Answer:
CG gh sure er go b vh pxuh FPI OO c AM h kh
What the velocity from the graph given above?
Answer:
i think its 4 or 35
Explanation:
its in the middle of 40 and 30
a sky driver jumps from an aircraft. the mass of sky driver sky driver is 70kg. state the equation linking weight,mass,and gravitional field strenght.
Answer:
Weight = Mass × Gravitational field strength
Explanation:
The mass of the skydiver = 70 kg
The weight of the skydiver, W, is given as follows;
W = m × g
Where-
g = The gravitational field strength ≈ 9.81 m/s²
Therefore, the equation linking weight mass and gravitational field strength is presented as follows;
Weight = Mass × Gravitational field strength
what will be the effect on the acceleration due to gravity of the earth if it is compressed to a the size the moon?
Answer:
it gravitional pull on earth will increased becauste it is compress to a form of moon which is comperatively smaller so the gravitonal pull on per cm of earth will incrased so we can say that there will be change in acceleration due to gravity
You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to the right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.
Please help me with this
Answer:
Explanation:
From your other post, the complete question is:
"You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to the right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.
Find the sum of the force on the sled.
Determine the acceleration of the sled .
If the sled has an initial velocity 2m/s to the right, how fast will it be traveling after 5 seconds?
"
Given the rope exerting a force of 50 N to the right and the snow exerts a friction force of 30 N to the left, the sum of forces
= 50 - 30
= 20N to the right
The mass of sled is 50 kg and force = mass * acceleration.
Acceleration = Force / mass
= 20 / 50 = 0.4 m/s^2 to the right
If the sled has an initial velocity 2m/s to the right, after 5 seconds it will be traveling at initial velocity + acceleration * time
= 2 + 0.4*5
= 4 m/s to the right
Answer:
Explanation:
net force = applied force - friction force
= 50-30
= 20N to right
acceleration = net force/mass
= 20/50
= 2/5m/s tp right
final velocity = initial velocity + acceleration*time
= 2+2/5*5
= 4m/s to right
If the final velocity is 0. third equation of motion will be
Answer:
vf²=vi²+2a∆x
Explanation:
The third equation of motion gives the final velocity of an object under uniform acceleration given the distance traveled and an initial velocity: v 2 = v 0 2 + 2 a d . v^2=v_0^2+2ad. v2=v02+2ad. The graph of the motion of the object.
A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an inclined plane. The surface of the incline is 2.75 m long, and is tilted at an angle of 22.0∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed v2 at the bottom of the ramp.
Answer:
The final translational seed at the bottom of the ramp is approximately 4.84 m/s
Explanation:
The given parameters are;
The radius of the sphere, R = 2.50 cm
The mass of the sphere, m = 4.75 kg
The translational speed at the top of the inclined plane, v = 3.00 m/s
The length of the inclined plane, l = 2.75 m
The angle at which the plane is tilted, θ = 22.0°
We have;
[tex]K_i[/tex] + [tex]U_i[/tex] = [tex]K_f[/tex] + [tex]U_f[/tex]
K = (1/2)×m×v²×(1 + I/(m·r²))
I = (2/5)·m·r²
K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²
U = m·g·h
h = l×sin(θ)
h = 2.75×sin(22.0°)
∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75×[tex]v_f[/tex]² + 0
7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93
∴ 77.93 ≈ 7/10 × 4.75×[tex]v_f[/tex]²
[tex]v_f[/tex]² = 77.93/(7/10 × 4.75)
[tex]v_f[/tex] ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84
The final translational seed at the bottom of the ramp, [tex]v_f[/tex] ≈ 4.84 m/s.
A. Tick (1) the best alternatives. 1. What is the acceleration due to gravity on the surface of moon ? (a) 9.8m/s (b)1.6m/s2 (c) 6.67x10-1m/s (d) 9.8m/s?
Answer:
[tex] \green{ \sf \: \: 1.6 \: m {s}^{ - 2} \: \: \: is \: the \: correct \: answer}[/tex]
Explanation:
[tex] \sf \: \huge{g } _{ \small{moon}} = \frac{ {\huge{g}}_{earth}}{6} \\ \\ \sf \implies \: \sf \: \huge{g } _{ \small{moon}} = \frac{ 9.8}{6} = 1.6 \: \: m {s}^{ - 2} [/tex]
A horse is running on a circular path with constant speed but its direction is changing at every point. It is
Answer:
accelerating.
