Answer:
The correct answer is (a)
Explanation:
A spring scale measures the weight of an object not the mass because according to hooke's law the extension of a spring is directly proportional to the load or force attached/applied to it. The force of gravity acting on the mass of any substance as it goes up actually reduces and increases as it comes down.
If F = ma, as a increases, F will also increase and vice versa
Where F = force
m = mass
a = acceleration (due to gravity in this case)
From the above explanation, it can be deduced that the scale will read more than mg as it gets to the ground because of an increase in the force of gravity (which also increases a) as it approaches the ground.
In which scenario is the least amount of work done on a crate?
A force of 120 N moves it 25 m.
A force of 115 N moves it 26 m.
A force of 130 N moves it 22 m.
A force of 125 N moves it 27 m.
Answer:
The answer is C. A force of 130 N moves it 22 m
The least amount of work done on a crate is when A force of 130 N moves it 22 m, the correct option is C.
What is work done?The total amount of energy transferred when a force is applied to move an object through some distance
The work done is the multiplication of applied force with the displacement.
Work Done = Force * Displacement
The work done for option A is
= 120*25= 3000 Nm
The work done for option B is
=115*26 = 2990 Nm
The work done for option C is
=130*22= 2860 Nm
The work don for option D is
= 125*27= 3375 Nm
Hence, the least amount of work is done on the crate when a force of 130 N moves it 22 m.
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Alejandro made 6.4 liters of punch using half apple juice and half orange juice. How many milliliters of apple juice are in the punch?
Answer:
3.2
Explanation:
I hope that this helps! Have a good day!!
Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at
Complete question is;
Allyson and Adrian have decided to connect their ankles with a bungee cord; one end is tied to each person's ankle. The cord is 40 feet long, but can stretch up to 120 feet. They both start from the same location. Allyson moves 10 ft/sec and Adrian moves 9 ft/sec in the directions indicated. Adrian stops moving at time t = 5.5 sec, but Allyson keeps on moving 10 ft/sec in the indicated direction. (If a coordinate system is used, assume that the girls' starting position is located at
(x, y) = (0, 0) and that Allyson and Adrian move in the positive y and negative x directions, respectively. Let one unit equal one foot.)
Compute the length of the bungee cord at t = 7 seconds. (Round your answer to three decimal places.)
Answer:
Length of bungee cord = 85.734 ft
Explanation:
We are told that Adrian moves 9ft/sec. Thus, at 5.5 seconds, distance he moved is; 9 ft/sec × 5.5sec = 49.5 ft in the negative x (-x) direction. Therefore, the coordinate is (-49.5, 0).
Now, Allyson has moved 10ft/sec. Thus, at 7 seconds, distance he moved would be; 10 ft/sec x 7sec = 70 feet in the positive (+y) direction. Therefore, the coordinate is (0, 70).
Now, since they started from the origin, it means (0, 0) is a coordinate. Thus, we now have 3 coordinates which are; (0, 0), (0, 70) & (-49.5,0). These 3 coordinates would therefore combine to form a right triangle.
The hypotenuse is the distance between Allyson and Adrian.
Thus, from pythagoras theorem, we can find the distance between them which is same as the length of the cord.
Thus;
(-49.5)² + 70² = D².
D² = 2450.25 + 4900
D = √7350.25
D = 85.734 ft
A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocity of -1.06 m/s. Her hands are in contact with the wall for 1.07 s. Ignore friction and wind resistance. Find the average force she exerts on the wall (which has the same magnitude, but opposite direction, as the force that the wall applies to her). Note that this force has direction, which you should indicate with the sign of your answer.
Answer:
F = 47.6 N
Explanation:
Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:[tex]F = \frac{\Delta p}{\Delta t}[/tex]
So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:[tex]F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N[/tex]
⇒ Fsk = 47.6 N (normal to the wall)
Given two vectors A= 4.00i + 7.00jand B= 5.00i-2.00ja.Find the magnitude of each vector.b.Find the vector differenceC= A –B, give its magnitude and direction. c.Draw the vector A, B and C.d.Find the scalar product of A.B e.Find the vector productD= A x B, draw D and give its magnitude.
