanation:
When the football reaches maximum height, then the vertical component of velocity will be zero and therefore the only component of velocity left will be the horizontal component, and acceleration of the object will be downward, due to gravity.
So at the maximum height, there is horizontal velocity only which means velocity is horizontal and acceleration is vertical thus, the velocity and acceleration are perpendicular to each other.
When electrons flow through wires from a terminal to a terminal a/an _____
created
Answer:
When electrons flow through wires from a terminal to a terminal circuit is created.
The Big Bang Theory states that
A. the universe is continuously
expanding
B. all of the above
C. the universe was created by an
explosion
D. all matter and energy in the
universe was created
Answer:
All of above
Explanation: