Answer:
Explanation:
From the given information:
The initial PE [tex](PE)_i[/tex] = m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = [tex]P.E_f -P.E_i[/tex]
ΔP.E = 0 - [tex](PE)_i[/tex]
ΔP.E = [tex]-P.E_i[/tex]
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
[tex]\Delta P.E + \Delta K.E + \Delta U = 0[/tex]
[tex]\Delta U = -\Delta P.E - \Delta K.E[/tex]
this can be re-written as:
[tex]\Delta U =- (-\Delta P.E_i) - \Delta K.E[/tex]
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
[tex]\Delta U =(70\%) \Delta P.E_i-0[/tex]
[tex]\Delta U =(0.70) (490.5)[/tex]
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
If you exert a force of 5 N into a nutcracker, and it outputs a force of 20 N, what is the mechanical advantage of the nutcracker. Show formula PLSSS HELPPPP!!! i'll make you brainliest
Answer: 4
Explanation:
MA = output force / input force
MA = 20 / 5
MA = 4
Hope this helps. Please mark brainliest.
Calculating Acceleration
Initial
velocity
Time to travel
0.25 m
Final
velocity
Acceleration
Time to travel
0.50 m
# of
washers
11
(m/s)
V2
(m/s)
ti
(s)
t₂
(s)
a = (v2 - v4)/(t2-tı)
(m/s)
1
0.11
0.28
2.23
3.13
0.19
2
0.13
0.36
1.92
2.61
The acceleration of the car with two washers added to the string would be
I can not even read this question.
What are you trying to even say?
The acceleration of the car with two (2) washers added is equal to 0.33 [tex]m/s^2[/tex].
Given the following data:
Initial velocity = 0.13 m/s.Final velocity = 0.36 m/s.Initial time = 1.92 seconds.Final time = 2.61 seconds.What is an acceleration?An acceleration can be defined as the rate of change of velocity of an object with respect to time and it is measured in meter per seconds square.
How to calculate average acceleration.In Science, the average acceleration of an object is calculated by subtracting its initial velocity from the final velocity and dividing by the change in time for the given interval.
Mathematically, average acceleration is given by this formula:
[tex]a = \frac{V\;-\;U}{t_f-t_i}[/tex]
Where:
V is the final velocity.U is the initial velocity.[tex]t_i[/tex]initial time measured in seconds.[tex]t_f[/tex] final time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]a = \frac{0.36\;-\;0.13}{2.61\;-\;1.92}\\\\a=\frac{0.23}{0.69}[/tex]
a = 0.33 [tex]m/s^2[/tex]
Read more on acceleration here: brainly.com/question/24728358
From the top of the leaning tower of Pisa, a steel ball is thrown vertically downwards with a speed of 3.00 m/s. if the height of the tower is 200 m, how long will it take for the ball to hit the ground? Ignore air resistance.
Answer:
66,7 seconds
Explanation:
the formula for height/distance is : S=v.t
According to Newton’s law of universal gravitation, which statements are true?
A hockey puck is sliding across the ice with an initial velocity of 25 m/s. If the coefficient of friction between the hockey puck and the ice is 0.08, how much time (in seconds) will it take before the hockey puck slides to a stop
Answer: 31.89seconds
Explanation:
Based on the information given, we are meant to calculate deceleration which will be:
t = V/a
where, a = mg
Therefore, t = V/mg
t = 25/0.08 × 9.8
t = 25/0.784
t = 31.89seconds
Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.
What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?
Answer: The energy released as thermal energy is 6.5 J
Explanation:
Energy stored by the spider when it relaxes is given by:
[tex]E_o=\text{Resilience}\times \text{Work}[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\text{Work done}-E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5J[/tex]
Hence, the energy released as thermal energy is 6.5 J
The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J
What is thermal energy?Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Energy stored by the spider when it relaxes is given by:
[tex]\rm E_o=Resilience \ \times Work[/tex]
We are given:
Resilience = 0.35
Work done = 10.0 J
Putting values in above equation, we get:
[tex]\rm E_o=0.35\times 10[/tex]
[tex]E_o=3.5\ J[/tex]
Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:
[tex]E_T=\rm Work done -E_o[/tex]
Putting values in above equation, we get:
[tex]E_T=(10-3.5)=6.5\ J[/tex]
Hence, the energy released as thermal energy is 6.5 J
To know more about thermal energy follow
https://brainly.com/question/19666326
A balloon pops, making a loud noise that startles you. What kind of energy best describes this experience?
A. Thermal Energy
B. Sound Energy
C. Gravitational Energy
D. Radiant Energy
2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time
Answer:
(c) at rest
Explanation:
Given
See attachment for the distance time graph
Required
What does the graph illustrate?
From the graph, we can see that the line of distance is a horizontal line.
