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fenix001 [56]
2 years ago
8

Imagine you derive the following expression by analyzing the physics of a particular system: v2=v20+2ax. The problem requires so

lving for x, and the known values for the system are a=2.55meter/second2, v0=21.8meter/second, and v=0meter/second. Perform the next step in the analysis.
Physics
2 answers:
lisabon 2012 [21]2 years ago
7 0

As per kinematics equation we are given that

v^2 = v_o^2 + 2ax

now we are given that

a = 2.55 m/s^2

v_0 = 21.8 m/s

v = 0

now we need to find x

from above equation we have

0^2 = 21.8^2 + 2(2.55)x

0 = 475.24 + 5.1 x

x = 93.2 m

so it will cover a distance of 93.2 m

bixtya [17]2 years ago
3 0

Answer:

x = -93.18 meters

Explanation:

The equation is given as :

v^2=v_o^2+2ax...........(1)

The known values are as follows:

a=2.55\ m/s^2

v_o=21.8\ m/s

v=0\ m/s

Putting all the values in equation (1) as :

0=(21.8\ m/s)^2+2\times 2.55\ m/s^2\times x

x = -93.18 meters

So, the vale of x is 93.18 meters. Hence, this is the required solution.

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How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
2 years ago
A 0.300kg glider is moving to the right on a frictionless, ­horizontal air track with a speed of 0.800m/s when it makes a head-o
e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

m_1 = Mass of first glider = 0.3 kg

m_2 = Mass of second glider = 0.15 kg

u_1 = Initial Velocity of first glider = 0.8 m/s

u_2 = Initial Velocity of second glider = 0 m/s

v_1 = Final Velocity of first glider

v_2 = Final Velocity of second glider

As momentum and Energy is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

{\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}

From the two equations we get

v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J

Final kinetic energy of first glider is 0.010935 J

K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J

Final kinetic energy of second glider is 0.0858675 J

6 0
2 years ago
Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of
schepotkina [342]
If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!
5 0
1 year ago
Two identical boxes are being pulled across a horizontal floor at a constant velocity by a horizontal pulling force of 176 N tha
MAXImum [283]

Answer:

the tension in the rope between the boxes is equal to 88 N

Explanation:

given,

the force applied on one body F = 176 N

When two bodies are moving on horizontal plane at constant velocity then their kinetic friction (f k) is equal to applied force F

According to newton third law the resultant force acting on one body is equal to the resultant force acting on the another body.

T is the tension in the rope

T- f_k = - (T- f_k)

T - F = - (T - F)

T - 176 = - (T - 0)

2 T = 176

T = 176/2 = 88 N

so, the tension in the rope between the boxes is equal to 88 N

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2 years ago
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What two factors determine the energy production of the Hoover Dam
swat32
Mar 28, 2011 · The Hoover Dam generates electricity using hydropower. Hydropower is the power generated by moving water. Dams are often built on a large river, so the water falls down from the top of the dam. At the bottom of the dam, the water rotates a large turbine

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6 0
2 years ago
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