Explanation:
You would think it should. But remember the Force is also determined by mass. The mass of the earth markes our mass like the smallest part of a mosquito leg. The earth will go on it its merry way without cosidering us at all.
Diwn unscramble the word
Answer:
WIND Is what you're looking for
Explanation:
The word is WIND
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
Imagine that you are standing on a spherical asteroid deep in space far from other objects. You pick up a small rock and throw it straight up from the surface of the asteroid. The asteroid has a radius of 9 m and the rock you threw has a mass of 0.113 kg. You notice that if you throw the rock with a velocity less than 45.7 m/s it eventually comes crashing back into the asteroid.
Required:
Calculate the mass of the asteroid.
Answer:
M = 1.409 10¹⁴ kg
Explanation:
In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.
Starting point. When you drop the stone
Em₀ = K + U
Em₀ = ½ m v² - G m M / r
where M and r are the mass and radius of the asteroid
Final point. When the stone is too far from the asteroid
Em_f = U = - G m M / R_f
as there is no friction, the energy is conserved
Em₀ = Em_f
½ m v² - G m M / r = - G m M / R_f
½ v² = G M (1 / r - 1 /R_f)
indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞
½ v² = G M (1 /r)
M = [tex]\frac{v^2 r}{2G}[/tex]
let's calculate
M = [tex]\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}[/tex]
M = 1.409 10¹⁴ kg
Baseball runner with a mass of 70kg, moving at 2.7m/s and collides head-on into a shortstop with a mass of 85kg and a velocity of 1.6m/s. What will be the resultant velocity of the system when they make contact with each other
Answer:
The speed of the combined mass after the collision is 2.1 m/s.
Explanation:
mass of runner, m = 70 kg
speed of runner, u = 2.7 m/s
mass of shortstop, m' = 85 kg
speed of shortstop, u' = 1.6 m/s
Let the velocity of combined system is v.
Use conservation of momentum
Momentum before collision = momentum after collision
m u + m' u' = (m + m') v
70 x 2.7 + 85 x 1.6 = (70 + 85) v
189 + 136 = 155 v
v = 2.1 m/s
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
An object produces a sound wave with a wavelength 75.0cm. if the speed of sound is 350.0m/s, the frequency of the sound is
Answer:
wavelength=75.0
speed of sound(v)=350 .0m/s
frequency(f)=?
we know,
v=f*wavelengh
350.0 =f*750
f. =350/75
=4.667
pls mark me brainlest
If the temperature stays constant, which change would decrease the amount
of thermal energy in an object?
A. Decreasing its density
B. Increasing its velocity
c. Decreasing its mass
D. Increasing its mass
Its Acceleration during the upward Journey ?
Speed of light in water
Answer:
225,000 kilometers per second
Explanation:
Have a nice day
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
Three forces are pulling on the same object such that the system is in equilibrium. Their magnitudes are F1 = 2.83 N.F= 3.35 N. and F3 = 3.64 N, and they make angles of 0, = 45.0°, 02 = -63.43 and 03 =164.05° with respect to the x-axis, respectively.
Required:
a. What is the x-component of the force vector F1?
b. What is the y-component of the force vector F1?
(a) 2.001N
(b) 2.001N
Explanation:A sketch of the scenario has been attached to this response.
Since only the force vector F₁ is required, the only force shown in the sketch is F₁.
As shown in the sketch;
The x-component of the force vector F₁ = [tex]F_{x}[/tex]
The y-component of the force vector F₁ = [tex]F_{y}[/tex]
The magnitude of F₁ as given in the question = 2.83N
The angle that the force makes with respect to the x-axis = 45.0°
Using the trigonometric ratio, we see that;
(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]
=> [tex]F_{x}[/tex] = F₁ cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 x 0.7071
=> [tex]F_{x}[/tex] = 2.001N
(b) Also;
sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]
=> [tex]F_{y}[/tex] = F₁ sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 x 0.7071
=> [tex]F_{y}[/tex] = 2.001N
Therefore, the x-component and y-component of the force vector F₁ is 2.001N
The x and y component of vector F1 is mathematically given as
F_x = 2.001N
F_y= 2.001N
What is the x and y component of vector F1?Question Parameters:
Generally, the equation for the x-component is mathematically given as
x=Fsin\theta
Therefore
F_x = F₁ cos 45.0°
F_x = 2.83 x 0.7071
F_x = 2.001N
For y component
x=Fcos\theta
F_y = F₁ sin 45.0
F_y = 2.83 x 0.7071
F_y= 2.001N
Read more about Cartesian
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What is the y component of a vector that is 673 m at -38o?