Answer:
444.99
Explanation:
Explanation:
C=A-B
=(4.00i+7.00j)-(5.00i-2.00j)
= -1.00i+9.00j
A.B=AxBx+AyBy
=(4 x 5)+(7 x -2)
=20-14
=6
What (rather remarkable!) equation relates the speed of light to other fundamental electromagnetic constants?
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equation is [tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
The value of c is [tex]c = 2.998 *10^{8} \ m/s [/tex]
Explanation:
From the question we are told that
Generally the equation that relates the speed of light to other fundamental electromagnetic constants is
[tex]c = \frac{1}{\sqrt{ \mu_o * \epsilon_o} }[/tex]
Here c is the speed of light
[tex]\mu_o[/tex] is the permeability of free space with value
[tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]
and [tex]\epsilon_o[/tex] is the permittivity of free space with value
[tex]\epsilon_o = 8.85*10^{-12} \ C/V \cdot m[/tex]
So
[tex]c = \frac{1}{\sqrt{ 4\pi *10^{-7} * 8.85*10^{-12}} }[/tex]
=> [tex]c = 2.998 *10^{8} \ m/s [/tex]
In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 4.7 kg book is pushed from rest through a distance of 0.85 m by the horizontal 42 N force from the broom and then has a speed of 1.0 m/s, what is the coefficient of kinetic friction between the book and floor
Answer:
μ_k = 0.851
Explanation:
We are given;
Mass of book; m_book = 4.7 kg
Horizontal force; F_horiz = 42 N
Distance; d = 0.85 m
Speed; v = 1 m/s
First of all let's find the acceleration using Newton's equation of motion;
v² = u² + 2ad
u is initial velocity and it's 0 m/s in this case.
Thus;
1² = (2 × 0.85)a
1 = 1.7a
a = 1/1.7
a = 0.5882 m/s²
Now, resolving forces along the vertical direction, we have;
W - N = 0
Thus,W = N
Where W is weight = mg and N is normal force
Thus; N = mg = 4.7 × 9.81 = 46.107 N
Now, resolving forces along the horizontal direction, we have;
F_horiz - ((μ_k)N) = ma
Where μ_k is coefficient of kinetic friction.
Thus;
42 - 46.107(μ_k) = 4.7 × 0.5882
42 - 46.107(μ_k) = 2.76454
μ_k = (42 - 2.76454)/46.107
μ_k = 0.851
What is an example of a scientific topics that requires knowledge of at least two branches of physics?
Explanation:
classical physics and modern physics.
Physics is the branch of science that deals with the interaction of energy and matter. Modern and Classical Physics are the two major branches of physics.
What is physics?Physics is the discipline of science that investigates the structure of matter and how the universe's fundamental constituents interact. It investigates objects ranging from the very small to the entire universe using quantum mechanics and general relativity.
One of the most crucial physics topics is kinematics. Kinematics is the study of an object 's direction of travel. Kinematics is only concerned with the motion of an object.
There is a vertical and a horizontal component to two-dimensional projectile motion, such as that of a football or other thrown object.
Throwing a rock or kicking a ball produces a projectile pattern of motion that includes both a vertical and a horizontal component.
Thus, these can be the topics that require the knowledge of physics.
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Question C) needs to be answered, please help (physics)
(a) Differentiate the position vector to get the velocity vector:
r(t) = (3.00 m/s) t i - (4.00 m/s²) t² j + (2.00 m) k
v(t) = dr/dt = (3.00 m/s) i - (8.00 m/s²) t j
(b) The velocity at t = 2.00 s is
v (2.00 s) = (3.00 m/s) i - (16.0 m/s) j
(c) Compute the electron's position at t = 2.00 s:
r (2.00 s) = (6.00 m) i - (16.0 m) j + (2.00 m) k
The electron's distance from the origin at t = 2.00 is the magnitude of this vector:
||r (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the x-y plane, the velocity vector at t = 2.00 s makes an angle θ with the positive x-axis such that
tan(θ) = (-16.0 m/s) / (3.00 m/s) ==> θ ≈ -79.4º
or an angle of about 360º + θ ≈ 281º in the counter-clockwise direction.
A body of rotational inertia 1.0 kg m2 is acted upon by a torque of 2.0 Nm. The angular acceleration of the body will be
Answer:
The angular acceleration is 2 rad/s².
