Answer:
work done=392J
Explanation:
work done =Mgh
DATA
M=5kg
g= 9.8
h=8m
W=Mgh
W=5*9.8*8
W=392J
The work done by the gravitational force is equal to -400 J.
What is gravitational force?The gravitational force can be described as a force that attracts a body towards the center of the earth or any physical object that has mass. Every object has mass and exerts a gravitational pull on another object with mass.
Given, the mass of the container, m = 5 Kg
The height from the ground level, h = 8 m
The work done can be calculated by using the below-mentioned formula:
[tex]W = m\times g\times h[/tex]
Where g is the gravitational acceleration.
As the container is lifted in the upward direction therefore the value of the g will be negative.
Substitute the values of m, g, and h in the above formula:
W = 5 × (-10) × 8
W = -400 J.
The negative sign shows that the work is done against the gravitational force.
Therefore, the work done by the gravitational force is equal to -400 J.
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Your maximum heart rate is the maximum number of times your heart can contract in one minute
True or false
Answer:
True
Explanation:
It is true i am right
A car drives 500m in 2 minutes.What is the cars speed? A.2m/s B.4m/s C.5 m/s
Answer:
B
Explanation:
1 minute = 60 seconds
2 minutes = 2 * 60 = 120 seconds.
d = r * t
t = 120 seconds
d = 500 meters
r = d/t
r = 500 / 120 = 4.167 m/s
The closest answer is B
th weight of an object is
Answer:
the weight of an object is the force acting on the object due to gravity. the gravitational force acting on the object. Others define weight as a scalar quantity, the magnitude of the gravitational force.
Explanation:
how big is the universe?
sorry this is so random
Answer:
Explanation: 46 billion light years I think
Aww I wanted brainliest:( be doggo, doggo sad :(
---Spring Question---
A 0.84 kg ball is held at rest against a spring (k = 142 N/m) that has been compressed 0.06 m. The ball and spring are on a table and the ball is 2.0 m above the ground. If friction between the ball and the table is negligible, then how far from the edge of the table (in meters) will the ball hit the ground?
Answer:
The distance from the edge of the table at which the ball will hit the ground is approximately 0.49 meters
Explanation:
The mass of the ball = 0.84 kg
The spring constant k = 142 N/m
The compression of the spring = 0.06
Therefore, by conservation of energy, we have;
The potential energy of the spring = The kinetic energy given to the ball
The potential energy of the spring = 1/2 × k × x² = 1/2 × 142 × 0.06² = 0.2556
The potential energy of the spring = 0.2556 J
The kinetic energy given to the ball = 1/2 × m × v²
Where;
v = The final velocity of the ball
Substituting gives;
1/2 × 0.84 × v² = 0.2556
∴ v = √(0.2556/(1/2 × 0.84)) ≈ 0.78
The final velocity of the ball = v ≈ 0.78 m/s
The height of the ball above the ground, h = 2.0 m
Therefore, we have;
The time, t, it takes the ball to touch the ground from the 2.0 m height, h, is given as follows;
h = 1/2·g·t²
t = √(2·h/g)
Where;
g = The acceleration due to gravity = 9.81 m/s²
∴ By substitution, we have;
t = √(2 × 2/9.81) ≈ 0.63855
t ≈ 0.63855 s
The distance from the edge of the table at which the ball will hit the ground = The horizontal distance covered during before the ball hits the ground
The horizontal distance covered during before the ball hits the ground = The (horizontal) velocity of the ball × The time it takes the ball to touch the ground
∴ The horizontal distance covered during before the ball hits the ground = 0.78 × 0.62855 ≈ 0.490269 ≈ 0.49
The distance from the edge of the table at which the ball will hit the ground = The horizontal distance covered during before the ball hits the ground ≈ 0.49 m
A sign is attached to the side of a building as seen below. The sign is attached to a steel rod. A wire connects to the end of the rod for support. The mass of the sign is 50 kg. Determine the supporting force exerted by the cable
Answer:
500N
Explanation:
50kgx10N/kg
= 500N
A crate is being pulled down an incline as shown in the figure. With respect to the crate's direction of motion, which of the following forces does only negative work on the crate?
