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Marianna [84]
2 years ago
13

The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration

Physics
1 answer:
a_sh-v [17]2 years ago
3 0

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
  • Force along x axis= mass × acceleration along x axis= 2N
  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

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Answer: Brownion motion is the erratic random movement of microscopic particles in a fluid, as a result of continuous bombardment from molecules of the surrounding medium.

Explanation:

Brownian motion is the random movement of particles in a fluid due to their collisions with other atoms or molecules.

4 0
4 years ago
HELP PLS i need this right now
N76 [4]

Answer:

V = 44.4 units.

Explanation:

In order to solve this problem, we must use the Pythagorean theorem. Which is defined by the following expression.

V = \sqrt{(V_{x})^{2} +(V_{y})^{2} }

where:

Vx = - 43 units

Vy = 11.1 units

Now replacing:

V=\sqrt{(-43)^{2} +(11.1)^{2} }\\V=44.4 units

7 0
3 years ago
When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

3 0
3 years ago
The height of the Washington Monument is measured to be 170 m on a day when its temperature is 35.0°C. What will the change in i
Alecsey [184]

Answer:

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m

Explanation:

Thermal coefficient of marble varies between (5.5 - 14.1) ×10⁻⁶/K = α

So, let us take the average value

(5.5+14.1)/2 = 9.8×10⁻⁶ /K

Change in temperature = 35-(-18) = 53 K = ΔT

Original length = 170 m = L

Linear thermal expansion

\frac{\Delta L}{L} = \alpha\Delta T\\\Rightarrow \Delta L=\frac{\alpha\Delta T}{L}\\\Rightarrow \Delta L=9.8\times 10^{-6}\times 53\times 170

The deformation is 0.088289 m

The final height of the monument is 170-0.088289 = 169.911702 m (subtraction because of cooling)

4 0
3 years ago
Use the circuit to answer the following questions.
Zanzabum

Answer:

1)ammeter

2)ised to check measure of current flow through a circuit

3)o.90 ambere

7 0
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