Answer:
Acceleration is 2.5 m/s^2
Explanation:
Let train 1 travel x meters and train 2 travels 1000-x meters
As per the las of acceleration
[tex]v^2 = u^2 + 2ax[/tex]
Substituting the given values, we get -
[tex]0 = 50^2 + 2ax\\ax = -1250\\[/tex]---Eq (1)
Similarly
[tex]0 = 50^2 + 2a(1000-x)\\a(1000-x) = 1250[/tex]-- Eq (2)
Substituting the value of ax from eq (1) into Eq(2), we get -
1000*a - 1250 = 1250
a = 2500/1000 = 2.5 m/s^2
A 20-g bullet is shot vertically into an 2.8-kg block. The block lifts upward 9 mm. The bullet penetrates the block and comes to rest in it in a time interval of 5 ms. Assume the force on the bullet is constant during penetration and that air resistance is negligible. What is the speed of the bullet just before the impact
Answer:
The speed of the bullet just before the impact is 701 m/s
Explanation:
Given;
mass of the bullet, m₁ = 20 g = 0.02 kg
mass of the block, m₂ = 2.8 kg
displacement of the block, d = 9 mm = 9 x 10⁻³ m
duration of motion of the bullet, t = 5 ms = 5 x 10⁻³ s
Apply the principle of conservation of energy;
The final kinetic energy of the bullet = maximum potential energy of the block
[tex]\frac{1}{2} m_1v^2 = m_2gh\\\\v^2 = \frac{2m_2gh}{m_1} \\\\v= \sqrt{\frac{2m_2gh}{m_1} } \\\\v = \sqrt{\frac{2 \times 2.8 \times 9.8 \times (9\times 10^{-3})}{0.02} } \\\\v = 4.97 \ m/s[/tex]
Apply the principle of conservation of linear momentum, to determine the initial velocity of the bullet before the impact.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
u₁ is the initial velocity of the bullet
u₂ is the initial velocity of the block = 0
m₁u₁ + 0 = v(m₁ + m₂)
m₁u₁ = v(m₁ + m₂)
0.02u₁ = 4.97(2.8 + 0.02)
0.02u₁ = 14.02
u₁ = 14.02 / 0.02
u₁ = 701 m/s
Therefore, the speed of the bullet just before the impact is 701 m/s
A red apple reflects ______
light and absorbs all other
colours.
A red apple reflects Red light and absorbs all other colours.
Please helppppppp I need it todayyyyyt!!!!!
An athlete training for an event does 6.53 104 J of work during a workout and gives off 5.97 105 J of heat. Consider the athlete to be like a heat engine. (a) Determine the magnitude of the change in internal energy of the athlete. J (b) What is the efficiency of the athlete
Answer:
(a) The magnitude of the change in internal energy is 6.623 x 10⁵ J
(b) the efficiency of the athlete is 10.94 %
Explanation:
Given;
work done by the athlete (system), W = 6.53 x 10⁴ J
the heat given off by the athlete (system), Q = 5.97 x 10⁵ J
The simple diagram below will be used to illustrate the direction of the energy flow assuming a heat engine.
Q← ⊕ →W
The work, W, points away from the system since the system does the work
The heat, Q, points away from the system since heat is given off
Apply first law of thermodynamic;
ΔU = Q + W
where;
q is the heat flowing into or out of the system
(+q if the heat is flowing into the system
(-q if the heat is leaving the system
w is the work done by or on the system
(+w if the work is done on the system by the surrounding
(-w if the work is done by the system to the surrounding
Thus, from the above explanation, the change in internal energy of the system is calculated as;
ΔU = -Q - W
ΔU = - 5.97 x 10⁵ J - 6.53 x 10⁴ J
ΔU = -6.623 x 10⁵ J
The magnitude of the change in internal energy = 6.623 x 10⁵ J
(b) the efficiency of the athlete;
[tex]Efficiency = \frac{W}{Q} \times 100\%\\\\Efficiency = \frac{6.53 \times 10^4}{5.97 \times 10^5} \times 100\%\\\\Efficiency = 10.94 \ \%[/tex]
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object. For an object to change its state of motion, a force must be applied to it.
A. Velocity
B. Inertia
C. Distance
D. Area
Answer:
B) Inertia is the resistance of any physical object
Two workers are sliding 450 kg kg crate across the floor. One worker pushes forward on the crate with a force of 380 NN while the other pulls in the same direction with a force of 230 NN using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor
Answer:
The coefficient of kinetic friction on the floor is 0.138
Explanation:
Given;
mass of the crate, m = 450 kg
force applied by the first worker, F₁ = 380 N
force applied by the second worker in the same direction as the first worker, F₁ = 230 N
frictional force opposing the motion of the box = -[tex]F_k[/tex]
Apply Newton's second law of motion;
∑F = ma
[tex]F_1 + F_2 - F_k = ma[/tex]
If the crate slides with constant speed, acceleration is zero (0).
