Answer:
m_2 = 2600kg
Explanation:
P_1 = P_2
P = (m_1)*(v_1)+(m_2)(v_2)
P_1 = (3250kg)(8.1m/s)+(m_2)(0m/s)
P_1 = 26,325 kg*m/s
P_2 = (3250kg)(4.5m/s)+(m_2)(4.5m/s)
P_2 = 14,625kg*m/s+(4.5m/s)m_2
26,325 kg*m/s = 14,625kg*m/s+(4.5m/s)m_2
11,700kg*m/s = (4.5m/s)m_2
m_2 = 2600kg
Vi
Vf t
Find the acceleration of a car that goes from 32 m/s to 96 m/s in 8s.
Select one:
O a. 16 m/s2
O b. 8 m/s2
O c. 62 m/s2
d. -8 m/s2
Answer:
8m/s^2
Explanation:
Acceleration=initial velocity-final velocity/time
initial velocity(u) is 32 final velocity (v) is 96
therefore
a=96-32/8
a=8m/s squared
hope it helps
A toy airplane is flying at a speed of 3 m/s with an acceleration of 1.1 m/s squared how fast is it flying after five seconds
Answer:
[tex]8.5\frac{m}{s}[/tex]
Explanation:
Use Kinematics:
[tex]v = v_0 + at[/tex]
[tex]v = 3 + 1.1 * 5[/tex]
[tex]v = 8.5 \frac{m}{s}[/tex]
In certain ranges of a piano keyboard, more than one string istuned to the same note to provide extra loudness. For example, thenote at 1.10 x 102 Hz has two strings at this frequency.If one string slips from its normal tension of 6.00 x102 N to 5.40 x 102 N, what beatfrequency is heard when the hammer strikes the two stringssimultaneously?
Answer:
The beat frequency is 5.5 Hz.
Explanation:
f = 110 Hz, T = 600 N , T' = 540 N
let the frequency is f'.
[tex]\frac{f'}{f}=\sqrt\frac{T'}{T}\\\\\frac{f'}{f}=\sqrt\frac{540}{600}\\\\\frac{f'}{f}=0.95\\\\f'= 104.5 Hz[/tex]
So, the beat frequency is f - f' = 110 - 104.5 = 5.5 Hz
it will be easier to lift a load in wheel barrow if the load is moved towards the wheel. Give Reason.
Answer:
Here's your answer: The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.
Explanation:
Answer:
Explanation:
The wheel barrows wheel and axle helps the wheelbarrow to move without friction this making it easier to push or pull.thats why it will be easier to lift a load in a wheelbarrow if it's transferred towards the wheel..
Hope it helps
you are pedaling forward on your bike. which of the following would decrease your acceleration?
A. A decrease in your mass
B. A increase in your applied force
C. An increase in your mass
D. An increase in your speed
Pedaling forward over the bike with the increase in your applied force would decrease your acceleration. Thus, option B is correct.
What is acceleration?Acceleration is define as the rate at which velocity changes with time, in terms of both speed and direction. The equation for acceleration is:
a=F/m
where, a = acceleration,
F = force applied
m = mass of the object
The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object.
As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Therefore, if you are pedaling forward on your bike then a increase in your applied force would decrease your acceleration. Thus, option B is correct.
Learn more about acceleration here:
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Answer:
I just did it. A is right.
Explanation:
An object 20mm high is placed 10cm from a sperical mirror and forms a virtual image which is 40mm high. what is the radius of curvature off the mirrors
Answer:
R = 40 cm
Explanation:
From the formulae of magnification:
[tex]\frac{q}{p}=\frac{image\ height}{object\ height}\\\\\frac{q}{10\ cm} = \frac{4\ cm}{2\ cm}\\\\q = (10\ cm)(2)\\\\q = 20\ cm[/tex]
where,
q = image distance from mirror
p = object distance from mirror
Using thin lens formula:
[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f}=\frac{1}{10\ cm}+\frac{1}{-20\ cm}\\\\\frac{1}{f} = 0.05\\\\f = 20\ cm[/tex]
q is negative for the virtual image.
