Explanation:
Mechanical advantage (MA) = Load/Effort. Velocity ratio (VR) = distance effort moves/ distance load moves in the same time
A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 22 m/s and the other ball, of mass 1.3 kg, is moving downward at 11 m/s. How high do the combined two balls of putty rise above the collision point
Answer:
The height balls rise above the collision point, is approximately 7.37 meters
Explanation:
The given parameters just before the collision are;
The mass, m₁ and velocity, v₁ of the ball moving upward are;
m₁ = 3.0 kg, v₁ = 22 m/s
The mass, m₂ and velocity, v₂ of the ball moving downward are;
m₂ = 1.3 kg, v₂ = -11 m/s (downward motion)
The type of collision = Inelastic collision
We note that the momentum is conserved for inelastic collision
Let, [tex]v_f[/tex], represent the final velocity of the balls after collision, we have;
∴ Total initial momentum = Total final momentum
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·[tex]v_f[/tex]
Therefore, we get;
m₁·v₁ + m₂·v₂ = 3.0 kg × 22 m/s + 1.3 kg × (-11) m/s = 51.7 kg·m/s
(m₁ + m₂)·[tex]v_f[/tex] = (3.0 kg + 1.3 kg) ×
∴ 51.7 kg·m/s = 4.3 kg × [tex]v_f[/tex]
[tex]v_f[/tex] = (51.7 kg·m/s)/4.3 kg ≈ 12.023 m/s
The final velocity, [tex]v_f[/tex] ≈ 12.023 m/s
The maximum height, h, the combined balls will rise from the point of collision, moving upward at a velocity of [tex]v_f[/tex] ≈ 12.023 m/s, is given from the kinetic equation of motion, v² = u² - 2·g·h, as found follows
At maximum height, we have;
[tex]h_{max} = \dfrac{v_f^2}{2 \cdot g }[/tex]
Therefore;
[tex]h_{max} \approx \dfrac{12.023^2}{2 \times 9.81 } \approx 7.37[/tex]
The height the combined two balls of putty rise above the collision point, [tex]h_{max}[/tex] ≈ 7.37 m.
An unstrained horizontal spring has a length of 0.40 m and a spring constant of 340 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.033 m relative to its unstrained length. Determine (a) the possible algebraic signs and (b) the magnitude of the charges.
Answer:
(a) Both the charges are positive or negative.
(b) Teh value of each charge is 1.53 x 10^-5 C.
Explanation:
Spring constant, K = 340 N/m
Natural length, L = 0.4 m
stretch, y = 0.033 m
(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.
(b) The electrostatic force is balanced by the spring force.
[tex]\frac{kq^2}{(L + y)^2}=Ky\\\\\\\frac{9\times 10^9 q^2}{(0.4 +0.033)^2} = 340\times0.033\\\\q= 1.53\times 10^{-5} C[/tex]
for the equation BaCI2 + Na2SO4 > BaSO4 + 2NaCI
A. reactants: 1 ;products: 1
B. reactants: 1 ;products: 2
C. reactants: 2 ;products: 1
D. reactants: 2 ;products: 2
How do space probes make it past the asteroid belt without crashing into asteroids?
Answer:
The thing is space is really vast like really big so even though the asteroid belt looks really cramped it isn't. There's a lot of space between asteriods and using simple navigation and maneuvering, space probes can easily make it through without the threat of crashing.
Explanation:
(a) What is the escape speed on a spherical asteroid whose radius is 301 km and whose gravitational acceleration at the surface is 0.412 m/s2
Answer:
[tex]V.E=498.02m/s^2[/tex]
Explanation:
From the question we are told that:
Radius [tex]r=301Km[/tex]
Gravitational acceleration [tex]g=0.412 m/s^2[/tex]
Generally the equation for Escape velocity is mathematically given by
[tex]V.E^2=2gr[/tex]
[tex]V.E^2=2*0.412m/s^2*301000[/tex]
[tex]V.E^2=248024[/tex]
[tex]V.E=\sqrt{248024}[/tex]
[tex]V.E=498.02m/s^2[/tex]
A student is investigating the affect of different salts on melting points. Four patches of ice of equal
size are roped off and a
different type of salt is poured on each, one receives table salt (NaCl), one
receives Calcium Chloride (CaCl2), one receives Potassium Carbonate (KCO3) and the fourth
receives inert sand instead. Each patch receivęs an equal amount of salt or sand. The student
measures the volume of ice remaining and subtracts it from the original volume of ice to see how
much melted away. What is a control variable in this experiment?