Explanation:
Given that,
Rotational inertia = 1.0 kg m²
Torque = 2.0 Nm
We need to calculate the angular acceleration
Using formula of torque
[tex]\tau=I\alpha[/tex]
[tex]\alpha=\dfrac{\tau}{I}[/tex]
Where, I = moment of inertia
[tex]\tau[/tex] = torque
Put the value into the formula
[tex]\alpha=\dfrac{2.0}{1.0}[/tex]
[tex]\alpha=2\ rad/s^2[/tex]
Hence, The angular acceleration is 2 rad/s².
Two equal forces act on two different objects, one of which has a mass ten times as large as the other. The larger object will have _________ acceleration that the less massive object.
Answer:
The larger object will have smaller acceleration that the less massive object.
Explanation:
Generally force is mathematically represented as
[tex]F = ma[/tex]
=> [tex]m = \frac{F}{a }[/tex]
at constant force we have
[tex]m \ \alpha \ \frac{1}{a}[/tex]
So if m is increasing a will be decreasing which means the object with the larger mass will have less acceleration
Challenge Question (This is supposed to be hard, think critically and take it one step at a time):
I am driving a car to a science lecture. The total distance from my house to the lecture is 75 miles. I get
halfway there before I must stop for gas. After filling up I realize I only have 15 minutes, before the
lecture starts (I know, bad planning on my part). How fast must I drive to be on time to the lecture?
Since it is a science lecture, report your answer in meters per second!!!
|
Answer:
You must travel at an average speed of 67.06 m/s to be on time to the lecture.
Explanation:
From the question, the total distance from your house to the science lecture is 75 miles.
Also, you get halfway there before you stop for a gas, that is, you have covered half of 75 miles, which 37.5 miles and you also have to cover 37.5 miles to get to the science lecture.
After filling up, you only have 15 minutes before the lecture starts,
To determine how fast you must drive to be on time to the lecture,
we will determine the average speed you need to travel.
From
Average speed = Distance / Time
Distance = 37.5 miles (Convert to meters)
(NOTE: 1 mile = 1609.344 meters)
Hence, 37.5 miles = 37.5 × 1609.344 miles = 60350.4 meters
∴ Distance = 60350.4 meters
Time = 15 minutes (Convert to seconds)
(NOTE: 1 minute = 60 seconds)
Hence, 15 minutes = 15 × 60 seconds = 900 seconds
Now, from
Average speed = Distance / Time
Average speed = 60350.4 m / 900 s
Average speed = 67.06 m/s
Hence, you must travel at an average speed of 67.06 m/s to be on time to the lecture.
How much work is required to move it at constant speed 5.0 m along the floor against a friction force of 290 N?
Answer:
The answer is 1450 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
force = 290 N
distance = 5 m
We have
workdone = 290 × 5
We have the final answer as
1450 JHope this helps you
Sound wave A has a lower frequency than sound wave B, but both waves
have the same amplitude. What must also be true of these two waves?
O A. Wave A is softer.
O B. Wave A is louder.
C. Wave A has a lower pitch.
D. Wave A has a higher pitch.
Answer: C
Explanation: Amplitude controls loudness, and frequency controls pitch. The more frequent the higher pitch.
The true statement regarding the two sounds is wave A has a lower pitch.
What is pitch of a sound?
The pitch of a sound is sensation of the sound frequency.
A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to low frequency.
Sound of the the same amplitude, have the same loudness.
Thus, the true statement regarding the two sounds is wave A has a lower pitch.
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Which of the following is not true about taxes? A. Mandatory sum of money by government so that it can operate B. Due on April 15th C. largely collected to support private businesses D. Collected by the Internal Revenue Service (IRS)
How much voltage (in terms of the power source voltage bV) will the capacitor have when it has started at zero volts potential difference, it is connected to the power supply and resistor and onehalf the characteristic time has passed (i.e. t= T(tau)/2)?