[tex]F_1 + F_2 - F_k = ma = 0\\\\F_1 + F_2 - F_k = 0\\\\F_k = F_1 + F_2\\\\\mu _kmg= F_1 + F_2\\\\\mu _k = \frac{F_1 + F_2}{mg} \\\\\mu _k = \frac{380 + 230}{450 \times 9.8} \\\\\mu _k = 0.138[/tex]
Therefore, the coefficient of kinetic friction on the floor is 0.138
What is the speed acquired by a freely falling object 4 seconds after being dropped from a rest position? Use units of meter per second (m/s) and assume acceleration from gravity is 10 m/s2.
speed = 40 m/s
Explanation:
Since the object is dropped, V0y = 0.
Vy = V0y - gt
= -(10 m/s^2)(4 s)
= -40 m/s
This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.
The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
What are the three equations of motion?There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,
By using the first equation of motion,
v = u + at
initial velocity(u) = 0 m/s
acceleration(a) = 10 m/s²
v = 0 + 10×4
v = 40 meters/seconds
Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.
Learn more about equations of motion from here,
brainly.com/question/5955789
#SPJ2
what's the difference between mass and inertia in a tabular form
Answer:
to be honest i dont know
Explanation:
^^
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Where does the extra energy go?
Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Then the extra energy goes in overcoming the gravitational acceleration and heating up of body etc
a force in the opposite direction
of the motion
forces: forces applied to an object in opposite directions that are not equal in size. Unbalanced forces result in a change in motion. friction: the force that opposes the motion or tendency toward motion of two objects that are in contact.
You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 9.50 m/sm/s. Part A What is the speed of the putty just before it strikes the ceiling
Answer: [tex]4.65\ m/s[/tex]
Explanation:
Given
Distance putty has to travel is 3.5 m
The initial speed of putty is 9.50 m/s
Using equation of motion to determine the velocity of putty just before it hits ceiling
[tex]v^2-u^2=2as[/tex]
[tex]\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s[/tex]
So, the velocity of putty just before hitting is [tex]4.65\ m/s[/tex]
A body initially at rest is accelerated at a constant rate for 5.0 seconds in the positive x direction. If the final speed of the body is 20.0 m/s, what was the body's acceleration?
Answer:
[tex]a=4\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of a body, u = 0
Final speed of the body, v = 20 m/s
Time, t = 5 s
We need to find the acceleration of the body. We know that the acceleration of an object is equal to the rate of change of velocity divided by time taken. So,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{20-0}{5}\\\\a=4\ m/s^2[/tex]
So, the body's acceleration is equal to [tex]4\ m/s^2[/tex].
A box attached to a spring is being pulled across a flat frictionless surface. The spring constant is 45 N/m, the box is accelerating at
1.3 m/s2, and the spring is stretched by 0.88 m. What is the mass of the box?
26 kg
O 38 kg
30 kg
O 33 kg
Plzzzzz help
Answer:
Mass = 30.46 kg
Explanation:
Given the following data;
Spring constant = 45 N/m
Acceleration = 1.3 m/s²
Extension = 0.88 m
To find the mass of the box;
First of all, we would determine the force acting on the spring.
Force = spring constant * extension
Force = 45 * 0.88
Force = 39.6 N
Next, we find the mass using Newton's second equation of motion.
Force = mass * acceleration
39.6 = mass * 1.3
Mass = 39.6/1.3
Mass = 30.46 kg
Probability of a woman wearing green to walk into a restaurant
If a fisherman applies a horizontal force with magnitude 47.0 NN to the box and produces an acceleration of magnitude 3.20 m/s2m/s2, what is the mass of the box
Answer:
The correct solution is "14.6875 kg".
Explanation:
Given values:
Force,
F = 47.0 N
Acceleration,
a = 3.20 m/s²
Now,
⇒ [tex]Force=Mass\times Acceleration[/tex]
or,
⇒ [tex]F=ma[/tex]
⇒ [tex]47.0=m\times 3.20[/tex]
⇒ [tex]m=\frac{47.0}{3.20}[/tex]
⇒ [tex]=14.6875 \ kg[/tex]
Jet aircraft maintenance crews are required to wear protective earplugs. Members of a particular crew wear earplugs that reduce the sound intensity by a factor of 305. If, when the jet is fired up, the sound intensity level experienced by the crew members wearing protective earplugs is 79 dB, determine the sound intensity level they would experience without the earplugs.
