We know, acceleration is given by :
a = F/m
a = 16.6/3.1 m/s²
a = 5.35 m/s²
Now, we need to find the speed after it has moved 3.8 m.
By equation of motion :
[tex]v^2 -u^2 = 2as\\\\v=\sqrt{2as}\\\\v = \sqrt{2\times 5.35 \times 3.8 }\ m/s\\\\v = 6.38\ m/s[/tex]
Therefore, the speed after it has moved 3.8 m is 6.38 m/s.
Which phenomena result from something other than the flow of thermal energy inside Earth? A-volcanic eruptions B-earthquakes C-thunderstorms D-valley formations
Answer:
D
Explanation:
Valley Formations
What is the magnitude of the force required to stretch or compress a spring of spring constant 50.0 N/m a distance 2.50 cm from its unstrained length?
a. 50.0 N
b. 125 N
c. 20.0 N
d. 1.25 N
Answer:
Magnitude of the force = 1.25 N
Explanation:
Given:
Spring constant k = 50.0 N/m
Distance s = 2.50 cm = 0.025 m
Find:
Magnitude of the force
Computation:
Magnitude of the force = Ks
Magnitude of the force = 50 x 0.025
Magnitude of the force = 1.25 N
On what factor does heat depend?
Answer:
the change in temperature..............
Jane is riding in a hot air balloon that is rising vertically at a constant speed of 3 m/s over a lake. She drops a rock from the balloon when the distance from the rock to the water is 50 m. Use g m/s^2, and let the up direction be positive. How long after Jane drops the rock will it splash into the water?
It is given that up direction be positive.
Ball velocity at the time is was dropped is 3 m/s.
g = -9.8 m/s²
s = -50 m
Now, using equation of motion for time :
s = ut + at²/2
3t - 4.9t² + 50 = 0
4.9t² - 3t - 50 = 0
Solving above equation, we get :
t = 3.52 s
Therefore, the rock will splash into the water after 3.52 s.
A solid disk and a hoop are simultaneously released from rest at the top of an incline and roll down without slipping. Which object reaches the bottom first?
A) The disk arrives first.
B) The hoop arrives first.
C) The one that has the largest radius arrives first.
D) The hoop and the disk arrive at the same time.
E) The one that has the largest mass arrives first.
Answer:
A) The disk arrives first
Explanation:
The disk arrives first, and the reason behind this is that the disk posses smaller moment of inertia, picking up quickly of angular speed of an object for a particular torgue can be determined by ts moment of inertia . So small quantity of energy will be used up as rotational energy, and with the rotational energy the disk will move faster that the hoop.
The option that illustrates the object that reaches the bottom first is A. The disk arrives first.
From the information given, we're told that a solid disk and a hoop are simultaneously released from rest at the top of an incline and rolled down without slipping.
It should be noted that the disk will arrive first due to the fact that it has smaller inertia. In such a case, the disk will arrive first while the hoop arrives later.
Read related link on:
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A scientist conducted research and plotted his results in the histogram shown below. Which of the
following is the best measure of the center of the data?
Median
Mode
Mean
They are all appropriate
Answer:
Median
Explanation:
we commonly use mean as the measure of the center of the data.But Median is better measure than mean.Extra information:-
Median formula:-
[tex]\boxed{\begin{minipage}{6cm}$\bigstar$\:\:\sf Median = l + $\sf\dfrac{\frac{n}{2}-C.f.}{f}\times h\\\\Here: \\1)\:n = \sum f =\\2)\:l=Lower\:limit\:of\:median\:class=\\3)\:C.f.=Cumulative\:frequency\:of\:class\\preceeding\:the\:median\:class=\\4)\:f= frequency\:of\:median\:class=\\5)\:h= Class\:interval = \end{minipage}}[/tex]
Answer:
The answer is above this
Explanation:
Light of wavelength 503 nm illuminates a round 0.2-mm diameter hole. A screen is placed 4.9 m behind the slit. What is the diameter of the central bright area on the screen in millimeters
Answer:
30 mm
Explanation:
[tex]\lambda[/tex] = Wavelength of light = 503 nm
d = Diameter of hole = 0.2 mm
D = Distance of screen from light source = 4.9 m
Diameter of the central bright area is given by
[tex]x=\dfrac{2.44\lambda D}{d}\\\Rightarrow x=\dfrac{2.44\times 503\times 10^{-9}\times 4.9}{0.2\times 10^{-3}}\\\Rightarrow x=0.03\ \text{m}=30\ \text{mm}[/tex]
The diameter of the central bright area is 30 mm.