A. The size of the ice patches.
B. The type of salt applied to the ice.
C. The amount of ice that melted.
D. None of these.
Answer:
A. The size of the ice patches
Explanation:
In an experiment, the control variable also known as the CONSTANT is the variable that must be kept uniform or the same for all groups throughout the experiment in order not to influence the outcome of the experiment.
According to the experiment described in this question, the effect of different salts on melting points is investigated by a student. Sodium chloride (NaCl), Calcium Chloride (CaCl2), Potassium Carbonate (KCO3) and inert sand are the four types of salt used. The volume of the ice used and melted was finally measured. This means that the SIZE OF THE ICE PATCHES USED is the control variable of the experiment as the same size was used for all groups throughout.
Answer:
the size of the ice patches
Explanation:
How energy is obtained due to flow of charges?
Q011) The Doppler effect a. occurs when the frequency of sound waves received is lower if the wave source is moving toward you than if it's moving away. b. occurs when the pitch of a sound gets lower if the source is receding. c. is the basic explanation for the blue shift of light in our Universe. d. can be applied only to sound waves.
Answer:
Option (c) is correct.
Explanation:
The apparent change in the frequency of light due to the relative motion between the source and the observer is called Doppler's effect.
When the source is moving towards the observer which is at rest, the apparent frequency increases and if the observer is moving away the frequency of sound decreases.
It occurs for both light and sound.
So, to explain the blue shift of light in the universe is due to the Doppler's effect of light.
If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final
temperature?
Answer:
T₂ = 305.17 K
Explanation:
Given that,
Heat, Q = 6000 J
Mass, m = 200 gram
Initial temperature, T₁ = 25° C
We need to find its final temperature. Let it is T₂.
We know that,
[tex]Q=mc\Delta T[/tex]
Where
c is the specific heat of water, c = 4.18 J/g°C
So,
[tex]6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K[/tex]
So, the final temperature is equal to 305.17 K.
A uniform metre ruler scale balanced at 40 cm mark, when weight 25 gf and 10gf are suspended at 10cm mark and 75 cm mark respectively.Calculate the weight of the metre scale.
Answer:
40 gf.
Explanation:
Please see attached photo for diagram.
In the attached photo, W is the weight of metre rule.
The weight of the metre rule can be obtained as follow:
Clockwise moment = (W×10) + (10×35)
Clockwise moment = 10W + 350
Anticlock wise moment = 25 × 30
Anticlock wise moment = 750
Clockwise moment = Anticlock wise moment
10W + 350 = 750
Collect like terms
10W = 750 – 370
10W = 400
Divide both side by 10
W = 400/ 10
W = 40 gf
Thus, the weight of the metre rule is 40 gf
Answer:
40 gf
Explanation:
The balance point of the uniform meter rule with the suspended weights = 40 cm = The pivot point
The location where the 25 gf weight is suspended = 10 cm
The location where the 10 gf weight is suspended = 75 cm
Let W represent the weight of the meter rule.
We have that the location of the application of the weight of the meter rule is at the center, 50 cm mark, point
Given that the meter rule is balanced, and taking moment about the pivot point, we have;
The moment om the left hand side, LHS, of the pivot point = The moment on the right hand side, RHS, of the pivot point
The moment on the LHS = 25 gf × (40 cm - 10 cm) = 750 gf·cm
The moment on the RHS = W × (50 cm - 40 cm) + 10 gf × (75 cm - 40 cm)
The moment on the RHS = W·(10 cm) + 350 gf·cm
∴ 750 gf·cm = W·(10 cm) + 350 gf·cm
W·(10 cm) = 750 gf·cm - 350 gf·cm = 400 gf·cm
W = (400 gf·cm)/(10 cm) = 40 gf
The weight of the meter scale (rule), W = 40 gf.