Answer:
the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
Explanation:
Given the data in the question;
sound intensity reduced by the factor, m = 305
the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB
Now, using the expression of sound intensity level;
L = 10log( [tex]I_0[/tex] )
where [tex]I_0[/tex] is the intensity at L level
so we substitute
79 = 10log( [tex]I_0[/tex] )
[tex]I_0[/tex] = [tex]10^{7.9[/tex]
Now, expression for actual intensity;
[tex]I[/tex] = m[tex]I_0[/tex]
where [tex]I[/tex] is the actual intensity
so we substitute
[tex]I[/tex] = 305 × [tex]10^{7.9[/tex]
Next, we write the expression of sound intensity level for reduced intensity;
L' = 10log( [tex]I[/tex] )
So we substitute
L' = 10log( 305 × [tex]10^{7.9[/tex] )
L' = 10log( 24227011159.09058 )
L' = 103.8 dB
Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB
One of the earliest vertebrate animal groups that evolved in the early Paleozoic Era
are
A red laser from the physics lab is marked as producing 632.8-nm light. When light from this laser falls on two closely spaced slits, an interference pattern formed on a wall several meters away has bright fringes spaced 6.00 mm apart near the center of the pattern. When the laser is replaced by a small laser pointer, the fringes are 6.30 mm apart.
Required:
What is the wavelength of light produced by the pointer?
Answer:
Wavelength = [tex]\lambda_p = 3.986 * 10^{-6}[/tex] m
Explanation:
As we know
Fringe width (w) = [tex]\frac{D*\lambda}{d}[/tex]
where
[tex]\lambda[/tex] is the wavelength
D is distance between source and screen
d is the distance between two slits
[tex]\frac{D}{d} = \frac{y}{\lambda}[/tex]
[tex]\frac{D}{d} = \frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\lambda_p = \frac{y_p* \lambda_r}{y_r} \\\lambda_p =\frac{6.30 * 10^{-3} * 632.8 * 10^{-9}}{6 *10^{-3}} \\\lambda_p = 3.986 * 10^{-6}[/tex]m
How is a continuous spectra is formed?
Answer:
Hello There!!
Explanation:
They are produced by the photodissociation of negatively charged hydrogen ions (H−).
hope this helps,have a great day!!
~Pinky~
A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the battery supplies energy at a rate of 25 W, how large is the resistance
Answer:
[tex]R=2.78\ \Omega[/tex]
Explanation:
Given that,
The current flowing in the circuit, I = 3 A
The power of the battery, P = 25 W
We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :
[tex]P=I^2R[/tex]
Put all the values to find R.
[tex]R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega[/tex]
So, the resistance is equal to [tex]2.78\ \Omega[/tex].
Walking at a brisk pace, you cover 10 m in 5.0 s . How many seconds will you need to cover 50 m ?
A circular loop of wire with radius 10.0 cm is located in the xy-plane in a region of uniform magnetic field. A field of 2 T is directed in the z-direction, which is upward. (a) What is the magnetic flux through the loop
Answer:
[tex]\phi=628.3[/tex]
Explanation:
From the question we are told that
Radius [tex]r=10.0 cm[/tex]
Magnetic field[tex]B=2T[/tex]
Generally the equation for area of circular path is mathematically given by
[tex]Area=\pi r^2[/tex]
[tex]A=\pi 10^2[/tex]
[tex]A=314.15m^2[/tex]
Generally the equation for Magnetic flux is mathematically given by
[tex]\phi=BA[/tex]
[tex]\phi=2*314.15[/tex]
[tex]\phi=628.3[/tex]
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 15 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 65 cm .
Required:
What is the wavelength of the sound?
Answer:
100 cm
Explanation:
It is given that there are two loudspeakers that produces [tex]$\text{sound waves }$[/tex] along x-axis.
The maximum intensity of the sound is [tex]$\text{when the speakers are}$[/tex] at a distance of = 15 cm apart.
The sound intensity becomes zero when the separation between the speakers are increased and becomes 65 cm.
Therefore, the sound waves are in the phase, [tex]$\Delta x_1=15 \ cm$[/tex]
The sound waves are out of phase when [tex]$\Delta x_2=65 \ cm$[/tex]
Therefore,
[tex]$\Delta x_2 - \Delta x_1 = \frac{\lambda}{2}$[/tex]
[tex]$\lambda= 2(\Delta x_2 - \Delta x_1)$[/tex]
= 2 (65 - 15)
= 2 x 50
= 100 cm
Hence the wavelength of the sound is 100 cm.