Answer:
The voltage is [tex]V = 0.993V_b[/tex]
Explanation:
From the question we are told that
The time that has passed is [tex]t = \frac{\tau}{2}[/tex]
Here [tex]\tau[/tex] is know as the time constant
The voltage of the power source is [tex]V_b[/tex]
Generally the voltage equation for charging a capacitor is mathematically represented as
[tex]V = V_b [1 - e^{- \frac{t}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{\tau}{2\tau} }][/tex]
=> [tex]V = V_b [1 - e^{- \frac{1}{2} }][/tex]
=> [tex]V = 0.993V_b[/tex]
The scientific method is the only way of learning about Nature used by scientist today *
A. true
B. false
Answer:
false
Explanation:
What is the car’s average velocity (in m/s) in interval between t=1.0s to t=1.5s?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula for velocity is;
Velocity (in m/s) = distance/time
The distance the car covered in the completed question is divided by the difference in the time interval
The difference in the time interval will be = 1.5s - 1.0s = 0.5s
NOTE: the distance must be in meters or be converted to meters
Select all the correct statements that are always true about conductors. a. A conductor in an electric field becomes polarized. b. In electrostatic equilibrium, the electric field inside a conductor should be zero. c. In electrostatic equilibrium, the electric field inside a conductor should be infinite. d. In electrostatic equilibrium, the electric field must point outward from the surface of the conductor. e. In electrostatic equilibrium, the electric field must point inward from the surface of the conductor.
Answer:
true b
Explanation:
They ask us to evaluate if the expressions are true
a) False. The concept of polarization refers to the direction of the oscillating electric field in a traveling wave, in this case the shocks cause the wave to disperse
b) True .. Yes, since no work is done inside the conductor, so all the load is on the surface
c) False. In electrostatic equilibrium the voltage must be the same everywhere
d) False. If the charges are positive point out
If the charges are negative point inwards
e) False. If the charge is positive, the pattern is outward
if negative largares points gavia inside.
What is the magnitude of the force exerted by the biceps FbicepsFbicepsF_biceps? What is the magnitude of the force exerted by the elbow FelbowFelbowF_elbow
Answer:
The force of forearm is 239.9 N
The force of elbow is 215.89 N
Explanation:
Suppose, When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a forearm is 1.50 kg. If the biceps is connected to the forearm a distance [tex]d_{b}[/tex] 2.50 cm from the elbow, how much force [tex]F_{b}[/tex] must the biceps exert to hold a 950 g ball at the end of the forearm at distance dball J 36.0 cm from the elbow, with the forearm parallel to the floor? How much force [tex]F_{l}[/tex] must the elbow exert,
Given that,
Mass of forearm = 1.50 kg
Distance of forearm = 2.50 cm
Mass of ball = 950 g
Distance of ball = 36.0 cm
We need to calculate the force of forearm
Using balancing torque about elbow
[tex]F_{b}\times d_{b}=w_{f}\times\dfrac{d_{f}}{2}+w_{ball}\times d_{ball}[/tex]
Put the value into the formula
[tex]F_{b}\times 0.025= 1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36[/tex]
[tex]F_{b}=\dfrac{1.50\times9.8\times\dfrac{0.36}{2}+0.95\times9.8\times0.36}{0.025}[/tex]
[tex]F_{b}=239.9\ N[/tex]
We need to calculate the force of elbow
Using balancing force
[tex]F_{b}=F_{l}+w_{f}+w_{b}[/tex]
[tex]F_{l}=F_{b}-w_{f}-w_{b}[/tex]
Put the value into the formula
[tex]F_{l}=239.9-(1.50\times9.8)-(0.95\times9.8)[/tex]
[tex]F_{l}=215.89\ N[/tex]
Hence, The force of forearm is 239.9 N
The force of elbow is 215.89 N
A horizontal spring-mass system has low friction, spring stiffness 163 N/m, and mass 0.3 kg. The system is released with an initial compression of the spring of 12 cm and an initial speed of the mass of 3 m/s. (a) What is the maximum stretch during the motion?
Complete Question
A horizontal spring-mass system has low friction, spring stiffness 163 N/m, and mass 0.3 kg. The system is released with an initial compression of the spring of 12 cm and an initial speed of the mass of 3 m/s.
(a) What is the maximum stretch during the motion?