A 0.54 kg particle has a speed of8.0 m/s at point A and kineticenergy of 7.5 J at point B.
(a) What is its kinetic energy at A? J
(b) What is its speed at point B?m/s
(c) What is the total work done on the particle as it moves fromA to B? J
Answer:
a) The translational kinetic energy of the particle at point A is 17.28 joules.
b) The speed of the particle at point B is approximately 5.270 meters per second.
c) The total work done on the particle as it moves from A to B is - 9.78 joules.
Explanation:
Let be this particle a conservative system, that is, that non-conservative forces (i.e. friction, viscosity) are negligible.
a) The translational kinetic energy of the particle ([tex]K[/tex]), measured in joules, is determined by the following formula:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Speed, measured in meters per second.
If we know that [tex]m = 0.54\,kg[/tex] and [tex]v = 8\,\frac{m}{s}[/tex], the translational kinetic energy at point A is:
[tex]K = \frac{1}{2}\cdot (0.54\,kg)\cdot \left(8\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 17.28\,J[/tex]
The translational kinetic energy of the particle at point A is 17.28 joules.
b) The speed of the particle is clear in (1):
[tex]v = \sqrt{\frac{2\cdot K}{m} }[/tex]
If we know that [tex]K = 7.5\,J[/tex] and [tex]m = 0.54\,kg[/tex], then the speed of the particle at point B:
[tex]v = \sqrt{\frac{2\cdot (7.5\,J)}{0.54\,kg} }[/tex]
[tex]v\approx 5.270\,\frac{m}{s}[/tex]
The speed of the particle at point B is approximately 5.270 meters per second.
c) According to the Work-Energy Theorem, the total work done on the particle as it moves from A to B ([tex]W_{A\rightarrow B}[/tex]), measured in joules, is equal to the change in the translational kinetic energy of the particle. That is:
[tex]W_{A\rightarrow B} = K_{B}-K_{A}[/tex] (2)
If we know that [tex]K_{A} = 17.28\,J[/tex] and [tex]K_{B} = 7.50\,J[/tex], then the change in the translational kinetic energy of the particle is:
[tex]W_{A\rightarrow B} = 7.50\,J-17.28\,J[/tex]
[tex]W_{A\rightarrow B} = -9.78\,J[/tex]
The total work done on the particle as it moves from A to B is - 9.78 joules.
A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the highest frequency associated with this portion of the spectrum
Answer:
1.5* 10¹⁵ Hz
Explanation:
In any electromagnetic wave (assuming it's propagating in vacuum), the propagation speed is just c, the speed of light.As in any wave, there exists a fixed relationship between speed (c), frequency (f) and wavelength (λ), as follows:[tex]c = \lambda * f (1)[/tex]
Since c is an universal constant, we can see that there exists an inverse relationship between λ and f, so if the lower the wavelength, the higher the frequency.So, in this case, the highest frequency will be the one associated with the lowest wavelength, i.e., 200 nm.Solving for f from (1), replacing by the values, we get:[tex]f = \frac{c}{\lambda} = \frac{3e8m/s}{200e-9m} = 1.5e15 1/sec = 1.5e15 Hz (2)[/tex]
A student lifts a box of books 2 meters with a force of 45 N. He then carries the box 10 meters to the living room. What is the total amount of work done in this situation?
PLEASE ANSWER FAST
Answer:
90J
Explanation:
The only time work is being done is when he lifts the box off the ground. Therefore, using the work formula, 2 x 45, you get 90J. Hope this helps someone.
A police car is moving to the left at 34m/sec while its siren is emitting a 600Hz tone. What tone would you hear if you were parked along side the road
Answer:
[tex]545.89\ \text{Hz}[/tex]
Explanation:
[tex]v_o[/tex] = Velocity of observer = 0
[tex]v_s[/tex] = Velocity of source = 34 m/s
v = Speed of sound in air = 343 m/s
f = Emitted frequency = 600 Hz
Doppler effect will be observed
[tex]f_o=f(\dfrac{v+v_o}{v+v_s})\\\Rightarrow f_o=600(\dfrac{343+0}{343+34})\\\Rightarrow f_o=545.89\ \text{Hz}[/tex]
The frequency heard will be [tex]545.89\ \text{Hz}[/tex].
can a body have two acceleration at a time?