Answer:
D_y = 414.38m
Explanation:
D_y = D*sin(x)
D_y = 673m*sin(38°)
D_y = 414.38m
A team of people who traveled to the North Pole by dogsled lived on butter because they needed to consume 6 000 dietitian's Calories each day. Because the ice there is lumpy and irregular, they had to help the dogs by pushing and lifting the load. Assume they had a 16-hour working day and that each person could lift a 500-N load. How many times would a person have to lift this weight 1.00 m upwards in a constant gravitational field, where (g = 9.80m/s2) where to do the work equivalent to 6 000 Calories?
Answer:
The right solution is "50200 days".
Explanation:
Given:
Calories intake,
= 6000 kcal,
or,
= [tex]2.52\times 10^7 \ J[/tex]
Force,
= 500 N
As we know,
⇒ [tex]Work \ done = Force\times distance[/tex]
Or,
⇒ [tex]distance = \frac{Work \ done}{Force}[/tex]
By putting the values, we get
[tex]=\frac{2.52\times 10^7}{500}[/tex]
[tex]=0.502\times 10^5[/tex]
[tex]=50200 \ m[/tex]
hence,
The number of days will be:
= [tex]\frac{50200}{1}[/tex]
= [tex]50200 \ days[/tex]
To accurately describe the wind, the measurement should include
A) a direction, but not a speed
B)a speed, but not a direction
C) both a speed and a direction
D) neither a speed nor a direction
Answer:
C. both a speed and a direction
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 7.56 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.22 rev/s.
Required:
a. Which rate of rotation gives the greater speed for the ball?
b. What is the centripetal acceleration of the ball at 8.16 rev/s?
c. What is the centripetal acceleration at 6.35 rev/s?
Answer:
a) [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b) [tex]a=30.7[/tex]
c) [tex]a=35.91[/tex]
Explanation:
From the question we are told that:
Initial angular velocity [tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
Initial Length [tex]L_1=0.600m[/tex]
Final angular velocity [tex]\omega _2=6.22rev/s=39rad/s[/tex]
Final Length [tex]L_2=0.900m[/tex]
a)
Generally the rotation with the greater speed is
[tex]\omega _1=7.56rev/s=>47.5rads/s[/tex]
b)
Generally the equation for centripetal acceleration at 8.16 is mathematically given by
[tex]a=\omega_1^2*L_1[/tex]
[tex]a=8.16 rev/s*0.6[/tex]
[tex]a=30.7[/tex]
c)
At 6.35 rev/s
[tex]a=6.35 rev/s*0.9[/tex]
[tex]a=35.91[/tex]
If R1 and R2 are in parallel and R3 is in series with them then equivalent resistance will be
Answer:
Refer to the attachment!~
An object is positively charged if it has more what
Answer:
An Object is positively charged if it has more Positive Electrons in that object
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
A student on a new planet wants to determine the value of gravity on that planet. Luckily for them they brought equipment that they can use to set up an oscillating spring or an oscillating pendulum. Which procedure would allow the student to determine the value of gravity on the new planet
Answer:
By measure the effective length and the time period of the pendulum.
Explanation:
Let the student take the oscillating pendulum at the planet.
He measure the time period of the pendulum by using the stop watch or the ordinary watch.
Then measure the effective length of the pendulum which is the distance between the center of gravity of the bob and the point of suspension of the pendulum.
Now, use the formula of the time period of the pendulum,
[tex]T =2\pi\sqrt\frac{L}{g}[/tex]
Here, L is the effective length of the pendulum, g is the acceleration due to gravity at the planet and T is time period of the pendulum.
By rearranging the terms, we get
[tex]T =2\pi\sqrt\frac{L}{g}\\\\T^{2}=4\pi^2\times\frac{L}{g}\\\\g =\frac{4\pi^2L}{T^2}[/tex]
Here, by substituting the values of L and T, the student get the value of acceleration due to gravity at that planet.