Explanation:
A horse is running on a circular path with constant speed but its direction is changing at every point. Thus, it is accelerating.
A force of 10 N is making an angle of 30° with the horizontal. Its horizontal component will be:
A. 4N
B. 5N
C. 7N
D. 8.7 N
Answer:The answer is A
Explanation:just did it on a test
a stone attached to 1m long string is moving with the speed of 5ms in a circle find the centripetal acceleration of the stone
Answer:
The centripetal acceleration of the stone is 5 m/s²
Explanation:
The length of the string to which the stone is attached, r = 1 m
The speed with which the string is rotated, v = 5 m/s
The centripetal acceleration, [tex]a_c[/tex], is given as follows;
[tex]a_c = \dfrac{v^2}{r}[/tex]
Therefore, the centripetal acceleration of the stone found as follows;
[tex]a_c = \dfrac{(5 \ m/s)^2}{1 \ m} = 5 \ m/s^2[/tex]
The centripetal acceleration of the stone, [tex]a_c[/tex] = 5 m/s².
Please help me with these I might need more than only 1 person to answer
Explanation:
a) copper
b) olive oil
Hope it helps✌✌
Electricity flows from
positive to positive
negative to positive
negative to negative
positive to negative
Answer:
is it from positive to negative am I right tell me if I am wrong
Answer:
positive to negative
The flow of electric current is similar to the flow of water which is from higher level to lower level.
The electric current flow from higher potential region(positive* low concentration of electron) to lower potential region(negative*high concentration of electron)
7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ball h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.
[tex] \\ \tt \longmapsto \: h = - 4.8 {t}^{2} + 8.2t + 1.8 \\ \\ \tt \longmapsto \: h = - 4.8(1.7) {}^{2} + 8.2 \times 1.7 + 1.8 \\ \\ \tt \longmapsto \: - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\ \tt \longmapsto \: 17.06[/tex]
Which of the following can be correct units for acceleration?
A. miles/hr/m
B. Km/s/hr
C. m/s/m
D. km/m/s
Answer:
B. Km/s/hr
Explanation:
A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radius of 40.0 m. The coefficient of static friction between the car tires and the road is 0.500. The car travels successfully around the turn at a constant speed of 10.0 m/s. Calculate the magnitude of the car's acceleration as it goes around the turn. _______ m/s^2
Answer:
2.5 m/s²
Explanation:
The given parameters are;
The mass of the car, m = 2,000 kg
The radius of the car, r = 40.0 m
The coefficient of friction between the car tires and the road, μ = 0.500
The constant speed with which the car moves, v = 10.0 m/s
The normal reaction of the road on the car, N = The weight of the car;
∴ N = m × g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
N ≈ 2,000 kg × 9.81 m/s² = 19,620 N
The frictional force, [tex]F_f[/tex] = μ × N
The centripetal force, [tex]F_c[/tex] = m·v²/r
The car moves without slipping when [tex]F_f[/tex] = [tex]F_c[/tex]
Therefore, [tex]F_f[/tex] = 0.500 × 19,620 N = 2,000 kg × [tex]v_{max}[/tex]²/40.0 m
∴ [tex]v_{max}[/tex] = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s
Therefore, the velocity with which the car moves, v < [tex]v_{max}[/tex]
The cars centripetal acceleration, [tex]a_c[/tex] = v²/r
∴ [tex]a_c[/tex] = (10.0 m/s)²/40.0 m = 2.5 m/s²
The cars centripetal acceleration as it goes round the turn, [tex]a_c[/tex] = 2.5 m/s².
(This is for other people with this Question i hope you find this when you need help) Need answer for this Help!!!
Question 7 of 20 You plan to use a slingshot to launch a ball that has a mass of 0.025 kg. You want the ball to accelerate straight toward your target at 19 m/s2. How much force do you need to apply to the ball? O A. 19.03 N OB. 0.48 N O C. 4.51 N D. 760.00 N
Answer:
0.48N
Explanation:
according to the second law of motion
force=mass×acceleration
the mass in this question is 0.025,the acceleration 19
therefore f=0.025×19
=0.48N
I hope this helps
In isoelectric focusing gel electrophoresis a. there is a pH gradient that parallels the electric field gradient b. particular care must be taken to ensure the same pH along the length of the gel c. the electric current is allowed to fluctuate d. the electric circuits of the apparatus must be very well insulated
Answer:
a. there is a pH gradient that parallels the electric field gradient.
