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2 years ago

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15. A body moving with a velocity of 20 m/s begins to accelerate at 3 m/s2. How far does the body move in 5 seconds? A. 137.5 m

B. 196 m C. 260 m D. 162.5 m

2 answers:

Rudik [331]2 years ago

7 0

Answer is B. According to the equation of motion s = vt + 1/2 at2 Where s is distance covered, v is velocity, a is acceleration and t is time taken. So, by putting all the values, we get s = (20)(5) + 1/2 (3)(5)2 s = 100 + 1/2 (3)(25) s = 100 + 1/2 75 s = 100 + 37.5 s = 137.5 meters

adelina 88 [10]2 years ago

5 0

The Correct answer to this question for Penn Foster Students is: 137.5 m

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A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, ca

viva [34]

Answer:

$a=0.2*10^{-5}g$

Explanation:

From the question we are told that:

Mass $M=180=>0.18kg$

Charge $Q=18mC=18*10^-^3C$

Velocity $v=2.2m/s$

Length of Wire $L=8.6cm=>0.086$

Current $I=30A$

Generally the equation for Magnetic Field of Wire B is mathematically given by

$B=\frac{\mu_0*I}{2\pi*l}$

$B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}$

$B=6.978*10^{-5}T$

Generally the equation for Force on the plane F is mathematically given by

$F=qvB$

Therefore

$ma=qvB$

$a=\frac{qvB}{m}$

$a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}$

$a=2.37*10^{-5}$

Therefore in Terms of g's

$a=\frac{2.37*10^{-5}}{9.8}$

$a=0.2*10^{-5}g$

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2 years ago

Steam enters the condenser of a steam power plant at 20kPa and a quality of 95% with a mass flow rate of 20,000kg/h. It is to be

avanturin [10]

Answer:

The mass rate of the cooling water required is: $1'072988.5\frac{kg}{h}$

Explanation:

First, write the energy balance for the condensator: The energy that enters to the equipment is the same that goes out from it; consider that there is no heat transfer to the surroundings and kinetic and potential energy changes are despreciable.

${m_{w}}*{h_{w}}^{in}+m_s{h_{s}}^{in}=m_w{h_{w}}^{out}+m_s{h_{s}}^{out}$

Where w refers to the cooling water and s to the steam flow. Reorganizing,

$m_w({h_{w}}^{out}-{h_{w}}^{in})=m_s({h_{s}}^{in}-{h_{s}}^{out})\\m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{({h_{w}}^{out}-{h_{w}}^{in})}$

Write the difference of enthalpy for water as Cp (Tout-Tin):

$m_w=\frac{m_s({h_{s}}^{in}-{h_{s}}^{out})}{C_{pw}({T_{w}}^{out}-{T_{w}}^{in})}$

This equation will let us to calculate the mass rate required. Now, let's get the enthalpy and Cp data. The enthalpies can be read from the steam tables (I attach the tables I used). According to that, ${h_{s}}^{out}=251.40\frac{kJ}{kg}$ and ${h_{s}}^{in}$ can be calculated as:

${h_{s}}^{in}={h_{f}}+x{h_{fg}}=251.40+0.95*2358.3=2491.8\frac{kJ}{kg}$ .

The Cp of water at 25ºC (which is the expected average temperature for water) is: 4.176 $\frac{kJ}{kgK}$ . If the average temperature is actually different, it won't mean a considerable mistake. Also we know that ${T_{w}}^{out}-{T_{w}}^{in}\leq 10$ , so let's work with the limit case, which is ${T_{w}}^{out}-{T_{w}}^{in}=10$ to calculate the minimum cooling water mass rate required (A higher one will give a lower temperature difference as a result). Finally, replace data:

$m_w=\frac{20000\frac{kg}{h}(2491.8-251.40)\frac{kJ}{kg} }{4.176\frac{kJ}{kgK} (10C)}=1'072988.5\frac{kg}{h}$

5 0

3 years ago

An object is dropped from a height of 75.0 m above ground level. (a) determine the distance traveled during the first second.

barxatty [35]

The relevant formula we can use in this case would be:

h = v0 t + 0.5 g t^2

where,

h = height or distance travelled

v0 = initial velocity = 0 since it was dropped

t = time = 1 seconds

g = 9.8 m/s^2

So calculating for height h:

h = 0 + 0.5 * 9.8 m/s^2 * (1 s)^2

<span>h = 4.9 meters</span>

6 0

3 years ago

Two point charges exert a 7.35 N force on each other. What will the force become if the distance between them is increased by a

I am Lyosha [343]

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

$q_1\ and\ q_2$ are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

$F'=\dfrac{kq^2}{r'^2}$

$F'=\dfrac{kq^2}{(2r)^2}$

$F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}$

$F'=\dfrac{1}{4}\times 7.35$

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.

3 0

3 years ago

a peice of titanium metal has a mass of 67.5 gm and a volume of 15 cm cubed. what is the density od titanium?

Nuetrik [128]

density=mass/volume

mass=67.5

volume=15cm cube

so

67.5/15

4.5g/cm cube

6 0

3 years ago

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