Answer:Fn
Explanation:
1. A student notices that wearing darker colors in sunlight makes him feel warmer, so he decides to conduct an experiment. He takes five pieces of different-colored cloth and wraps
each one around a water bottle. He then places all five bottles in direct sunlight and measures the temperature of the water in each bottle an hour later.
What should be kept constant throughout this experiment?
O the color of the cloth
O the temperature of the water
the amount of water in each bottle
O the temperature of the outside air
Answer: the color of the cloth
Explanation:
A 10 kg object experiences an acceleration of 2 m/s squared. What net force was applied to the object?
F = ma
= 10kg × 2m/s
= 20N
Answer : 20N
Explanation :
F = Force (N)
m = mass (kg)
a = acceleration (m/s)
which term describe the substance that forms when two or more types of atoms join?
A. element
B. compound
C. composite
D. mixture
Answer:
A. element
Explanation:
A molecule is formed when two or more atoms join together chemically. If atoms combine that are of two or more different elements, we call that a compound. All compounds are molecules, but not all molecules are compounds.
help me with this please :((
Answer:
huh?
Explanation:
a car travelling at 20 ms-1 accelerates uniformly for 8 s until its velocity reaches 36 ms-1.what are the acceleration and the average velocity of the car during this period
Answer:
Acceleration of the car = 2 m/s²
Average velocity of the car =28 m/s
Explanation:
Initial velocity = 20 m/s
Final velocity = 36 m/s
Time taken = 8 s
Acceleration = Δ velocity / time taken
Acceleration = 36-20 / 8 = 16/8 = 2 m/s²
Average velocity of the car =[ Initial velocity + Final velocity ] /2
Average velocity of the car = [ 20+36] / 2 = 28 m/s
In one of his many action movies Jackie Chan jumped off a building by wrapping a rope around his waist and then allowed it to unwind as he fell to the ground, much the same as a yo-yo. Assuming his acceleration toward the ground was a constant much less than g, the tension in the rope would be
Directions: Summarize the main ideas of this lesson by answering the question below.
How might a person acquire a fear of or aversion to something even though it has no negative effect
on him or her?
Answer:
Explanation:
Trypophobia is a fear or disgust of closely-packed holes. People who have it feel queasy when looking at surfaces that have small holes gathered close together.A person may start by imagining what they fear, then looking at pictures of the fear object, and then finally being near or even touching the source of their anxiety. In the case of trypophobia, a person with symptoms may start by simply closing his eyes and imagining something such as a honeycomb or seed poThey can help you find the root of the fear and manage your symptoms. Last medically reviewed on July 20, 2017 Medically reviewed by Timothy J. Legg, Ph.D., CRNP — Written by Annamarya Scaccia ...
Es muy común que cuando se viaja hacia un río o lago se juegue "ranita", el cual consiste en lanzar una piedra horizontalmente hacia adelante para que cuando ésta toque la superficie del agua haga varios "saltos" sobre el agua. Durante un juego de estos, un desocupado nota que una de las piedras que arroja se demora 0,4 s en tocar la superficie del agua y la toca a 2,5 m de la orilla del lago, desde donde fue lanzada. Encuentre: a) La altura de la que fue lanzada la piedra. b) La velocidad con la que fue lanzada.
Answer:
a) La piedra es lanzada desde una altura de 0,785 metros.
b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
Explanation:
a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:
[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Donde:
[tex]y[/tex] - Altura final, medida en metros.
[tex]y_{o}[/tex] - Altura inicial, medida en metros.
[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.
Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:
[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]
[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]
[tex]y_{o} = 0,785\,m[/tex]
La piedra es lanzada desde una altura de 0,785 metros.
b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:
[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)
Donde:
[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.
[tex]t[/tex] - Tiempo, medido en segundos.
Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:
[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]
[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]
La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.
Four measurements of the file card's width are made. Of the measurements given below, the LEAST precise.
O 7.62
O 7.63
O 7.34
O 7.6
!!!!!
While following directions on a treasure map, a person walks 45.0 m south, then turns and
walks 7.50 m east. What is the direction of the straight line that the treasure hunter could have
walked directly from the starting to reach the same spot?