Answer:
No.
Explanation:
Acceleration of an object is defined as the rate of change of velocity per unit time. It can be given by :
a = dv/dt
dv is the change in velocity i.e. final velocity-initial velocity
dt is change in time
At a particular time, its change in velocity can be one only. It would imply that a body cannot have two acceleration at a particular time.
Calculate the magnitude of the normal force on a 15.2 kg block in the following circumstances. (Enter your answers in N.) (a) The block is resting on a level surface. __________ N (b) The block is resting on a surface tilted up at a 35.8 angle with respect to the horizontal. __________ N (c) The block is resting on the floor of an elevator that is accelerating upward at 3.53 m/s ___________ N (d) The block is on a level surface and a force of 155 N is exerted on it at an angle of 35.8 below the horizontal. _________N
Answer:
[tex]149.112\ \text{N}[/tex]
[tex]120.94\ \text{N}[/tex]
[tex]302.77\ \text{N}[/tex]
[tex]58.44\ \text{N}[/tex]
Explanation:
m = Mass of block = 15.2 kg
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
[tex]\theta[/tex] = Angle
a = Acceleration
F = Applied force
Normal force is given by
[tex]N=mg\\\Rightarrow N=15.2\times 9.81\\\Rightarrow N=\boldsymbol{149.112\ \mathbf{N}}[/tex]
[tex]\theta=35^{\circ}[/tex]
[tex]N=mg\cos\theta\\\Rightarrow N=15.2\times 9.81\times \cos35.8^{\circ}\\\Rightarrow N=\mathbf{120.94\ N}[/tex]
[tex]a=3.53\ \text{m/s}^2[/tex]
[tex]N=m(g+a)\\\Rightarrow N=15.2(9.81+3.53)\\\Rightarrow N=\boldsymbol{302.77\ \mathbf{N}}[/tex]
[tex]F=155\ \text{N}[/tex]
[tex]N=mg-F\sin35.8^{\circ}\\\Rightarrow N=15.2\times 9.81-155\sin35.8^{\circ}\\\Rightarrow N=\boldsymbol{58.44\ \mathbf{N}}[/tex]
Problem 7:__deletededc326999bef85931fda5d1ab0d68e21218f27b46f4f11ab80872358251b584ddeleted__ 0N86-C1-52-40-A837-22820 If object A is twice the mass of object B what can be said about their kinetic energies just as they hit the ground
Answer:
the kinetic energy of body B is twice the kinetic energy of body A
Explanation:
The kinetic energy of a body is given by
K = ½ m v²
If two objects leave the same point, suppose that at the same height when they reach the ground they have the same velocity.
Therefore if the mass of body b is twice the mass of body A
[tex]m_{b} = 2 m_{a}[/tex]
[tex]K_{b}[/tex] = ½ (2 [tex]m_{a}[/tex]) v²
K_{b} = 2 (½ m_{a} v²)
K_{b} = 2 K_{a}
therefore the kinetic energy of body B is twice the kinetic energy of body A
A car accelerates from rest at 1.0 m/s2 for 20.0 second along a straight road. It then moves at a constant speed for half an hour and then decelerates uniformly to a stop in 30.0 s. Find the total distance covered by the car and the average velocity of the car for the entire trip.
Total distance = 36500 m
The average velocity = 19.73 m/s
Further explanationGiven
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration
[tex]\tt d=vo.t+\dfrac{1}{2}at^2\\\\d=\dfrac{1}{2}\times 1\times 20^2\rightarrow vo=0\\\\d=200~m[/tex]
[tex]\tt vt=vo+at\\\\vt=at\rightarrow vo=0\\\\vt=1\times 20\\\\vt=20~m/s[/tex]
State 2 : constant speed
[tex]\tt d=v\times t\\\\d=20\times 1800\\\\d=36000~m[/tex]
State 3 : deceleration
[tex]\tt vt=vo+at\rightarrow vt=0(stop)\\\\vo=-at\\\\20=-a.30~s\\\\a=-\dfrac{2}{3}m/s^2(negative=deceleration)[/tex]
[tex]\tt d=vot+\dfrac{1}{2}at^2\\\\d=20.30-\dfrac{1}{2}.\dfrac{2}{3}.30^2\\\\d=300~m[/tex]
Total distance : state 1+ state 2+state 3
[tex]\tt 200 + 36000 + 300=36500~m[/tex]
the average velocity = total distance : total time
[tex]\tt avg~velocity=\dfrac{36500}{20~s+1800~s+30~s}=19.73~m/s[/tex]
The total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.