Rick works off commission. He earns 10 percent of all manufacturing equipment he sells. if he made a sale of $9,000 how much was his commission
Answer:
$900
Explanation:
Step 1: Our output value is 9000.
Step 2: We represent the unknown value with x.
Step 3: From step 1 above,$9000=100\%$
Step 4: Similarly, x=10%
Step 5: This results in a pair of simple equations:
$9000=100
Step 6: By dividing equation 1 by equation 2 and noting that both the RHS (right hand side) of both
equations have the same unit (%); we have
\frac{9000}{x}=\frac{100\%}{10\%}
Step 7: Again, the reciprocal of both sides gives
\frac{x}{9000}=\frac{10}{100}$
\Rightarrow x=900$
Therefore, $10\%$ of $9000$ is $900$
unknown value with x
step 1 above,$9900=100%
similarly ,x=10%
Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
The softball experiences
force as a result of Amy’s throw. As the ball moves, it experiences
from the air it passes through. It also experiences a downward pull because of
.
Answer:
1.the friction of air, gravity2.gravity
Answer:
The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.
Explanation:
Plato
Which statement is the best interpretation of the ray diagram shown?
Answer:
Explanation:
First off, this lens is concave. Second, the image is obviously smaller, and third, the only thing that is NOT obvious, is the fact that real images are always upside down and virtual images are always right-side-up. So the choice you're looking for is D.
Answer:
D - A concave lens forms a smaller, virtual image
Explanation:
Which would be used to measure the distance between the earth and a planet ,meter ruler or a measuring tape? Why?
Answer:
parallax
Due to foreshortening, nearby objects show a larger parallax than farther objects when observed from different positions, so parallax can be used to determine distances. To measure large distances, such as the distance of a planet or a star from Earth, astronomers use the principle of parallax.
show that pv=nrt has the same unit with energy
Answer:
It is proved.
Explanation:
PV = n R T
where, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, T is the absolute temperature.
Unit of pressure is Pascal= Newton/square meter = [tex][ML^{-1}T^{-2}][/tex]
Unit of volume is cubic meter = [tex][L^{3}][/tex]
So, the unit of P V
[tex][ML^{-1}T^{-2}] \times [L^{3}]\\\\[ML^2T^{-2}}[/tex]
[tex][ML^{2}T^{-2}][/tex]
It is same as the unit of energy.
hence proved.
What is the magnitude of the electric field at a point 0.0055 m from a 0.0025
C charge?
kg
Use E = and k=9.00 x 10 N.m²/C2.
O A. 7.4 x 1011 N
O B. 2.0 x 1010 N
O C. 4.1 x 10°N
OD. 7.9 x 1012 N
Answer:
Explanation:
The equation for the electric field is
[tex]E=\frac{kQ}{r^2}[/tex] so filling in:
[tex]E=\frac{9.00*10^9(.0025)}{(.0055)^2}[/tex] which in the end gives you
E = 7.4 × 10¹¹, choice A
What is the enthalpy change, AH, for this reaction? Show your work to receive full credit. URGENT PICTURE INCLUDED !!!
Answer:
150 kJ
Explanation:
Applying,
ΔH = Energy level of Product(E') - Energy level of reactant(E)
Where ΔH = enthalpy change, E' and E = energy level of the product and the reactant respectively
ΔH = E'-E............. Equation 1
From the diagram,
Given: E' = 200 kJ, E = 50 kJ
Substitute these values into equation 1
ΔH = 200-50
ΔH = 150 kJ
Hence the enthalpy change for the reaction is 150 kJ
A 1.25 kg block is attached to a spring with spring constant 17.0 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 49.0 cm/s . What are You may want to review (Pages 400 - 401) . Part A The amplitude of the subsequent oscillations
Answer:
The amplitude of the subsequent oscillations is 13.3 cm
Explanation:
Given;
mass of the block, m = 1.25 kg
spring constant, k = 17 N/m
speed of the block, v = 49 cm/s = 0.49 m/s
To determine the amplitude of the oscillation.
