A ball experiences forces of 14 N [N] and 9.2 N [W]. A Free Body diagram is required. What is the acceleration of the ball if its mass is 225 g?
Answers
Answer:
First, we can define the North as the positive y-axis, and the East as the positive x-axis.
Here we have two forces:
F₁ = 14N to north
F₂ = 9.2 N to west.
Then the free body diagram will be the one that can be seen below:
Now we want to find the acceleration that the ball experiences if the mass is 225g
Knowing that 1k = 1000g, we can rewrite the mass as:
M = 0.225g
Here we also need to remember the second Newton's law:
F = M*a
Net force equals mass times acceleration.
First, let's find the net force:
F = F₁ + F₂ = 14N*(0, 1) + 9.2N*(-1, 0 )
F = (-9.2N, 14N)
The module of the force is then:
I F I = √( (-9.2N)^2 + (14N)^2) = 16.75 N
Then the acceleration is given by
16.75 N = 0.225kg*a
(16.75 N)/(0.225 kg) = a = 74 m/s^2
![A Ball Experiences Forces Of 14 N [N] And 9.2 N [W]. A Free Body Diagram Is Required. What Is The Acceleration](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/XYQepwlTMSPOCduihT5CrmTUz07tOKhJ.png)
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Answers
Answer: 0.04139m
Explanation:
First, we need to calculate the weight of the man which will be:
Weight = mass × acceleration due to gravity
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Answers
Answer:
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Explanation:
For an ideal machine:
[tex]Work\ In = Work\ Out\\(P)(L)=(W)(H)[/tex]
where,
P = Effort = Spring Scale Reading = 70 N
L = Effort Arm = Length of inclined plane = 12 m
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Answer:
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Answer:
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To get the displacement, we will make use of the expression;
[tex]D = \sqrt{(\sum d_x)^2+(\sum d_y)^2}[/tex]
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dx = 1km - 4km
dx = -3km
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dy = 3km - 3km
dy = 0km
Substitute into the formula above;
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Hence his displacement is 3km
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B.) It provides an accurate scale of the concept it is modeling.
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D.) It provides a useful understanding of the concept it is modeling.
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Answers
Answer:
Final velocity of ball = 20 m/s
Explanation:
Given:
Height when ball released = 20meter
Find;
Time taken to fall
Final velocity of ball
Computation:
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S = ut + (1/2)(g)(t²)
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Question 8
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Answer:
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Explanation:
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couple = force x length
From the given question, the couple acting on the ring can be determined as;
couple = 25 x 0.65
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The magnitude of the couple is 16.25 Nm.
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Answer:
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2)Covection
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120N
N8'0
ON
Answers
Answer:
120N
Explanation:
Newton's second law formula: F= ma
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The mathematical equation of the second law of motion can be written as:
F = ma
Given, the mass of the box, m = 10 Kg
The acceleration of the box, a = 12m/s²
We can find the force acting on the box by putting the values a and m:
Force acting on the box, F = 10 ×12 = 120 N
Therefore, the force acting on the box is equal to 120 newtons.
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Answer:
3.185×10^-29 kgm/s
Explanation:
Momentum(p)=mass×velocity
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4.- Una vagoneta de 1000 kg de peso parte del reposo en el punto 1 y desciende, sin rozamiento, por la vía indicada en la figura. A) Calcular la fuerza que la vía ejerce sobre la vagoneta en el punto 2, donde el radio de curvatura es de 6 m. B) Determinar el mínimo valor del radio de curvatura en el punto 3 para salvar dicho punto
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Answer:
A) 49,050 N
B) 16 m
Explanation:
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El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.
La masa dada del vagón, m = 1,000 kg
La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m
A) El radio en el punto 2, el punto más bajo, R = 6 m
La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;
N = El peso del vagón + La fuerza de movimiento del vagón
∴ N = m × g + m × a
Dónde;
g = La aceleración debida a la gravedad ≈ 9,81 m / s²
a = La aceleración del vagón
Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R
∴ m × a = m × v² / R
Dónde;
v² = La velocidad del vagón en el punto 2 = 2 · g · h₁
Por lo tanto;
N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R
∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050
La fuerza que ejerce el vagón en el punto 2, N = 49,050 N
B) En el punto 3, tenemos;
N = m · g - m · a₃
La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃
∴ La altura en el punto 3, h₃ = 4 m
El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)
Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando
Por lo tanto;
N = m · g - m · a₃ = 0
m · g = m · a₃ = m · v₃² / R₃ = m · (2 · g · (h₁ - h₃)) / R₃
∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16
El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.
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A. An electron in an atom jumping from a lower energy state to a higher one.
B. Two electrons from the same atom colliding
C. An electron in an atom jumping from a higher energy state to a lower one.
D. Two electrons from two different atoms colliding.
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Answer:
Im not 100% sure but i think the answer is A. An electron in an atom jumping from a lower energy state to a higher one.
Explanation:
lmk if its wrong
How do the properties of the planets change as you move away from the sun?
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Answer:
as we move away from the sun the velocity of the planet decreases
a 20 A fuse is connected in series with a circuit containing a 240 V source. What is the minimum resistance required to prevent the fuse from melting
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Answer:
12 ohms
Explanation:
you just divide 240 and 20
240 V/ 20 A = 12 ohms
A set of charged plates have an
area of 8.22*10^-4 m^2 and
separation 2.42*10^-5 m. The
plates are charged with
5.24*10^-8 C. What is the
potential difference V between
the plates?
(Unit = V)
pls hurry <3
Answers
Answer:
175
Explanation:
its for Acellus
The potential difference is V=1.743*10^-7
The voltage is generally referred to as the electric potential difference and it can be measured using a voltmeter. The electrical potential difference between the two plates is expressed as q=CV.
CALCULATION
we know,
q=CV
q={EA/d}*V
here q= 5.24*10^-8C
A=8.22*10^-5m^2
d=2.42*10^-8m
therefore , C={( 8.85*10^-12)*(8.22*10^-5)}/(2.42*10^-8)
C=3.006*10^-1
q=CV
q/C=V
V=1.743*10^-7
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9. How is spring potential energy determined from a force versus position graph?
Answers
Answer: The potential energy associated with a mass attached to a spring depends on how much the spring is stretched or compressed. ... The gravitational force on the mass is −mg (“−” because the force points down). The force is the negative of the slope on the potential energy versus position graph.
Explanation:
:)
By getting the data from the graph of force versus displacement we can calculate the Potential energy of the spring.
What will be the potential energy?As we know that potential energy is defined as the energy possessed by the body due to its position with respect to other positions.
For an example the potential energy of an object if it is placed at height h will be due to the gravitational acceleration of the earth and is given by [tex]PE=mgh[/tex]
Similarly, for a spring-mass system, the potential energy will be due to the weight of the mass that will be the force exerted by the mass on the spring so there will be an extension in the spring.
So this extension will be found out by the force and displacement graph as the position of the mass attached to the spring goes down so the potential energy can be calculated as the
[tex]PE= F\times X[/tex]
[tex]F=mg[/tex] weight of the mass
X= deflection of the spring
Thus by getting the data from the graph of force versus displacement we can calculate the Potential energy of the spring.
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Which one of the following properties apply to both waves and particles?
A) Photoelectric Effect
B) Diffraction
C) Refraction
Answers
Answer:
b
Explanation:
sorry if im wrong