Now, the radius of the spherical mirror is double the focal length (f):
R = 2f
R = 2(20 cm)
R = 40 cm
A bowling ball is pushed with a force of 22.0 N and accelerates at 5.5 m/s2. What is the mass of the bowling ball?
Answer:
4 kg
Explanation:
F (force) = 22.0 N
m (mass) = ? (in Kg)
a (acceleration) = 5.5 m/s²
We apply the data to the Resultant Force formula, we have:
f=ma
22=m x 5.5
5.5m=22
m=22/5.5
m=4kg
Answer:
[tex]\boxed {\boxed {\sf 4 \ kilograms}}[/tex]
Explanation:
According to Newton's Second Law of Motion, force is the product of mass and acceleration.
[tex]F= m*a[/tex]
The bowling ball was pushed with a force of 22.0 Newtons and the acceleration was 5.5 meters per second squared.
1 Newton is equal to 1 kilogram meter per second squared. Therefore, the force of 22.0 N is equal to 22 kg*m/s² (we do this conversion so we can easily cancel the units later).
F= 22.0 kg*m/s²a= 5.5 m/s²Substitute the values into the formula.
[tex]22.0 \ kg*m/s^2 = m * 5.5 \ m/s^2[/tex]
We are solving for the mass, so we must isolate the variable m. It is being multiplied by 5.5 meters per second squared. The inverse of multiplication is division, so we divide both sides by 5.5 m/s²
[tex]\frac { 22.0 \ kg*m/s^2}{5.5 \ m/s^2} = \frac{m * 5.5 \ m/s^2}{5.5 \ m/s^2}[/tex]
[tex]\frac { 22.0 \ kg*m/s^2}{5.5 \ m/s^2}=m[/tex]
The units of meters per second squared (m/s²) cancel.
[tex]\frac { 22.0 \ kg}{5.5 }=m[/tex]
[tex]4 \ kg =m[/tex]
The mass of the bowling ball is 4 kilograms.
The graph shows the motion of a car. Which segment
shows that the car is slowing down?
O A
OB
Ос
OD
Answer: its C
Explanation:
When developing an experiment design, which could a scientist take to improve the quality of the results?
The options missing are;
A. Make only a few changes to the manipulated variable.
B. Identify any biases about the answer to the scientific question.
C. Include as many details as possible when writing the conclusion.
D. Share her ideas through peer review.
Answer:
B. Identify any biases about the answer to the scientific question.
Explanation:
Since the question is asking what can be done to improve the quality of the results, it means what can the scientist do to make sure that the result answers the question properly and has little or no biases.
Thus, looking at the options, the only one that comes close to actually improving on the quality of the results before finalizing is option B where the scientist is to identify any biases.
The search for black holes involves searching for Group of answer choices single stars that emit large amounts of X-rays. a stellar core greater than 3 solar masses large spherical regions from which no light is detected. pulsars with periods less than one millisecond. pulsars that are orbited by planets.
Answer:
A stellar core greater than 3 solar masses
Explanation:
The process of the evolution of a star in which the star changes form with time, includes phases of evolution which are determined by the mass of the core of the star, which involves a process of conversion of elements and the contraction of the star
Where the mass of the core of the star is more than the mass of three Suns, (3 solar masses) the contraction of the star continues, due to the gravitational force being larger than the pressure exerted by neutron, such that the star collapses to a black hole
Therefore, the search for black holes involves searching for a stellar core greater than 3 solar masses.
An electron enters the magnetic field at right angle from left B into paper. The electron will be deflected?
Answer:
According to this direction of force is perpendicular to the direction of current and magnetic field. Therefore force is opposite to electron into the paper at 90°.
Determine the mass of the object below to the correct degree of precision.
433.0 g
433.45 g
433.40 g
434.50 g
Answer:
You did not include the picture but I found a similar question that I can explain and you can then use to answer yours.