(1 Point)
9.5° south of east
9.5° east of south
21° south of east
21° east of south
Answer:
46 m at 9.5 degrees east of south
Explanation:
Which best represents the appearance of the moon on July 25, 2011
Answer:
Its the third quarter or waning quarter
Hope this helps :3
A fish swimming at a rate of .6 m/s notices a huge shark. Three seconds later, the fish is swimming at a speed of 3 m/s. What is the fish's acceleration?
0.8 m/s/s
-0.8 m/s/s
12.5 m/s/s
-12.5 m/s/s
Answer:
C
Explanation:
???
i think
a 10 kilogram steel ball is dropped from the top of a tower 100 meters high the kinetic energy of the ball just before it sttikes the ground is most nearly
Explanation:
If we assume negligible air resistance and heat loss, we can assume that all of the Gravitational potential energy of the ball will turn into Kinetic energy as it falls toward the ground.
Therefore our Kinetic energy = mgh = (10kg)(9.81N/kg)(100m) = 9,810J.
1. *A car is going over the top of a hill whose curvature approximates a circle of radius 200 m. At
what velocity will the occupants of the car appear to weigh 20% less than their normal weight (or
their normal weight times 0.8)?
Answer:
The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s
Explanation:
The given parameters are;
The curvature of the hill, r = 200 m
Due to the velocity, v, the occupants weight = 20% less than the normal weight
The outward force of an object due to centripetal (motion) force is given by the following equation;
[tex]F_c = \dfrac{m \times v^2}{r}[/tex]
Where;
r = The radius of curvature of the hill = 200 m
Given that the weight of the occupants, W = m × g, we have;
[tex]F_c = 0.2 \times W = 0.2 \times m \times g[/tex]
[tex]\therefore 0.2 \times m \times g = \dfrac{m \times v^2}{r}[/tex]
v = √(0.2 × g × r)
By substitution, we have;
v = √(0.2 × 9.81 × 200) ≈ 19.81
The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.
The velocity is 19.81 m/sec at which the occupants of the car appear to weigh 20% less than their normal weight and this can be determined by using the formula of centripetal force.
Given :
A car is going over the top of a hill whose curvature approximates a circle of radius 200 m.
The centripetal force acting on the car is given by the formula:
[tex]\rm F_c = \dfrac{mv^2}{r}[/tex] ---- (1)
where m is the mass of a car, v is the velocity of the car and r is the radius of curvature.
Now, it is given that the occupants of the car appear to weigh 20% less than their normal weight that means:
[tex]\rm F_c = 0.2\times W[/tex]
[tex]\rm F_c = 0.2 \times m \times g[/tex] --- (2)
Now, equate equation (1) with equation (2).
[tex]\rm \dfrac{mv^2}{r}=0.2\times m \times g}{}[/tex]
[tex]\rm v = \sqrt{0.2\times g \times r }[/tex]
[tex]\rm v = \sqrt{0.2\times 9.81 \times 200 }[/tex]
v = 19.81 m/sec
The velocity is 19.81 m/sec at which the occupants of the car appear to weigh 20% less than their normal weight.
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Indica en cada caso si estas fuerzas producen un efecto deformador o dinámico: a) Empujar un carro de la compra c) Atraer mediante un imán un clavo que estaba en reposo b) Modelar una figura de arcilla d) Lanzar una pelota de tenis
Answer:
Dynamic a, b and d
Deformer c and d
Explanation:
When a force is applied to an object, the acceleration may or may not change, which is analyzed from Newton's second law, this effect is a dynamic effect on the system; The environment can change the shape of the object, in this case the force creates a deformation in it.
With these clarifications let's analyze the different situations.
a) Dynamic. When pushing a car initially an acceleration is created in it, shortly after this decreases to zero and continues with constant speed.
b) Dynamic. When the nail moves it does not change the shape of the body that contains it, nor the shape itself, therefore the system can be analyzed with Newton's second law and the force is of the Dynamic type
c) Deformer. In this case the body (clay) has no net movement, but it does change its shape, so the effect of the force translates into a deformation of the system
d) Deformar and Dynamic. The launch of tennis balls has an initial deformation stage during the acceleration process, in this part the force is of the DEFORMING type, but during the trajectory the deformation disappears and the ball has (gravity) a DYNAMIC action
In a place covered by shadow of cloud sun cannot be seen . Explain with reasons .