This suggests that a time increases, the distance remains unchanged
When distance remains unchanged over time, then it means the object is at rest.
Hence, (c) is correct
A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV
Answer:
The period the field must be reduced to zero is 9.81 x 10⁻⁵ s
Explanation:
Given;
initial value of the magnetic field, B₁ = 0.276 T
number of turns of the solenoid, N = 517 turns
diameter of the solenoid, d = 10.5 cm = 0.105 m
induced emf, = 12.6 kV = 12,600 V
when the field becomes zero, then the final magnetic field value, B₂ = 0
The induced emf is given by Faraday's law;
[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]
Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s
A laser emits a single 3.0-ms pulse of light that has a frequency of 2.83E11 Hz and a total power of 65000 W. How many photons are in the pulse? Please provide all equations and work.
6.0E23
1.0E24
2.4E25
3.6E25
4.8E26
Answer:
The number of photons in the pulse is 1.04 x 10²⁴
Explanation:
Given;
frequency of the emitted photons, f = 2.83 x 10¹¹ Hz
duration of the incident light, t = 3 ms = 3 x 10⁻³ s
power of the incident light, P = 65,000 W
The energy of each photon emitted is calculated as;
E = hf
where;
h is Planck's constant, = 6.626 x 10⁻³⁴ Js
E = 6.626 x 10⁻³⁴ x 2.83 x 10¹¹
E = 1.875 x 10⁻²² J
let the number of photons in the pulse = n
n(E)= Power x time
[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]
A mass of 4 kg is traveling over a quarter circular ramp with a radius of 10 meters. At the bottom of the incline the mass is moving at 21.3 m/s and at the top of the incline the mass is moving at 2.8 m/s. What is the work done by all non-conservative force in Joules?
Answer:
499.7 J
Explanation:
Since total mechanical energy is conserved,
U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.
So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
Substituting the values of the variables into the equation, we have
mgh₁ + 1/2mv₁² + W₁ = mgh₂ + 1/2mv₂² + W₂
4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)² + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)² + W₂
0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s² + W₂
907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s² + W₂
907.38 kgm²/s² = 407.68 kgm²/s² + W₂
W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²
W₂ = 499.7 kgm²/s²
W₂ = 499.7 J
Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J
The power in an electrical circuit is given by the equation P= RR, where /is the current flowing through the circuit and Ris the resistance of the circuit. What is the current in a circuit that has a resistance of 100 ohms and a power of 15 watts?
[pleas ee helpppp)
I= 0.39 A
OPTION B is the correct answer.
A force of 3 newtons moves a 10 kilogram mass horizontally a distance of 3 meters. The mass does not slow down or speed up as it moves. Which of the following must be true?
a) 9 joules of kinetic energy were produced
b) 9 joules of gravitational potential energy were produced
c) 9 joules of heat energy were produced
d) 9 joules of kinetic energy and heat were produced
Answer:
9 joules of heat energy was produced
Explanation: there is no acceleration therefore its not a kinetic energy
Energy= force × distance
= 3×3
=9
An electric drill starts from rest and rotates with a constant angular acceleration. After the drill has rotated through a certain angle, the magnitude of the centripetal acceleration of a point on the drill is 8.2 times the magnitude of the tangential acceleration. What is the angle?
Answer:
The angle is 4.1 rad.
Explanation:
The centripetal acceleration (α) is given by:
[tex] \alpha = \omega^{2} r [/tex] (1)
Where:
ω: is the angular velocity
r: is the radius
And the tangential acceleration (a) is:
[tex] a = \alpha r [/tex] (2)
Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:
[tex] \omega^{2} r = 8.2\alpha r [/tex]
[tex] \omega^{2} = 8.2\alpha [/tex] (3)
Now, we can find the angle with the following equation:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]
Where:
[tex] \omega_{f}[/tex]: is the final angular velocity [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)
[tex]\Delta \theta[/tex]: is the angle
[tex] \omega^{2} = 2\alpha \Delta \theta [/tex] (4)
By entering equation (3) into (4) we can calculate the angle:
[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]
[tex] \Delta \theta = 4.1 rad [/tex]
Therefore, the angle is 4.1 rad.
I hope it helps you!
20 pts.
A man forgets that he set his coffee cup on top of his car. He starts to drive and the coffee CUP rolls off the car onto the road. How does this scenario demonstrate the first law of motion? Be specific and use the words from the law in your answer.
Answer:
The cup is acted upon by an unbalanced force which is the acceleration of the car, but before it was an object at rest that stayed at rest.
Explanation:
Newton's first law of motion states, "if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force."
Since the cup is at rest while sitting on top of the car, it stays at rest as the car begins to move. Since the car is accelerating and the cup is not, the cup falls off of the car.