The middle scale is in hundreds. The arrow here is pointing to 100 so the mass of the object is more than 100g but less than 200g.
The highest scale is in 10s. The arrow is pointing to 0. This means that the object weighs more than 100 but less than 110 g. If the arrow was pointing to 10 then the object would weigh more than 110g.
The bottommost scale is in 1s. Notice how the scale is pointing to 5.7 grams. The object is therefore 105.7 grams.
Look for the option closest to this because the 105.7g might be rounded up.
You will then find 105.65g as your answer.
Answer:433.0 g
Explanation:i’m not 100% sure
How can we measure electromotive force of a battery illustrated answer with the help of circuit diagram
Answer:
the emf of the cell can be determined by measuring the voltage across the cell using a voltmeter and the current in the circuit using an ammeter for various resistances.
The net charge difference across the membrane, just like the charge difference across the plates of a capacitor, is what leads to the voltage across the membrane. How much excess charge (in picocoulombs, where 1 pc = 1x10-12 C) must lie on either side of the membrane of an axon of length 2 cm to provide this potential difference (0.07 V) across the membrane? You may consider that a net positive charge with this value lies just outside the axon cell wall, and a negative charge with this value lies just inside cell wall, like the equal and opposite charges on capacitor plates. Again, make sure to include LaTeX: \kappa=7κ = 7 for the lipid bilayer.
Answer:
a) Q = 1.24 10⁻² pC, b) Q = 8.68 10⁻² pC
Explanation:
a) the capacitance is defined
C = [tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]
Q = ε₀ [tex]\frac{A}{d} \ \Delta V[/tex]
let's calculate
Q = 8.85 10⁻¹² 0.07 [tex]\frac{A}{d}[/tex]
Q = 0.6195 10⁻¹² [tex]\frac{A}{d}[/tex]
where a is the area of the membrane
A = d L
Q = 0.6195 10⁻¹² Ll
Q = 0.6195 10⁻¹² 0.02
Q = 1.24 10⁻¹⁰ C
Q = 1.24 10⁻² pC
B) the membrane is full of fat with k = 7
C = [tex]\frac{Q}{\Delta V} = k \epsilon_o \ \frac{A}{d}[/tex]
Q = k ε₀ [tex]\frac{A}{d} \ \Delta V[/tex]
Q = k Q₀
Q = 7 1.24 10⁻²
Q = 8.68 10⁻² pC
Use the information below the answer the following 3 questions.
A 50 kg crate is being dragged across a floor by a force of 225 N at an angle of 40o from the horizontal. The crate is dragged a distance of 5.0 m and the frictional force is 60 N.
Question 2 (2 points)
Question 2 options:
The work done on the crate by the applied force is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 3 (2 points)
Question 3 options:
The work done on the crate by the frictional force is -___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Question 4 (2 points)
Question 4 options:
The net work done on the crate is ___x102 Nm. (Give your answer with the correct number of sign digs and do not include units).
Hint: Do not use rounded answers in subsequent calculations
Answer:
2. 8.62×10² Nm
3. 2.30×10² Nm
4. 6.32×10² Nm
Explanation:
2. Determination of the work done by the applied force.
Force (F) = 225 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fd × Cos θ
Wd = 225 × 5 × Cos 40
Wd = 8.62×10² Nm
3. Determination of the work done by the frictional force.
Frictional Force (Fբ) = 60 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fբd × Cos θ
Wd = 60 × 5 × Cos 40
Wd = 2.30×10² Nm
4. Determination of the net work done.
We'll begin by calculating the net force acting on the crate
Force applied (F) = 225 N
Frictional Force (Fբ) = 60 N
Net force (Fₙ) =?
Fₙ = F – Fբ
Fₙ = 225 – 60
Fₙ = 165 N
Finally, we shall determine the net Workdone. This can be obtained as follow:
Net force (Fₙ) = 165 N
Distance (d) = 5 m
Angle (θ) = 40°
Workdone (Wd) =?