Answer:
Because even though our eyes have a huge dynamic range (ability to pick out details in sharply lit and lesser lit areas simultaneously) than any camera, there's a limit.
When there's strong sunlight, your pupils contract and let less light in, which makes the shadows look darker.
When it's cloudy, your pupils widen and let more light in, which makes the shadows look less dark.
Do some experiments with a camera and you'll soon get the hang of it.
NOTE: Also test HDR (high dynamic range) photography, where the camera takes three or more pictures in quick succession, with different exposure settings, and combines them to get the most detail of both bright and dark areas. The result is more or less what we percieve.
If u drop a bar of soap on the floor then is the floor clean or is the bar of soap dirty? Lol I’m so bored i dont even know what I’m doing.
Answer:
CLEAN
Explanation:
Answer:
I just wash mine with water when I drop it . That is a good question tho :)
Explanation:
How does seeing an object compare with seeing a shadow?
Answer: shadows are not parts of their objects—we can change a shadow without changing the object
Answer:
So we can see objects by seeing silhouettes. By contrast, shadows are not parts of their objects—we can change a shadow without changing the object (say, by putting another object in between the object and the light source). So we cannot see objects by seeing shadows.
marcus is on a merry-go-round holding a string. you are standing next to him and holding the other end of the string so that the string forms a tangent line to the edge of the merry-go-round. if marcus is 2 meters from the center of the ride, what torque are you applying on him if you pull 30 N? _Nm
Answer: 60 nm
Multiply 30 and 2
1 point
A leak in a swimming pool allows 2.50 liters to flow out every minute. How
many gallons leak out in week? (There are 0.946 liters per quart and 4
quarts per gallon.) *
O 910. gallons
O 12 700 gallons
0 7 140 gallons
6 660 gallons
Answer:
66.6 G
Explanation:
Why is a teacher's feedback valuable?
Answer:
Because they could help you move on better with your strategy and ultimately get you better scores on tests
Answer:
Through feedback, teachers can provide the students with suggestions for development, learning strategies and corrections for errors. The importance of constructive feedback allows for many positive opportunities. This component also enhances a student's self-efficacy and provides an avenue for motivation.
According to the Natural Resources Defense Council, what is the largest contributor to land pollution?
According to the Natural Resources Defense Council, activities related to
production are the largest contributor to land pollution.
Answer: Food
Explanation:
Food is the largest contributor to land pollution.
What is the Natural Resources Defense Council?
NRDC strives to rid our food, air, water, and household products from toxic chemicals.
40 percent of food produced in the US is trashed each year according to the Natural Resources Defense Council.
Food decomposes anaerobically and releases methane emissions, a greenhouse gas that is 28 times more potent than carbon dioxide over a 100-year period and 84 times more on a 20-year scale , when food waste ends up in landfill.
Worldwide, some percent of greenhouse gas emissions result from food waste alone.
Therefore,
Food is the largest contributor to land pollution.
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A race car makes one lap around a track with a radius of 41 meters in 8 seconds. What is the velocity?
Answer:
31.21 m/s
Explanation:
radius=41m
time = 8 sec
circumference=2πr
from formula
circumference=2×22/7×41
=257.71m
now
velocity=total distance( circumference)/total time
=257.71/8
= 32.21m/s
The velocity of race car makes one lap around a track with a radius of 41 meters in 8 seconds is 32.19 m/s
What is velocity ?Velocity is "rate of change of displacement with respect to time".
i.e. v= dx/dt
it is also defined as displacement over time. i.e. v=Displacement/Time.
Velocity shows how much distance can be covered in unit time. It's SI unit is m/s. It is vector quantity ( having both direction as well as momentum). where displacement is distance from mean position.
Given,
Radius of the track r= 41m
time taken by car to complete one lap T = 8s
velocity v = ?
we know that ω = 2π÷T
ω = 2×3.14 ÷ 8
ω = 0.785 rad/s
v=rω
v= 41×0.785
v= 32.185 ≅32.19 m/s
Hence velocity of the car is 32.19 m/s
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