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goldfiish [28.3K]
2 years ago
14

What is the rate of acceleration of a 2,000 kilogram truck if a force of 4,200 N is used to make it start moving forward?

Physics
2 answers:
gizmo_the_mogwai [7]2 years ago
7 0
Answer: 2.1

Explain:
F = ma
a = F/m = 4,200/ 2,000
a = 2.1
Svetlanka [38]2 years ago
3 0
2.1m/s^2 is the answer to
Your question. I hope that helps.
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A capacitor consists of a set of two parallel plates of area a separated by a distance
MakcuM [25]
When a dielectric material is inserted between two plates of capacitor that are connected to a battery, you would observe that both the charge and the capacitance of the capacitor would change. This is due to the dielectric material which is able to transmit electric force.
8 0
3 years ago
A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o
Ierofanga [76]

Say we have a cylinder that has a height of dx, we see that the cylinder has a volume of: <span>

<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx

Then, the weight of oil in this cylinder is: 

Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx. 

Then, since the oil x feet from the top of the tank needs to travel x feet to get the top, we have: 

Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx. 

<span>Integrating from x1 to x2 ft gives the total work to be: (x1 = distance from top liquid level to ground level; x2 = distance from bottom liquid level to ground level)</span>

<span>W = ∫ 1250π x dx  
<span>W = 1250π ∫ x dx
W = 625π * (x2 – x1)</span></span></span></span>

<span>x2 = 14 ft  + 15 ft = 29 ft</span>

x1 = 14 ft + 1 ft = 15 ft

<span>
W = 625π * (29^2 - 15^2) 
<span>W = 385,000π ft-lbs = 1,209,513.17 ft-lbs</span></span>

3 0
3 years ago
A pendulum consists of a 2.7 kg stone swinging on a 4.0 m string of negligible mass. The stone has a speed of 8.1 m/s when it pa
Brut [27]

Answer:

a). v=4.77 \frac{m}{s}

b). β=80.61

c). E=88.5J

Explanation:

a).

m1=2.7 kg\\L=4m\\v_{1}=8.1 \frac{m}{s}\\\beta=63

\frac{m*v_{1} ^{2} }{2}+m*g*y_{1}=\frac{m*v_{2}^{2} }{2}+m*g*y_{2}\\\frac{m*v_{1} ^{2} }{2}=\frac{m*v_{2}^{2} }{2}+m*g*y_{2}\\v_{2}=\sqrt{v_{1}^{2}-2*g*y_{2}}\\y_{2} =L-L*cos(\beta)\\v_{2}=\sqrt{v_{1}^{2}-2*g*(L-L*cos(\beta))}\\v_{2}=\sqrt{8.1^{2}-2*9.8*(4(1-*cos(63))}\\v_{2}=\sqrt{8.1^{2}-42.8}\\v_{2}=\sqrt{64-42.8}\\v_{2}=\sqrt{21.19}=4.77 \frac{m}{s}

b).

\frac{m*v_{1}^{2}}{2} +m*y_{1} =\frac{m*v_{2}^{2}}{2} +m*y_{2}\\\frac{m*v_{1}^{2}}{2} +0 =0 +m*y_{2}\\y_{2}=\frac{v_{1}^{2}}{2*g}=\frac{(8.1^\frac{m}{s}){2}}{2*9.8\frac{m}{s^{2} }}=3.34 m\\ 3.34m=L(1-cos(\alpha)\\3.34m=4*(1-cos(\alpha ))\\cos(\alpha )=\frac{4-3.34}{4}=0.16\\(\alpha )= cos^{-1}*(0.163)\\ (\alpha )= 80.61

c).

E=\frac{m*v_{1} ^{2} }{2}\\ E=\frac{(2.7kg)*(8.1\frac{m}{s})^{2}}{2}=\frac{177.147\frac{kg*m^{2}}{s^{2}}}{2}\\\\E=88.5 J

6 0
3 years ago
What do you think will happen to the potential energy of the ball if the mass of the ball remains the same and placed in a highe
lapo4ka [179]

Answer:

Potential energy of the ball would increase

Explanation:

The formula for potential energy of an object is:

Potential Energy (object) = mass x gravity x height

In your question, the value for height is increased when the object (ball) is placed in a higher position. This creates a net positive change in the right-hand side of the formula and since the value of mass and gravity doesn't change because gravity is a constant (meaning it's always 9.8N/kg) and the mass is said to be kept the same, the ball's potential energy would increase.

4 0
2 years ago
Calculate the pressure in pascals if a 400N force is applied to an area of 0.5m2.
Zinaida [17]

Answer:

0.00789pa

Explanation:

since pressure=force/area

=400/0.5

=800N/m2

1.013*10^5N/m2=1pascal

800N/m2=xpascal

therefore it becomes,800/1.013*10^5

=0.00789pascal

7 0
2 years ago
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