Wd = Fₙd × Cos θ
Wd = 165 × 5 × Cos 40
Wd = 6.32×10² Nm
The speed of sound on steel is about 5000 m/s. A friend of yours, 2000 m away, strikes a railroad track which you have your ear on. How long after she strikes it will the sound arrive?
Answer:
t = 0.4 s
Explanation:
Given that,
The speed of sound on steel is about 5000 m/s.
A friend of yours, 2000 m away, strikes a railroad track which you have your ear on.
We need to find the time after she strikes it will the sound arrive. Let the time is t. We know that,
Speed = distance/time
So,
[tex]t=\dfrac{d}{v}\\\\t=\dfrac{2000}{5000}\\\\t=0.4\ s[/tex]
So, the required time is 0.4 seconds.
8. A copper container of 84g mass contains 84g of water at 20°C. 46g of water at 200°C is mixed with water in the copper ontainer. What is the final temperature of the water? Specific heat capacity of water = 4200 J kg-1 °C-1, Specific heat capacity of copper = 400 J kg-1 °C-1
Answer:
80 °C
Explanation:
The heat transfer parameters for the water and copper container are;
Mass of the copper container, m₁ = 84 g
Mass of the water in the container, m₂ = 84 g
Initial temperature of the water in the container, T₂ = 20°C
Mass of the hot water added, m₃ = 46 g
Initial temperature of the hot water, T₃ = 200°C
Specific heat capacity of water, c₂ = 4,200 J·kg⁻¹·°C⁻¹
Specific heat capacity of copper, c₁ = 400 J·kg⁻¹·C⁻¹
The formula for the specific heat, ΔQ = m·c·ΔT
The heat lost by the hot water = The heat gained by the container the and the cold water
The formula for the specific heat of the mixture is presented as follows;
m₃ × c₃ × (T₃ - T) = m₁ × c₁ × (T - T₁) + m₂ × c₂ × (T - T₂)
Where T represents the final temperature of the water
Therefore, by plugging in the values, we get;
46 × 4200 × (200 - T) = 84 × 400 × (T - 20) + 84 × 4200 × (T - 20)
38640000 - 193200·T = 386400·T - 7728000
38640000 + 7728000 = 46368000 = 386400·T + 193200·T = 579,600·T
∴ T = 46368000/579,600 = 80
The final temperature of the water, T = 80°C
Draw a closed circuit diagram of the battery of 2 cells arranged in series, connecting wire, switch and bulb; mark the direction of the current.
pls ill give brainly
(d) Below shows a hydraulic press with a pump piston of area 4 cm2 and a ram area of
12 cm2 . IF A force of 300 N is applied on the pump piston, find:
(i)the maximum load on the ram.
(ii) the velocity ratio of the machine.
(iii) the mechanical advantage of the machine.
(iv)the efficiency of the machine.
Answer:
Explanation:
hi
absolute potential difference ,due of point charge of 1C at a distance of 1 m is given by
Answer:
[tex] \implies U = \dfrac{kq}{r} [/tex]
[tex]\implies U = \dfrac{9 \times {10}^{9} \times 1}{1} [/tex]
[tex]\implies U = 9 \times {10}^{9} \: J[/tex]
We have that absolute potential difference ,due of point charge of 1C at a distance of 1 m is given by
[tex]\rho=9x10 ^{10}J[/tex]
From the question we are told that
point charge of 1C at a distance of 1 m
Generally the equation for the Electrostatic potential energy is mathematically given as
[tex]\rho=\frac{kq_1q_2}{r}[/tex]
Where k is a constant
[tex]k=9*10^9Nm^2/c^2[/tex]
Therefore
[tex]\rho=\frac{(9*10^9)1*10^(-6)*1*10^{-6}}{1}[/tex]
[tex]\rho=9x10 ^{10}J[/tex]
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Tammy leaves the office, drives 26 km due
north, then turns onto a second highway and
continues in a direction of 30.0° north of east
for 62 km. What is her total displacement from
the office?
please explain properly
Answer:
72.98 km
Explanation:
Her displacement is simply the distance from her final position to her initial position.