What is average velocity?
The average velocity is the ratio of the total distance traveled to the total time taken by the body. Its unit is m/sec.
The given data in the problem is;
v₀ is the initial velocity=0(from rest)
a is the acceleration= 1 m/s²
t₁ is the time period 1= 20 s
t₂ is the time period 2= 0.5 hr = 1800 s
t₃ is the time period 3= 30 s
For part 1
The distance travelled is found as;
[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=\frac{1}{2}\times 1 \times (20)^2 \\\\ \rm d=200\ m[/tex]
The initial velocity is found as ;
[tex]\rm v_t=v_0+at\\\\ \rm v_t=0+1 \times 20\\\\ \rm v_t=20 \ m/sec[/tex]
For part 2
The distance traveled is found as;
[tex]\rm d= v \times t \\\\ \rm d= 20 \times 1800 \\\\ \rm d= 36000 \ m[/tex]
For part 3
The acceleration is found as;
[tex]\rm v_t=v_0+at\\\\ \rm 0=v_0+a \times 30\\\\ \rm 20=-30 a \\\\ a= - \frac{2}{3}\ m/sec^2[/tex]
The distance traveled is found as;
[tex]\rm d=v_0t+\frac{1}{2} at^2 \\\\ \rm d=20.30 +\frac{1}{2} \times \frac{2}{3} (30)^2 \\\\ \rm d=300 \ m[/tex]
Total distance travelled= distance travelled( part 1+ part 2+ part3)
Total distance travelled=200+3600+300 = 36500 m
The average velocity is found as;
[tex]v_{avg}= \frac{36500}{20+1800+30} \\\\ v_{avg}= 19.73 \ sec[/tex]
Hence the total distance covered by the car will be 36500m and the average velocity of the car for the entire trip will be 19.73 m/sec.
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What is force? Define with an example.
Explanation:
A force involves an interaction between two or more objects, and it causes a push or pull between the objects. An example of opposing force include drag due to interaction with an air mass and the force due to friction between two objects.
hope it helps!
If you push a 50 kg mass with a force of 80 N, what will be the objects
acceleration?
Answer:
acceleration=1.6m/s²
Explanation:
F=m×g
g=F/m
g=80N/50=1.6m/s²
You notice electric field lines going towards an isolated charged object in a radial manner. What is the sign of the net charge on the object?a. negative.b. positive.c. neutral.
Answer:
The correct answer is a
Explanation:
The electric field is given by the relation
F = q E
where The force is the Coulomb force and q is a positive test charge, therefore the electric field has the same direction as the electric force.
Consequently if the charge is positive the field must go out of the charge, if the charge is negative the electric field must be directed towards the charge.
Consequently, in this exercise we are told that the lines are directed towards the load, therefore the load must have a negative sign.
The correct answer is a
Two particles are separated by 0.38 m and have charges of -6.25 x 10-°C and 2.91 x 10-°C. Use Coulomb's law to predict the force between the particles if the distance is cut in half. The equation for Coulomb's law is F = kqi 42, and the constant, k, equals 9.00 x 109 Nm2/C2 2
Answer:
-4.35 × 10^-6 N
Explanation:
i just answered it on ap3x :)
A rock with a mass of 10 kg sits at the top of a hill 20 m high. What is the potential energy? PE= mghPE= hwKE= 1/2mv2
Answer:
1962 joules
Explanation:
m = 10 kg
h = 20 m
g = 9.81 ms^-2
PE = ?
PE = MGH
PE = 10 x 9.81 x 20
PE = 1962 joules
Which of the following is an example of kinetic energy? answers as follows
Which of the following is an example of kinetic energy?
A.
a boy resting on a park bench
B.
a book sitting on a shelf
C.
a rubber band stretched as far as it can go
D.
a swing moving back and forth
List the four outer planets from smallest to largest.
