Answer:
We know that both objects fall at with the same acceleration, let's define that acceleration as g = 9.8m/s^2
Then the acceleration of both objects will be:
A(t) = -9.8m/s^2
Where the minus sign is because the gravity pulls the objects down.
Now, to get the velocity of the objects, we integrate over time, and get:
V(t) = (-9.8m/s^2)*t + v0
Where V0 is the initial velocity of the objects, because we do not have information about this, we can assume that both objects have the same value of V0, so for now, the movement equations are the same for both of them
To get the position we integrate again over time and get:
P(t) = (1/2)*(-9.8m/s^2)*t^2 + v0*t + P0
Where P0 is the initial position and same as above, we can assume that is the same for both objects.
Then both objects have the same motion equations, and the mass does not appear in any of them, then we can conclude that they will hit the ground at exactly the same time, because their motion equations are exactly the same.
B) By Newton's second law we know that:
F = m*a
Force equals mass times acceleration.
We know that both objects suffer the same acceleration, 9.8m/s^2
Then the force for the feather will be:
(remember that 1kg = 1000g, then 74g = (74/1000) kg = 0.074kg)
F = 0.074kg*9.8m/s^2 = 0.7252 N
And for the bowling ball will be:
F = 7.26kg*9.8m/s^2 = 71.148 N
Then you can see that the bowling ball experiences a greater force.
what are the 3 main states of matter in water?/
Answer:
ice (solid), water (liquid) and vapor (gas)
Explanation:
A 5-newton force directed east and a 5-newton force directed
north act concurrently on a point. Calculate the resultant of the two forces.
Answer:
[tex]5\sqrt{2} \ N[/tex]
Explanation:
It is given that,
A 5-newton force directed east and a 5-newton force directed north.
We need to find the resultant of the two forces.
As the two forces are acting in perpendicular to each other. The resultant of two such forces is given by :
[tex]F=\sqrt{5^2+5^2} \\\\=5\sqrt{2} \ N[/tex]
So, the resulatnt force is [tex]5\sqrt{2} \ N[/tex]
If you can answer my last post ill give you 75 points pls its very important and please make sure it correct!!!!!!!
Answer:
I can't see the post :/
Explanation:
Answer:
ok
Explanation:
ok how many points is this?????
Please help me with this question guys.
Answer:
The average speed is 22.2 km/h
Explanation:
Average Speed
Given an object travels a total distance d and took a total time t, then the average speed is:
[tex]\displaystyle \bar v=\frac{d}{t}[/tex]
The mailman first drives d1=7 km at v1=15 km/h. The time taken to drive is:
[tex]\displaystyle t1=\frac{d1}{v1}=\frac{7}{15}=0.467\ h[/tex]
Then he drives d2=7 km at v2=43 km/h taking a time of:
[tex]\displaystyle t2=\frac{d2}{v2}=\frac{7}{43}=0.163\ h[/tex]
The total time is
t=0.467 h + 0.163 h = 0.63 h
The total distance is
d = 7 km + 7 km = 14 km
The average speed is:
[tex]\displaystyle \bar v=\frac{14}{0.63}=22.2\ km/h[/tex]
The average speed is 22.2 km/h
Taking your hands off the wheel is an example of what type of driver distraction?
manual
visual
cognitive
adaptation
Answer:
manual
Explanation:
A manual distraction is when a driver takes their hand(s) off the wheel for any reason, for any amount of time. Without both hands on the wheel, your reaction time suffers, and so does your ability to steer.
How do you calculate the net force when there are multiple forces in different directions?
To find [tex]F_{net}[/tex] we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force [tex]F_{net} = \sqrt{(3)^2+(4)^2} \\[/tex] which gives [tex]F_{net} = 5[/tex]. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using [tex]\theta = \tan^{-1} (\frac{4}{3})[/tex] where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.
Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?
This is my question
Answer:
baby yoda
Explanation:
1.) There was an earthquake in Salt Lake City, Utah, on March 18, 2020, in the morning at 9 hours, 9 minutes, and 45 seconds Mountain Standard Time (9:9:45 MST). If the velocity of the p-wave is 7.3 km/sec and the velocity of the s-wave is 5.1 km/sec and the s-p time lag is 16 seconds, what is the distance in kilometers from Salt Lake City to the focus of the earthquake? Explain how you calculated the answer.