(b) What is the maximum speed during the motion?
m/s
(c) Now suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Answer:
a
[tex] x_m = 0.18 \ m [/tex]
b
[tex]v_m = 4.10 \ m/s[/tex]
c
[tex]P = 0.085 \ W [/tex]
Explanation:
From the question we are told that
The spring stiffness is [tex]k = 163 \ N/m[/tex]
The mass is [tex]m = 0.3 \ kg[/tex]
The initial compression is x = 12 cm = 0.12 m
The initial speed is v = 3 m/s
Generally the total energy stored in the spring at maximum stretch is equivalent to the kinetic energy and the energy stored at the initial compression
This can be mathematically represented as
[tex]E_{max} = K + E_{i}[/tex]
Here [tex]E_{max}[/tex] is the energy at maximum stretch which is mathematically represented as
[tex]E_{max} = \frac{1}{2} * k * x_m^2[/tex]
and
[tex]E_{max} = \frac{1}{2} * k * x^2[/tex]
So
[tex]\frac{1}{2} * k * x_m^2 = \frac{1}{2}m * v^2 + \frac{1}{2} * k * x^2 [/tex]
=> [tex]\frac{1}{2} * 163 * x_m^2 = \frac{1}{2}* 0.3 * 3^2 + \frac{1}{2} * 163 * 0.12^2 [/tex]
=> [tex]\frac{1}{2} * 163 * x_m^2 = \frac{1}{2}* 0.3 * 3^2 + \frac{1}{2} * 163 * 0.12^2 [/tex]
=> [tex] x_m = 0.18 \ m [/tex]
Gnerally the maximum kinetic energy when the stretch is zero(i.e during motion), according to the law of energy conservation, is equivalent to the kinetic energy when the spring is compressed plus the energy stored in the spring before it was released. , this can be mathematically represented as
[tex]\frac{1}{2} * k * x^2 + \frac{1}{2} * m * v^2 = \frac{1}{2} * m * v_m ^2[/tex]
Here [tex]v_m[/tex] is the maximum velocity when the stretch is zero
So
[tex]0.5 * 163 * 0.12^2 + \frac{1}{2} * 0.3 * 3^2 = \frac{1}{2} * 0.3 * v_m ^2[/tex]
=> [tex]v_m = 4.10 \ m/s[/tex]
Generally the period of the spring is mathematically represented as
[tex]T = 2 * \pi * \sqrt{\frac{m}{k} }[/tex]
=> [tex]T = 2 * 3.142 * \sqrt{\frac{0.3}{163} }[/tex]
=> [tex]T = 0.2696 \ s [/tex]
From the question we are told that the energy is 0.02 J
Generally the power is
[tex]P = \frac{E}{T}[/tex]
=> [tex]P = \frac{0.02}{0.2696}[/tex]
=> [tex]P = 0.085 \ W [/tex]
Exercise releases chemicals in our bodies called
Answer:
endorphins
Explanation:
What are two ways that an object can have kinetic energy?
Answer:
The object has to have mass and speed
Explanation:
You can increase both speed and mass to increase the kinetic energy, hope this answers your question.
Happy Halloween!
The posted speed limit on the road heading from your house to school is 45 mi/h which is about 20 m/s if you live 8 km from school how long will it take you to get to school if there’s no traffic convert your answer in to minutes
Answer:
t = 6.7 min.
Explanation:
Assuming there's no traffic, and neglecting the time needed to start from rest, and the one needed to stop, all the path is covered at the same constant speed of 20 m/s.Applying the definition of average velocity, taking as the origin your house , you get the time used for the travel from the following expression:[tex]t = \frac{d}{v_{avg} } = \frac{8000 m}{20 m/s} = 400 s[/tex]
In order to convert the answer to minutes, we need to do the following, multiplying by something equal to one, as follows:[tex]400 s * \frac{1 min}{60 s} = 6.7 min.[/tex]
an initially stationary 15.0kg crate of cheese wheels is pulled via a cable a distance of 5.70m up a frictionless ramp to a height of 2.5m, where it stops. What is the potential energy of the crate after it is lifted? How much work is done during the lift? If it takes 10 seconds to lift the crate, what is the power rating of the lifting device?
Answer:
The potential energy is 367.5 J and power is 36.75 W.
Explanation:
Given that,
Mass of a crate, m = 15 kg
It is lifted to a height of 2.5 m
We need to find the potential energy of the crate after it is lifted. It is given by the formula as follows :
[tex]E=mgh\\\\E=15\ kg\times 9.8\ m/s^2\times 2.5\ m\\\\E=367.5\ J[/tex]
If time, t = 10 s, Power is given by :
[tex]P=\dfrac{E}{t}\\\\P=\dfrac{367.5}{10}\\\\P=36.75\ W[/tex]
So, the potential energy is 367.5 J and power is 36.75 W.