WILL MARK BRAINLIEST! ASAP
Answer: Mercury, Mars, Venus, Earth, Neptune, Uranus, Saturn, and Jupiter.
Explanation:
That's all of the planets if you need them. Hope this helps!
Which phase of matter does line segment CD represent?
A. plasma
B. liquid
C. gas
D. solid
Answer:
C. Gas
Explanation:
Segment CD represents gaseous phase of matter as segment CD shows highest energy which is in case of gas.
What is matter?
Matter in chemistry, is defined as any kind of substance that has mass and occupies space that means it has volume .Matter is composed up of atoms which may or not be of same type.
Atoms are further made up of sub atomic particles which are the protons ,neutrons and the electrons .The matter can exist in various states such as solids, liquids and gases depending on the conditions of temperature and pressure.
The states of matter are inter convertible into each other by changing the parameters of temperature and pressure.The intermolecular forces of attraction are different in different states of matter.Particles of matter have different sizes.
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The density of mercury is 13.6 g/cm? What is this value in kg/m??
Answer:
68 kg/m
Explanation:
bro
A neutron has
A. an alternating charge.
B. a negative charge.
C. a positive charge.
D. a neutral charge.
Answer:
D
Explanation:
a neutron does not have a positive nor negative charge it remains neutral
You work at a garden store for the summer. You lift a bag of fertilizer with a force of 112 N, and it moves upward with an acceleration of 0.790 m/s^2.
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Given :
Force provided, F = 112 N.
Acceleration of the bag, a = 0.79 m/s².
To Find :
a. What is the mass of the fertilizer bag?
b. How much does the fertilizer bag weigh?
Solution :
We know, force is given by :
F = ma
m = F/a
m = 112/0.79 kg
m = 141.77 kg
Now, weight is given by :
W = mg
W = 141.77 × 9.8 N
W = 1389.35 N
Therefore, the mass of fertilizer bag is 141.77 kg and weight us 1389.35 N.
If the elevator car in Figure 9-3A weighs 3,000 newtons, what is the tension in each supporting cable?
Answer:
500 N
Explanation:
1. What is the gravitational force between a 100 kg football player and the Earth at 5.98 x 1024 kg, with a
distance of 6.38 x 106 m?
Answer:
981N
Explanation:
The gravitational force between the football player and the Earth at the given distance is 979.91 N.
The given parameters:
Mass of the player, m = 100 kgMass of the Earth, M = 5.98 x 10²⁴ kgDistance between the bodies, r = 6.38 x 10⁶ mThe gravitational force between the football player and the Earth at the given distance is calculated by applying Newton's law of universal gravitation as follows.
[tex]F = \frac{GmM}{r^2} \\\\F = \frac{6.67 \times 10^{-11} \times 100 \times 5.98 \times 10^{24}}{(6.38 \times ^6)^2} \\\\F = 979.91 \ N[/tex]
Thus, the gravitational force between the football player and the Earth at the given distance is 979.91 N.
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An iron ball weighs 80 N on the surface
of Earth. How much will it weigh after
being moved to a height equal to the
radius of the Earth?
Answer:
2√5N will be the weight.
Explanation:
A man is pushing a box of mass m along the ground with a force F. The coefficient of kinetic friction between the box and the ground is μk. The box is accelerating, but then the person stops pushing and the box slides to a halt. The person then starts pushing on the box again with a force F, but the box doesn't budge. The maximum coefficient of static friction between the box and the ground is μs. Identify which of the following statements is true?
a. F > μk/μs
b. μk > μs
c. μkmg < F <= μsmg
d. μsmg= F
Answer: Option C.
Explanation:
When the box is moving, he pushes with a force F.
in this case, the coefficient is μk, and the box is accelerating, this means that the net force is not zero, so F is larger than the friction force.
When the box is still, the man pushes again with a force F, but now the box does not move, so there is no acceleration, which means that the net force is zero, then F is not greater than the maximum static friction force.
Now, the friction force between an object of mass M, and a surface with a coefficient of friction μ is: μ*m*g
where g is the gravitational acceleration.
Then, from the first part, we can conclude that:
μk*m*g < F
(the force F is larger than the kinetic friction force)
and from the second part, we know that:
F ≤ μs*m*g
(The force F is not greater than the static friction force)
If we write those two together, we have:
μk*m*g < F ≤ μs*m*g
Then the correct option is c.