Answer:
7 because salt lake and Southis weat
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]
So, the mass of the cart is 150 kg.
A piston has an external pressure of 15.0 atmatm. How much work has been done in joules if the cylinder goes from a volume of 0.120 liters to 0.450 liters
Answer: Work done is - 501.56 J
Explanation:
Given that;
External pressure P = 15.0 atm
Volume V1 = 0.120 liters
Volume V2 = 0.450 liters
Work done = ?
we know that; Work = -Pdv
where P is pressure and dv is change in volume
so we substitute our values into the equation
Work = -15.0 × ( 0.450 - 0.120)
= -15 × 0.33
= - 4.95 atm/L
we know that;
1 atm.L = 101.325 J
so
- 4.95 atm/L = 101.325 J × -4.95 atm/L ÷
= - 501.56 J
Therefore Work done is - 501.56 J
A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?
Answer:
a = 0.62 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.
[tex]v_{f} =v_{o} -a*t[/tex]
[tex]55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ][/tex]
where:
Vf = final velocity = 2,77 [m/s]
Vo = initial velocity = 15.27 [m/s]
t = time = 20 [s]
a = acceleration [m/s²]
Now replacing:
[tex]2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}][/tex]
Answer:
It would be 10.00
Explanation:
Hope this helps its different for everyone what was it for u it was D for me
we feel cold in winter when we come out from the quilt but the same room becomes warmer after coming back from outside the room
Answer:
Yes
Explanation:
I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.
HOPE THIS HELPED
A 600 kg rocket sled can be accelerated at a constant rate from rest to 1400 km/h in 2.1 s. What is the magnitude of the required net force
Answer:
7.2 × 10^5 N
Explanation:
The first step is to convert 1400 km/hr to m/s
= 1,400×1000/3600
= 1,400,000/3600
= 388.88 m/s
The acceleration can be calculated as follows
a= v-u/t
= 388.88/2.1
= 185.18
Therefore the required net force can be calculated as follow
= 388.88 × 185.18
= 7.2 × 10^5 N
A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Answer:
Explanation:
To get the focal length, we will use the lens formula;
1/f = 1/u + 1/v
f is the focal length
u is the object distance
v is the image distance
Given
since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.
u = 35cm
v = -2.0
1/f = 1/35-1/2
1/f = 2-35/70
1/f = -33/70
f = -70/33
f = -2.12 cm
f = -0.0212m
Power of a lend is the reciprocal of its focal length
Power of the lens = 1/f
P = 1/-0.0212
P = -47.17dioptres
The power of the lens is -47.17D
Ancient peoples both feared and revered Solar and Lunar Eclipses as they happened so rarely. Explain the following about eclipses:
1. Why are SOLAR eclipses more rare then lunar eclipses.
2. Explain what must occur for a SOLAR eclipse to occur.
3. Explain what must occur for a LUNAR eclipse to occur.
Answer:
Lunar and solar eclipses occur with about equal frequency. Lunar eclipses are more widely visible because Earth casts a much larger shadow on the Moon during a lunar eclipse than the Moon casts on Earth during a solar eclipse. As a result, you are more likely to see a lunar eclipse than a solar eclipse.
For a total eclipse to take place, the sun, moon and Earth must be in a direct line. The second type of solar eclipse is a partial solar eclipse. This happens when the sun, moon and Earth are not exactly lined up. The sun appears to have a dark shadow on only a small part of its surface.
For a lunar eclipse to occur, the Sun, Earth, and Moon must be roughly aligned in a line
Running with an initial velocity of 13 m/s , a horse has an average acceleration of -1.88 m/s2 . How long does it take for the horse to decrease its velocity to 6.3 m/s
pls :/
Answer:
3.56s
Explanation:
Ya that's the right answer.
A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water?
magnitude _________ N
The force required to hold it completely submerged under water is 0.252 N
As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.
Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation
F = Buoyant force - weight of sphere
Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m
Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³
Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg
Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N
Volume of water displaced = 4/3 π r³ = 2.805 e-5
Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N
F = 0.275 - 0.023 = 0.252 N
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The force required to hold it completely submerged under water is 0.25 N
The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³
The density of water [tex]\rho_w[/tex] = 1000 kg/m³
Diameter = 3.77 cm = 0.0377 m
radius of ball = 0.0377/2 = 0.01885 m
The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]
Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:
The force is required to hold it completely submerged under water (F) is:
[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]
F = 0.25 N
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an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying after traveling a distance of 1 kilometer?
Answer:
331.7m/s
Explanation:
Given parameters:
Initial velocity = 100m/s
Acceleration = 50m/s²
Distance = 1km = 1000m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to apply the right motion equation shown below;
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Now insert the parameters and solve;
v² = 100² + (2 x 50 x 1000)
v² = 110000
v = √110000 = 331.7m/s
What property of a wave remains unchanged when a wave enters a different medium? 5 points amplitude frequency wavelength velocity
Answer:
frequency
Explanation:
When a wave enters a different medium, the property that remains unchanged is the frequency of the wave.
The frequency of a wave is defined as the number of wave cycles passing a given point per unit time. It represents the rate at which the wave oscillates or repeats its pattern.
The frequency of a wave remains constant when it enters a different medium, regardless of any changes in other properties such as amplitude, wavelength, or velocity.
Therefore, When a wave enters a different medium, the property that remains unchanged is the frequency of the wave.
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Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just slides down the alley, while the other rolls down the alley. Which ball has more energy?
a. Sliding ball
b. Rolling ball
c. They both have the same energy
Answer:
Rolling Ball
Explanation
Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just slides down the alley, while the other rolls down the alley, rolling ball has more energy.
what is velocity ?Velocity can be referred to the rate of change of the position of the object with respect to time which is basically speeding the object in a specific direction.
Velocity is vector quantity, as it is present both magnitude and direction and The SI unit is meter per second (ms-1). The change in magnitude or the direction of velocity of a body, the object will accelerate.
the final velocity of the object is simple but few calculations and basic conceptual knowledge are needed.
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Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 Newtons on earth, which has more mass?
a. Object a
b. Object b
c. Same mass
d. Other
Answer:
a. Object A
Explanation:
The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.
The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.
So that,
For object A on the moon, mass = 25 lbs
For object B on the earth, W = 25 N,
W = m x g
25 = m x 10 (g = 10 m/[tex]s^{2}[/tex])
m = [tex]\frac{25}{10}[/tex]
= 2.5 lbs
Mass of object B is 2.5 lbs.
Therefore, the mass of the object A is more than that of B.
An amateur golfer swings a golf club, striking a golf ball that has a mass of 55.0 g. The club is in contact with the ball for only 0.00360 s. After the collision, the ball leaves the club at a speed of 34.0 m/s. What is the magnitude of the average force (in N) exerted on the ball by the club?
Answer:
The average force ≅ 519.44 N.
Explanation:
Impulse = change in momentum of a body
i.e Ft = m(v - u)
where F is the force, t is the time, m is the mass of the body, v is the final velocity and u is the initial velocity.
m = 55.0 g (0.055 Kg), t = 0.00360 s, v = 34.0 m/s, since the ball was initially at rest; u = 0 m/s
So that,
F x 0.00360 = 0.055(34 - 0)
F x 0.00360 = 0.055 x 34
= 1.87
F = [tex]\frac{1.87}{0.0036}[/tex]
= 519.4444
The average force exerted on the ball by the club is approximately 519.44 N.
The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point located 0.40 L from the end
Answer:
The value is [tex]I = 0.0932 ML ^2[/tex]
Explanation:
From the question we are told that
The rotational inertia about one end is [tex]I_R = \frac{1}{3} ML^2[/tex]
The location of the axis of rotation considered is [tex]d = 0.4 L[/tex]
Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is [tex]0.4 M[/tex]
Generally the length of the rod from the its beginning to the axis of rotation consider is
[tex]k = 1 - 0.4 L = 0.6L[/tex]
Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is
[tex]m = 1- 0.4 M = 0.6 M[/tex]
Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is
[tex]I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
[tex]I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2[/tex]
Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is
[tex]I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is
[tex]I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
=> [tex]I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ][/tex]
=> [tex]I = 0.0932 ML ^2[/tex]
How many seconds did it take (after starting his descent) for the worker to hit the ground? Answer in units of s.