Apply the principle of conservation of energy;
maximum kinetic energy of the stone when hit = maximum potential energy of spring when displaced
[tex]K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm[/tex]
Therefore, the amplitude of the subsequent oscillations is 13.3 cm
State any ten reasons why students in Uganda need to study physics.(use examples of the possible technological advancements which can be made using the knowledge of physics)
Answer:
The answer is "physics ".
Explanation:
Physics is the branch of science that addresses the properties of crystalline and its interaction with the fundamental elements of the universe. It covers subjects ranging in quantum mechanics with extremely little ones with quantum mechanics to the whole cosmos. You must be constant whether you like it or not, thus everyone must learn physics, irrespective of whether they're in Uganda, and plenty of other countries should have physics to dare study.
a body is moving along a circular path 'r'. what will be the distance and displacement of the body when it completed half a revolution?
After half of a revolution ...
==> Distance = π•r
==> Displacement = 2•r
a ball is thrown straight up into the air while the ball is traveleling upwards what are the magnitue and direction
Answer: hi your question is incomplete attached below is the complete question
answer :
magnitude of acceleration : | a | = g = 9.81 m/s^2
direction : a = - g j
Explanation:
Neglecting Air resistance
magnitude of acceleration :
| a | = g = 9.81 m/s^2
Direction of acceleration
a = - g j ( given that the direction of acceleration is against the acceleration due to gravity i.e. in the opposite direction )
URGENT
A student runs at 4.5 m/s [27° S of W] for 3.0 minutes and then he turns and runs at 3.5 m/s [35° S of E] for 4.1 minutes.
a. What was his average speed?
b. What was his displacement?
PLEASE SHOW ALL WORK
Answer:
(a) 3.93 m/s
(b) 861.66 m
Explanation:
A = 4.5 m/s [27° S of W] for 3.0 minutes
B = 3.5 m/s [35° S of E] for 4.1 minutes
Distance A = 4.5 x 3 x 60 = 810 m
Distance B = 3.5 x 4.1 x 60 = 861 m
(a) The average speed is defined as the ratio of the total distance to the total time.
Total distance, d = 810 + 861 = 1671 m
total time, t = 3 + 4.1 = 7.1 minutes = 7.1 x 60 = 426 seconds
The average speed is
[tex]v=\frac{1671}{426}=3.93 m/s[/tex]
(b)
[tex]\overrightarrow{A} = 810(- cos 27 \widehat{i} - sin 27 \widehat{j})=- 721.7 \widehat{i} - 367.7 \widehat{j}\\\\\overrightarrow{B} = 861( cos 35 \widehat{i} - sin 35 \widehat{j})= 705.3 \widehat{i} - 493.8 \widehat{j}\\\\\overrightarrow{C} = (- 721.7 + 705.3) \widehat{i} - (367.7 + 493.8) \widehat{j} \\\\\overrightarrow{C}= - 16.4 \widehat{i} - 861.5 \widehat{j}[/tex]
The magnitude is
[tex]C =\sqrt{16.4^2+861.5^2} = 861.66 m[/tex]
20 swings takes 5 seconds in the pendulum. calculate the periodic time of the swing
Answer:
0.25s
Explanation:
5/20 = 0.25s
This might be correct
Complete the sentence below using one of the following
words: equilibrium, flux, motion.
If supported at its center of gravity, an object will remain in ______ any position.
Answer:
equilibrium
Explanation:
acceleration will remain constant
How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead is Lv = 871 × 103 J/kg.
Answer:
Heat required = mass× latent heat Q = 0.15 × 871 ×
The heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.
What is heat?The movement of energy from a hot to a cold item is characterized as heat. Heat energy flows from a hot material to a cold one.
This occurs because faster-vibrating molecules transmit their energy to slower-vibrating ones.
The given data in the problem is;
m is the mass of lead = 0.15 kg
T is the temperature = 1750°C,
The latent heat of vaporization for lead is, [tex]\rm L_V[/tex] = 871 × 10³ J/kg.