Explanation:
Gel electrophoresis is a technique used to separate DNA fragments according to their size. DNA samples are loaded in the wells and then electric current is applied to pull the DNA fragments out from the gel. Isoelectric gel electrophoresis is a process in which negatively charged DNA fragments are separated from the gel.
20.In case the conductor is a heating appliance, then this energy(w) is converted into heat(H) i.e
A. w=H C. w=I2Rt
B. w=VIt D. all of above
Answer:
not sure just need points
Explanation:
a+b+c
Hai điện tích đặt cách nhau một khoảng R trong không khí thì lực tương tác
giữa chúng là 2.10−3N. Nếu khoảng cách đó mà đặt trong môi trường điện môi thì
lực tương tác giữa chúng là 10−3N. Để lực tương tác giữa hai điện tích đó khi đặt
trong môi trường điện môi bằng lực tương tác giữa hai điện tích đó khi đặt trong
không khí thì khoảng cách giữa 2 điện tích là bao nhiêu?
An electromagnetic wave has a frequency of 6.0 x 10^18 Hz. What is the
wavelength of the wave? Use the equation 2 = and the speed of light as 3.0
x 108 m/s.
Answer:
Wavelength = 5 * 10^{-11} meters
Explanation:
Given the following data;
Frequency = 6.0 x 10^18 Hz
Speed = 3 * 10⁸ m/s
To find the wavelength of the wave;
Mathematically, the wavelength of a wave is given by the formula;
[tex] Wavelength = \frac {speed}{frequency} [/tex]
Substituting into the formula, we have;
[tex] Wavelength = \frac {3 * 10^{8}}{6.0 x 10^{18}} [/tex]
Wavelength = 5 * 10^{-11} meters
why do black holes have a large gravitational pull that even light cannot escape from
Answer:
Because matter has been squeezed into a tiny space.
Explanation:
According to NASA, this can happen when a star is dying.
A black hole has no more gravity than the same amount of matter in any other form.
But remember that the gravitational forces are stronger as you get closer to the center of the body. The mass of a black hole is packed into such a small size (theoretically zero !) that you can get very close to its center. THAT'S where its gravity is hugely strong.
Algunos estudiantes de tu promoción consideran que se debería aumentar el número de horas semanales de Educación Física. Otros consideran que el número de horas debería mantenerse. Escribe un artículo de opinión en el que presentes tu punto de vista sobre este tema y al menos tres razones que justifiquen tu postura. Tu artículo se publicará en el periódico escolar La Voz.
una ayuda plis :c
La respuesta correcta para esta pregunta abierta es la siguiente.
Te compartimos las tres ideas que te pueden ayudar para hacer tu artículo.
Título:
La Educación Física, una prioridad para las escuelas.
Existe una frase milenaria que se le atribuye a los griegos que dice "Mente sana en cuerpo sano."
Y es muy cierta.
Las escuelas deberían considerar seriamente aumentar el número de horas semanales para la impartición de la educación física por las siguientes razones.
1.- La educación física es vital para el desarrollo físico del estudiante. Desde los grados más básicos, las escuelas deberían fortalecer la enseñanza de la educación física por motivos de salud, y dejar este buen hábito en los alumnos para toda la vida.
2.- La educación física, mejora el rendimiento académico ya que ayuda oxigenando los músculos, fortaleciendo el cerebro para que pueda concentrase mejor, despeja a los alumnos de tal forma que puedan regresar a los salones e clase más "despiertos."
3.- La educación física sirve para introducir la importancia de practicar deporte toda la vida, y hacerlo en la etapa escolar a través de equipos deportivos como el Futbol Americano, el beisbol, el basquetbol, el volibol, el soccer, y otros tantos.
Este idea del trabajo en equipo a través del deporte es muy importante para desarrollar habilidades como el liderazgo, el compromiso, la constancia y el superar la adversidad.