Now, I've drawn and attached a triangle diagram to depict this her movement.
Point O is her initial starting point.
Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.
From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.
Thus;
x² = 62² + 26² - 2(62 × 26)cos 120
x² = 4520 + 806
x² = 5326
x = √5326
x = 72.98 km
Two astronomy students travel to South Dakota. One stands on Earth’s surface and enjoys some sunshine. At the same time, the other descends into a gold mine where neutrinos are detected, arriving in time to detect the creation of a new radioactive argon nucleus. Although the photon at the surface and the neutrinos in the mine arrive at the same time, they have had very different histories. Describe the differences.
Answer:
The photon takes millions of years to reach the Surface of the sun while the Neutrinos travelling at the speed of light reaches the surface of the sun in approximately 2 seconds
The Photon is million year old while the neutrino is just some minutes old as observed by the student .
Explanation:
Although The Photon ( sunshine from the sun's surface ) heating up the student standing on the Earth's surface and the neutrinos discovered by the other student inside the gold mine are both formed in the Sun's core.
The difference between both are
The photon takes millions of years to reach the Surface of the sun while the Neutrinos travelling at the speed of light reaches the surface of the sun in approximately 2 seconds
The Photon is million year old while the neutrino is just some minutes old as observed by the student .
help asap
i) Two sides of a parallelogram are given as i + 2j - 3k and 5i+j-k. Find area of parallelogram and also find the area of triangle formed by the diagonal of the parallelogram.
ii) A vector makes an angle of 75 degrees with vertical. Find its rectangular components.
explain step by step
magnitude is area. hope it helps.
.
What is the equation for work?
Work = force / distance
Work = force - distance
Work = force + distance
Work = force * distance
Answer:
Work = F X D
Explanation:
Work (J) equals Force (N) multiplied by Distance (M)
Help
The Euler buckling load of a 160-cm-long column will be _____ times the Euler buckling load of an equivalent 120-cm-long column.
1.78
0.56
0.75
2.37
1.33
0.42
Answer:
Length in times = 0.75 times
Explanation:
Given:
Length of Euler buckling load = 160 cm
Equivalent load length = 120 cm
Find:
Length in times
Computation:
Length in times = Equivalent load length / Length of Euler buckling load
Length in times = 120 / 160
Length in times = 12 / 16
Length in times = 3 / 4
Length in times = 0.75 times
I am struggling with the question below, pls help: What is the distance between the moon and the earth if the mass of the moon is 7.34 x 10²² kg and the force of attraction between the two is 2.00 x 10 ^ 20.
Thank you in advance!
Answer:
Explanation:
What you forgot to include is the mass of the earth, which is 5.98 × 10²⁴ kg. NOW we can do the problem:
[tex]F=\frac{Gm_1m_2}{r^2}[/tex] where m1 and m2 are the masses of the objects experiencing this force of gravity, F. G is the universal gravitational constant. Filling in:
[tex]2.00*10^{20}=\frac{(6.67*10^{-11})(7.34*10^{22})(5.98*10^{24})}{r^2}[/tex]
We are going to rearrange and solve for r before we do any math on this thing:
[tex]r=\sqrt{\frac{(6.67*10^{-11})(7.34*10^{22})(5.98*10^{24})}{2.00*10^{20}} }[/tex] and when we plug all that mess into our calculators we will do it just like that and then round to 3 significant digits at the very end.
Doing all of that gives us that
r = 3.83 × 10⁸ m
A car is stationary. It accelerates at 0.8 ms^2 for 10 s and then at 0.4 ms^2 for a further 10 s. Use the equations of motion to deduce the car's final displacement. You will have to split the journey into two parts, since the acceleration changes after 10 s.
Answer:
The final displacement of the car is 140 meters.