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is the magnitude
Answer:
Centripetal acceleration = 0.79 m/s²
Explanation:
Given the following data;
Radius, r = 2.6 km
Time = 360 seconds
Conversion:
2.6 km to meters = 2.6 * 1000 = 2600 meters
To find the magnitude of centripetal acceleration;
First of all, we would determine the circular speed of the car using the formula;
[tex] Circular \; speed (V) = \frac {2 \pi r}{t}[/tex]
Where;
r represents the radius and t is the time.Substituting into the formula, we have;
[tex] Circular \; speed (V) = \frac {2*3.142*2600}{360} [/tex]
[tex] Circular \; speed (V) = \frac {16338.4}{360} [/tex]
Circular speed, V = 45.38 m/s
Next, we find the centripetal acceleration;
Mathematically, centripetal acceleration is given by the formula;
[tex] Centripetal \; acceleration = \frac {V^{2}}{r}[/tex]
Where;
V is the circular speed (velocity) of an object.r is the radius of circular path.Substituting into the formula, we have;
[tex] Centripetal \; acceleration = \frac {45.38^{2}}{2.6}[/tex]
[tex] Centripetal \; acceleration = \frac {2059.34}{2600}[/tex]
Centripetal acceleration = 0.79 m/s²
PLEASE HELP! Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0° with the
horizontal. What is the height of the building?
A. 174 m
B. 110 m
C. 98 m
D. 50 m
Answer:
The height of building should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.
Explanation:
A cart weighing 40 pounds is placed on a ramp incline 15 degrees to the horizon. The cart is held in place by a rope inclined 60 degrees to the horizontal. find the force that the rope must exert on the cart to keep it from rolling down the ramp.
Answer: [tex]14.64\ N[/tex]
Explanation:
Given
Inclination of ramp is [tex]\theta=15^{\circ}[/tex]
Rope is inclined [tex]\phi=60^{\circ}[/tex] to the horizontal
Weight of cart [tex]W=40\ lb[/tex]
from the diagram, rope is at angle of [tex]45^{\circ}[/tex] w.r.t ramp
Sine component of weight pulls down the cart Cosine component of force applied through rope held it at the position
[tex]\Rightarrow 40\sin 15^{\circ}=F\cos 45^{\circ}\\\\\Rightarrow F=40\cdot \dfrac{\sin 15^{\circ}}{\cos 45^{\circ}}\\\\\Rightarrow F=40\times 0.366\\\Rightarrow F=14.64\ N[/tex]
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W
Complete question:
How many x-ray photons per second are created by an x-ray tube that produces a flux of x rays having a power of 1.00 W. Assume the average energy per photon in 78.0 keV.
Answer:
The number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
Explanation:
Given;
power of the flux produced, P = 1 W = 1 J/s
energy per photon, E = 78 keV
Convert the energy per photon to J
E = 78 x 10³ x 1.6 x 10⁻¹⁹ = 1.248 x 10⁻¹⁴ J / photon
let the number of photons = n
n(1.248 x 10⁻¹⁴ J / photon) = 1 J/s
[tex]n = \frac{1 \ J/s}{1.248 \times 10^{-14}\ J/photon } = 8.01 \times 10^{13} \ photons/s[/tex]
Therefore, the number of x-ray photons per second created by the x-ray tube is 8.01 x 10¹³ photons/sec
What is the essential concept of the photon theory of light?
Answer:
Cause it gives us internet
The brake shoes of your car are made of a material that can tolerate very high temperatures without being damaged. Why is this so
they have rubber and meatal so its going to get hot of course cause it a engine pushing and the gasoline is just all coming together but damage wise I still it the protection on that part to keep it from damage and life risk
I hope this may help
Suppose an astronomer observes a binary star system where the stars are separated by 2.0 AU , and they have an orbital period of 7.0 years . Using Newton's version of Kepler's Third Law, find the combined mass of the stars.
Answer:
4.408 [tex]\mathsf{M_{sun}}[/tex]
Explanation:
According to Kelper's Third Law, the equation of the combined mass (m₁+m₂) can be expressed as:
[tex](m_1 + m_2) = \dfrac{\text{(distance between stars)}^3}{\text{(orbital period)}^2}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{(6.0)^3}{(7)^2} \ M_{sun}[/tex]
[tex]\text{combined mass}(m_1+m_2)} =\dfrac{216}{49} \ M_{sun}[/tex]
combined mass (m₁+m₂) = 4.408 [tex]\mathsf{M_{sun}}[/tex]