A particle is moving with a constant velocity in the +x
+
x
-direction. When the particle reaches the origin, a constant force is exerted on it in the +y
+
y
-direction. Which of the following graphs best shows the path the particle will follow?
Answer:
The options are not shown, so i will answer in a general way:
We can separate the problem in two cases.
The movement in the x-axis, that is unperturbed, so the particle will keep moving in the +x direction with the same speed.
The movement in the y-axis, as a constant force is applied, by the second Newton's law:
F = m*a
a = F/m = constant
v = a*t + v0
p = (a/2)*t^2 + v0*t + p0
Where v0 = initial velocity in the y-axis, p0 = initial position (in this case the origin) and t is the time. (In this case, these values really do not matter, i only want to show the general equations of movement in the y-axis)
We will also have a constant acceleration, which means that the velocity in this axis will be linear, and the position quadratic, then after the particle reaches the origin, it will start moving up and to the right, and the upward movement will increase in magnitude as time passes, a graph in the x-y plane would be something like:
Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message encourages us to send a radio answer, which we decide to do. Suppose our governing bodies take 2 years to decide whether and how to answer. When our answer arrives there, their governing bodies also take two of our years to frame an answer to us. How long after we get their first message can we hope to get their reply to ours? Enter your answer in years.
Answer:
The duration is [tex]T =72 \ years /tex]
Explanation:
From the question we are told that
The distance is [tex]D = 35 \ light-years = 35 * 9.46 *10^{15} = 3.311 *10^{17} \ m [/tex]
Generally the time it would take for the message to get the the other civilization is mathematically represented as
[tex]t = \frac{D}{c}[/tex]
Here c is the speed of light with the value [tex]c = 3.08 *10^{8} \ m/s[/tex]
=> [tex]t = \frac{3.311 *10^{17} }{3.08 *10^{8}}[/tex]
=> [tex]t = 1.075 *10^9 \ s[/tex]
converting to years
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 1.075 *10^9 * 3.17 *10^{-8} [/tex]
[tex]t = 34 \ years [/tex]
Now the total time taken is mathematically represented as
[tex]T = 2* t + 2 + 2[/tex]
=> [tex]T = 2* 34 + 2 + 2[/tex]
=> [tex]T =72 \ years /tex]
How can you increase the amount of kinetic energy in an object
The amount of kinetic energy in an object can be increased by two ways:
by adding mass on the object.by accelerating the object.What is kinetic energy?The energy an object has as a result of motion is known as kinetic energy in physics. It is described as the effort required to move a mass-determined body from rest to the indicated velocity.
The body holds onto the kinetic energy it acquired during its acceleration until its speed changes. The body exerts the same amount of effort when slowing down from its current pace to a condition of rest.
Mathematically kinetic energy can be expressed as:
Kinetic energy = 1/2 × mass × speed²
Hence, the amount of kinetic energy in an object can be increased by two ways:
by adding mass on the object.by accelerating the object.Learn more about kinetic energy here:
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If 2000 kg cannon fires 2 kg projectile having muzzle velocity 200 m/s then the KINETIC ENERGY of the CANNON will be ("E4" means "*10^4") *
1 point
4 J
40 J
400 J
None of the above
Answer:
M V = m v conservation of momentum (Caps-cannon Small-projectile)
V = m / M * V = 2 / 2000 * 200 m/s = .2 m/s recoil velocity of cannon
KE = 1/2 M V^2 = 2000 / 2 kg * (.2 m/s)^2 = 40 kg m^2/s^2 = 40 J
Two small boats are in a lake. An observer on the shore sees the boats moving with the same velocity, 1 m/s due east. The mass of boat A is 10 kg; the mass of boat B, 5 kg. The momentum of Boat A is
Answer:
The momentum of the boat A is 10 kg.m/s.
Explanation:
Given;
velocity of the two boats, V = 1 m/s
mass of boat A, mₐ = 10 kg
mass of boat B, mb = 5 kg
Linear momentum is given by the product of mass and linear velocity.
The momentum of the boat A is given by;
Pₐ = mₐv
Pₐ = 10 x 1
Pₐ = 10 kg.m/s
Therefore, the momentum of the boat A is 10 kg.m/s.