Explanation:
The final displacement of the car ([tex]s[/tex]), in meters, is the sum of the change in displacement associated with each part of the journey, which is derived from the following kinematic formulas:
[tex]s = s_{1} + s_{2}[/tex] (1)
[tex]s_{1} = \frac{1}{2}\cdot a_{1}\cdot t_{1}^{2}[/tex] (2)
[tex]v_{o,2} = a_{1}\cdot t_{1}[/tex] (3)
[tex]s_{2} = v_{o,2} + \frac{1}{2}\cdot a_{2}\cdot t_{2}^{2}[/tex] (4)
Where:
[tex]s_{1}[/tex] - Traveled distance of the first part, in meters.
[tex]s_{2}[/tex] - Traveled distance of the second part, in meters.
[tex]a_{1}[/tex] - Acceleration in the first part, in meters per square second.
[tex]a_{2}[/tex] - Acceleration in the second part, in meters per square second.
[tex]v_{o,2}[/tex] - Initial speed of the car in the second part, in meters per second.
[tex]t_{1}[/tex] - Time taken in the first part, in seconds.
[tex]t_{2}[/tex] - Time taken in the second part, in seconds.
If we know that [tex]a_{1} = 0.8\,\frac{m}{s^{2}}[/tex], [tex]t_{1} = 10\,s[/tex], [tex]a_{2} = 0.4\,\frac{m}{s^{2}}[/tex] and [tex]t_{2} = 10\,s[/tex], then the distance traveled by the car is:
By (2):
[tex]s_{1} = \frac{1}{2}\cdot \left(0.8\,\frac{m}{s^{2}} \right)\cdot(10\,s)^{2}[/tex]
[tex]s_{1} = 40\,m[/tex]
By (3):
[tex]v_{o,2} = \left(0.8\,\frac{m}{s^{2}} \right)\cdot (10\,s)[/tex]
[tex]v_{o,2} = 8\,\frac{m}{s}[/tex]
By (4):
[tex]s_{2} = \left(8\,\frac{m}{s} \right)\cdot (10\,s) + \frac{1}{2}\cdot \left(0.4\,\frac{m}{s^{2}} \right)\cdot (10\,s)^{2}[/tex]
[tex]s_{2} = 100\,m[/tex]
By (1):
[tex]s = 140\,m[/tex]
The final displacement of the car is 140 meters.
what is the equation of preassure
Select the definition that best describes the term “strain.”
A measure of the degree of plastic deformation that a material can withstand before failure.
The elongation (or change in length) of a specimen divided by the original length, sometimes referred to as percent elongation.
Indication of how much energy a material can absorb before failure.
Also known as the modulus of elasticity, this is the constant of proportionality as defined by Hooke’s Law.
The instantaneous force applied to a specimen divided by the cross-sectional area.
Answer:
The elongation (or change in length) of a specimen divided by the original length, sometimes referred to as percent elongation.
Explanation:
Strain is defined mathematically as follows .
strain = Δ L/ L
Δ L is change in length of a wire when some force is applied on it to stretch it along its length and L is original length of a wire .
Stress is The elongation (or change in length) of a specimen divided by the original length, sometimes referred to as percent elongation.
Answer:
The elongation (or change in length) of a specimen divided by the original length, sometimes referred to as percent elongation.
Explanation:
Every object has some elastic and inelastic properties.
The strain is defined as the ratio of change in dimensions to the original dimensions.
There are three types of strains:
1. Longitudinal strain
2. Volumetric strain
3. Shear strain
So, it is defined as elongation (or change in length) of a specimen divided by the original length, sometimes referred to as percent elongation.
URGENT
A river flows toward 90°. Mark, a riverboat
pilot, heads the boat at 297º and is able to go
straight across the river at 6.0 m/s.
a. What is the velocity of the current?
b. What is the velocity of the boat as seen
from the river bank?
If the boat's speed is s, and the river's speed is r, and the boat is traveling east (0 degrees),
(0,r) + (s cos297,s sin297) = (6,0)
now just solve for r and s.
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