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Alexxx [7]
3 years ago
13

9. True or False: Sawmills and gristmills are machines that do not need

Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:

true can i get brainliest :)

Explanation:

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The energy carried by sound waves is called -
goldenfox [79]

Answer:

kinetic mechanical energy

8 0
2 years ago
Read 2 more answers
A ball is thrown upward. As it passes 5.0 m height it is traveling at 4.0 m/s up. What was its initial upward velocity? (a) 7.0
sveticcg [70]

Answer:

c) 10.7m/s

Explanation:

From the exercise we know that at 5m the ball  is traveling at 4m/s

To calculate its initial velocity we need to solve the following equation:

v_{y}^{2}=v_{oy}^{2}+2g(y-y_{o})

Since the initial height is 0

Solving for v_{o}

v_{oy}=\sqrt{v_{y}^{2}-2gy}=\sqrt{(4m/s)^2-2(-9.8m/s^2)(5m)}=10.7m/s

5 0
3 years ago
A electric heater that draws 13.5 a of dc current has been left on for 10 min. how many electrons that have passed through the h
kicyunya [14]
By definition, Ampere is a unit of current which is a measure of the amount of charge passing through a point in a circuit per unit  of time, with an equivalent charge of 1.602 x 10^(-19) Coulomb per electron. To determine the number of electrons passing through the heater, we use the definition of the current. We calculate as follows:

13.5 A = 13.5 C per second
Charge = 13.5 C/s (10 min) ( 60 s / 1 min)
Charge = 8100 C 

Number of electrons = 8100 C / 1.602 x 10^(-19) C per electron
Number of electrons = 5.1 x 10^22 electrons

Therefore, there are 5.1 x10^22 electrons that assed through the heater for 10 minutes.
3 0
4 years ago
Find the equivalent resistance of this parallel circuit with two strands.
svlad2 [7]
In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then 
1/Rt = 1/7 + 1/49 
1/Rt = 1/6.125 = 1/ 49/8
Rt = 49/8 <span>Ω

The total resistance hence is </span>49/8 Ω
6 0
3 years ago
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur
muminat

Answer:

a) E = 0

b) E =  \dfrac{k_e \cdot q}{ r^2 }

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

From which we have;

E \cdot  A =  \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}}  = \dfrac{0}{\varepsilon _{0}} = 0

E = 0/A = 0

E = 0

b) \phi_E = \oint E \cdot  dA =  \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}

E \cdot  A  = \dfrac{+q }{\varepsilon _{0}}

E  = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}

By Gauss theorem, we have;

E\oint dS =  \dfrac{q}{\varepsilon _{0}}

Therefore, we get;

E \cdot (4 \cdot \pi \cdot r^2) =  \dfrac{q}{\varepsilon _{0}}

The electrical field outside the spherical shell

E =  \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }=  \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }

k_e=  \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }

Therefore, we have;

E =  \dfrac{k_e \cdot q}{ r^2 }

5 0
3 years ago
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