The heat is found as;
[tex]\rm Q= m \times L_V \\\\ \rm Q= 0.15 \times 871 \times 10^3 \\\\ Q=130,650 \ J[/tex]
Hence the heat required to evaporate 0.15 kg of lead at 1750°C will be 130,650 J.
To learn more about the heat refer to the link;
brainly.com/question/1429452
How many more neutrons are in a I SOTOPE of copper-14 than in standard carbon atom
Answer:
2 more neutrons
Explanation:
To obtain the answer to the question, let us calculate the number of neutrons in carbon–14 and standard carbon (i.e carbon–12). This can be obtained as follow:
For carbon–14:
Mass number = 14
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
14 = 6 + Neutron
Collect like terms
14 – 6 = Neutron
8 = Neutron
Neutron number = 8
For carbon–12:
Mass number = 12
Proton number = 6
Neutron number =?
Mass number = Proton + Neutron
12 = 6 + Neutron
Collect like terms
12 – 6 = Neutron
6 = Neutron
Neutron number = 6
SUMMARY:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Finally, we shall determine the difference in the neutron number. This can be obtained as follow:
Neutron number of carbon–14 = 8
Neutron number of carbon–12 = 6
Difference =?
Difference = (Neutron number of carbon–14) – (Neutron number of carbon–12)
Difference = 8 – 6
Difference = 2
Therefore, carbon–14 has 2 more neutrons than standard carbon (i.e carbon–12)
In which region is there most likely to be a volcano
A
B
C
D
please help me with my question due tomorrow morning,
Answer:
D)7 1/2 or 15/2
Explanation:
Let's calculate Combined resistance of the parallel first
1/Rt= 1/2+1/6=4/6
Rt=6/4 which is also equal with 3/2
Now let's add it with the series one
Rt= 6+3/2
=15/2 And when we put that un a mixed fraction 7 1/2
Juanita ran one mile around her school track in six minutes. What is
her average speed, and what is the magnitude of her average velocity?
10 mph, 0 mph
6 mph, 0 mph
6 mph, 6 mph
10 mph, 10 mph
Answer:
The correct option is a) 10 mph, 0 mph.
Explanation:
1. The average speed (S) is a magnitude given by:
[tex] S = \frac{D}{T} [/tex]
Where:
D: is the total distance = 1 mi
T: is the total time = 6 min
[tex] S = \frac{D}{T} = \frac{1 mi}{6 min}*\frac{60 min}{1 h} = 10 mph [/tex]
Hence, the average speed is 10 mph.
2. The average velocity is a vector:
[tex] V = \frac{\Delta d}{\Delta t} = \frac{d_{f} - d_{i}}{t_{f} - t_{i}} [/tex]
Where:
[tex]d_{f}[/tex]: is the final distance
[tex]d_{i}[/tex]: is the initial distance
[tex]t_{f}[/tex]: is the final time
[tex]t_{i}[/tex]: is the initial time
Since Juanita ran one mile around her school track, the final position is the same that the initial position, so the magnitude of the average velocity is zero.
Therefore, the correct option is a) 10 mph, 0 mph.
I hope it helps you!
A cannon sitting on level ground is aimed at 45.0 degrees relative to the horizontal. It fires a test shot at a target located 100.0 meters away from the cannon on the same level ground. The test overshoots the target by 20.0 meters. Which of the following angles can the cannon be adjusted to to hit the target. You may neglect air resistance and assume the cannon always delivers the same initial velocity to the cannonball .
A. 35.9 deg
B. 49.1 deg
C. 28.2 deg
D. 52.8 deg
E. 22.7 deg
Answer:
C. 28.2 deg
Explanation:
The horizontal range of a projectile is given as:
[tex]R = \frac{v^2Sin2\theta}{g}[/tex]
where,
R = Range
v = speed
θ = angle of launch
g = acceleration due to gravity = 9.81 m/s²
First, we will find the launch speed (v) by using the initial conditions:
R = 120 m
θ = 45°
Therefore,
[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]
Now, consider the second scenario to hit the target:
R = 100 m
Therefore,
[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]
Hence, the correct option is:
C. 28.2 deg