El profesor de educación física debe enseñarte la forma correcta de respirar cuando haces ejercicio, la importancia de la relajación del cuerpo para iniciar una actividad. El profesor debe hacer énfasis en la etapa del calentamiento y el estiramiento para preparar al cuerpo antes de realizar el ejercicio. Así como muchas otras enseñanzas.
Por esa razón, las escuelas deberían considerar seriamente la posibilidad de aumentar las horas de educación física por semana.
f an object has a mass of 200 kg and a weight of 1000 N, what is g?
Answer:
g = 5 m/s square
Explanation:
Weight(W), Mass(m), Gravity(g)
W = mg
1,000N = 200g
g = 1000/200
g = 5 m/s square
A car increase its speed steadily from 30km/hr to 60km/hr in 1 min A what is the average speed during this time
Explanation:
initial velocity(u)=30km/hr = 30*1000/60*60=8.33 m/s
final velocity(v) =60km/hr = 60*1000/60*60 =16.67 m/s
time taken(t) = 1 minutes
= 60 seconds
Now,
Average velocity = u+v/2
= 8.33 m/s + 16.67m/s÷2
=12.5 m/s
which species are near extinctions and where do they live now?
Answer:
Amur Leopard, Southeastern Russia, and northern china
Explanation:
Answer:
Amur Leopard China/Russia
Hawks bill Turtle live in Oceans
Question 3 of 10
Which image shows an example of the strong nuclear force in
action?
A.
B.
C.
D.
Answer: The answer is B
Explanation:
There are 4 fundamental forces that hold matter together.
- Gravitational Force
- Electromagnetic Force
- Strong Nuclear Force
- Weak Nuclear Force
We have barely just scratched the info about nuclear forces but the reason why B is the answer to the question is that Strong nuclear force actually holds the protons and neutrons together in the nucleus of an atom, much like the picture in B.
Answer: B
Explanation:
What do you mean by unit?
Unit is the quantity of a constant magnitude which is used to measure the magnitudes of other quantities of the same nature.
Answer:
The standard known quantity which is used to measure a physical quantities is known as unit.
3 An un calibrated mercury in glass thermometer immersed in melting ice. The length of the mercury thread is 25 mm when the thermometer immersed in steam from pure water boiling under a pressure of 1 atmosphere the length of the thread is 200 mm what is the temperature in degree centigrade when the length of the thread is 95mm.
Answer:
25 mm = 0 deg C
200 mm = 100 deg C
200 - 25 = 175 = change in thread per 100 deg C
95 - 25 = 70 mm - change in thread from 0 deg C
70 / 175 * 100 = 40 deg C final temperature at 95 mm
.........도움......
Is this correct? (the option is marked)
Answer:
sorry I don't understand this language
Please please helpjejejnebwbww
Answer:
C. 540 N
Explanation:
Let suppose that system formed by the athlete and the load are in equilibrium. By Newton's Laws of Motion, we use the moment equation with respect to the feet of the athlete to determine the upward force exerted by his two arms:
[tex]\Sigma M = -F\cdot r_{1} + m_{L}\cdot g \cdot r_{2} + m_{A}\cdot g \cdot r_{3} = 0[/tex]
[tex]F = \frac{(m_{L}\cdot r_{2} + m_{A}\cdot r_{3})\cdot g}{r_{1}}[/tex] (1)
Where:
[tex]m_{L}[/tex] - Mass of the load, in kilograms.
[tex]m_{A}[/tex] - Mass of the athlete, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]r_{1}[/tex], [tex]r_{2}[/tex], [tex]r_{3}[/tex] - Distances with respect to the feet, in meters.
[tex]F[/tex] - Upward force exerted by his two arms, in newtons.
If [tex]m_{L} = 6\,kg[/tex], [tex]r_{2} = 1.20\,m[/tex], [tex]m_{A} = 70\,kg[/tex], [tex]r_{3} = 0.90\,m[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]r_{1} = 1.30\,m[/tex], then the upward force is:
[tex]F = \frac{[(6\,kg)\cdot (1.20\,m)+(70\,kg)\cdot (0.90\,m)]\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{1.30\,m}[/tex]
[tex]F = 529.578\,N[/tex]
The upward force exerted by his two arms is 529.578 newtons. (Right answer: C)