Answer:
This question is incomplete
Explanation:
The question is incomplete. However, to determine the time (in seconds) it took a worker to hit the ground from an elevated point. The speed the worker was coming with to the ground and the distance between the elevated point and the ground will have to be considered. Thus the formula to be used here will be
Speed (in meter per second) = distance (in meters) ÷ time (in seconds)
time (in seconds) = distance (in meters) ÷ speed (in meter per seconds)
The above photo shows Plates A and B at an mid ocean ridge. If you were to draw arrows on the plates showing plate movement, which way would you draw arrows?
A . The arrows on plates A and B would show the plates moving in the same direction
B . The arrows on plates A and B would show the plates moving towards eachother
C . The arrows on plates A and B would show the plates moving in random directions
D . The arrows on plates A and B would show the plated moving away from each other
Answer:
D . The arrows on plates A and B would show the plated moving away from each other
Explanation:
The correct arrow in this instance will show that the plates A and B would be moving away from each other.
At the mid-ocean ridge, two plates are moving apart and pulling away from one another.
In short, at the mid-oceanic ridge, we have a divergent margin. At a divergent margin, the plates are spreading or moving apart.What famous equation describes the energy of stars?
Answer:
E=mc2
Explanation:
It explains that in their interiors, atoms fuse together, creating energy.
To solve this we must be knowing each and every concept related to energy and its different types. Therefore, the famous equation that describes the energy of stars is E=mc².
What is energy?In physics, energy is the capacity to accomplish work. It can be potential, kinetic, thermal, electric, chemical, radioactive, or in other forms.
There is also heat and work—energy inside the process of being transferred by one body to the other. Energy is always classified according to its type once it has been transmitted. As a result, heat transported could become thermal energy, whereas labor done may emerge as mechanical energy. The famous equation that describes the energy of stars is E=mc²
Therefore, the famous equation that describes the energy of stars is E=mc².
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Terell lives in Washington, D.C., where he can see the Potomac River from his bedroom window. When he sees this river, Terell thinks of swimming, boating, fun, and friends. What type of mental image is this?
A. analogical
B. symbolic
C. both analogical and symbolic
D. representative
Answer:
B. symbolic
Explanation:
The mental image formed in Terell's mind is a symbolic mental image.
A symbolic mental image helps to bring perception to imagery formed on the mind.
This process is symbolic because Terell can see that river ahead. If he cannot see the river then it is more of an analogical fantasy. The thoughts are predicated on the image of the River seen. Therefore, it is right to conclude that this is a symbolic mental image.Answer:
Terell lives in Washington, D.C., where he can see the Potomac River from his bedroom window. When he sees this river, Terell thinks of swimming, boating, fun, and friends. What type of mental image is this?
symbolic
Explanation:
which statement is true of the particles that make up a substance? a. particles in a liquid have more energy than particles in a gas. b. particles in a liquid and particles in a solid have the same amount of energy. c. particles in a gas have more energy than particles in liquid. d. particles in a solid have moe energy than particles in a gas.
Explanation:
so sorry
don't know but please mark me as brainliest please
A rock climber is scaling rock wall at 10 m above the ground, and is moving downward at a speed of 0.25 m/s. If their phone falls out of their back pocket, how long before it shatters onto the ground?
Answer:
The time taken before the phone hits the ground is 1.4 s.
Explanation:
Given;
height of the rock wall, h = 10 m
speed of the climber, v = 0.25 m/s
The time taken before the phone hits the ground is given;
h = ut + ¹/₂gt²
10 = 0.25t + ¹/₂(9.8)t²
20 = 0.5t + 9.8t²
9.8t² + 0.5t - 20 = 0
solving this quadratic equation;
t = 1.4 s
Therefore, the time taken before the phone hits the ground is 1.4 s.