Convert the following:
1) 367.5 mg = _______ g
2) 367 mL = _______ L
3) 28.59 in =______ cm
4) 8 0z =_______lb
5) 0.671 mm =_____m
Answer:
1) 0.3675
2) 0.367
3) 72.6186
4) 0.5
5) 0.000671
Answer:
1) 367.5 mg = 0.3675 g
2) 367 mL = 0.367 L
3) 28.59 in = 72.61 cm
4) 8 0z = 0.5 lb
5) 0.671 mm = 0.0000671 m
3.1Chất điểm chuyển động thẳng với phương trình: x = – 1 + 3t2
– 2t
3
(hệ SI, với t ≥ 0). Chất điểm dừng lại để
đổi chiều chuyển động tại vị trí có tọa độ:
Answer:
eqcubuohwehuuc
Explanation:
the the the the the the the the the the the the the the the the dhueirrjhrhjrirjrheh3jeiiruj fnrhfjjjrjrj fjfiirrjejrjejej jrkrjrjrjrjrjdjdjfjrhruriruru rjridjhwjjsjd
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
Two small silver spheres, each of mass m=6.2 g, are separated by distance d=1.2 m. As a result of transfer of some fraction of electrons from one sphere to the other, there is an attractive force F=900 KN between the spheres. Calculate the fraction of electrons transferred from one of the spheres: __________
To evaluate the total number of electrons in a silver sphere, you will need to invoke Avogadro's number, the molar mass of silver equal to 107.87 g/mol and the fact that silver has 47 electrons per atom.
Answer:
4.60 × 10⁻⁸
Explanation:
From the given information;
Assuming that q charges are transferred, then:
[tex]F = \dfrac{kq^2}{d^2}[/tex]
where;
k = 9 ×10⁹
[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]
[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]
q = 0.012 C
No of the electrons transferred is:
[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]
[tex]= 7.5 \times 10^{16} \ C[/tex]
Initial number of electrons = N × 47 × no of moles
here;
[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]
no of moles = 0.0575 mol
∴
Initial number of electrons = [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]
= 1.63 × 10²⁴
The fraction of electrons transferred [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]
= 4.60 × 10⁻⁸
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
Which of the following is acceleration toward the center of a circular motion? O A. Centripetal acceleration O B. Uniform circular motion O C. Centrifugal force D. Centripetal force
PLEASE HELP ASAP!!
We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
To solve more questions on Work, Energy and Power, visit the link below -
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Choose the CORRECT statements. The superposition of two waves.
I. refers to the effects of waves at great distances.
Il. refers to how displacements of the two waves add together.
Ill. results into constructive interference and destructive interference
IV. results into minimum amplitude when crest meets trough.
V. results into destructive interference and the waves stop propagating.
A. I and II
B. II and III
C. I, II and III
D. II, III and IV
E. III, IV and V
F. II, III, IV and V
Answer:
A
Explanation:
I guess not that much confidential!
Which one of the following pairs of units may not be added together, even after the appropriate unit conversions have been made?
a. grams and milligrams
b. miles and kilometers
c. kilometers and kilograms
d. centimeters and yards
Answer:
c. kilometers and kilograms
Explanation:
If two units represent different things, like for example grams and joules, where the first one is a unit for mass, and the second one a unit for energy we can not add them together. (
Let's analyze each one of the given options:
a: grams and milligrams
both grams and milligrams are units of mass, so after the appropriate unit conversions, we can add them.
b: miles and kilometers
Both are units for distance, after the appropriate unit conversions, we can add them.
c: kilometers and kilograms
"Kilometer" is a unit for distance, and "kilogram" a unit for mass, we can not add distance and mass, so these two can not be added together.
d: centimeters and yards
Both are units of distance, after the appropriate unit conversions, we can add them.
Then the only one that we can not add together is option c: kilometers and kilograms
Answer: The correct option is C (kilometers and kilograms).
Explanation:
UNITS is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature. The magnitude of a physical quantity is expressed as, Physical quantity = (numerical value) × (unit). There are six basic units of measurement which includes:
--> metre (m) - unit of length.
--> kilograms (kg) - unit of mass.
--> second (s) - unit of time.
--> ampere (A) - unit of electrical current.
--> kelvin (K) - unit of temperature and
--> mole (mol) - unit of the amount of substance.
Mass is the quantity of matter present in a body which is constant and is measured in KILOGRAMMES or GRAMS. While position is the distance and direction of an object from a particular reference point which is known to vary and is measured in METERS OR KILOMETRES.
From the above definition, KILOGRAMMES is used to measure the magnitude of a different quantity from KILOMETERS and therefore should not be added up after conversation.
Kirchhoff's junction rule is a statement of: Group of answer choices the law of conservation of momentum the law of conservation of charge the law of conservation of energy. the law of conservation of angular momentum. Newton's second law
Answer:
the law of conservation of energy.
Explanation:
An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.
Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.
In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.
The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.
Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.
One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.
Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.
This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.
The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.
This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.
Question: A car of mass 500kg travelling at 12m/s enters a stretch of road where there's a constant resistive force of 8000N. The car comes to a stop due to this resistive force. Calculate the distance travelled by the car before stopping.
Answer:
ans: 2.25 meter
explanation
use following equations
F = ma
V = U + aT
S = UT + 1/2 aT^2
