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Alexandra [31]
3 years ago
7

A particular wire has a resistivity of 6.47×10-8 Ωm and a cross-sectional area of 2.32 mm2. A length of this wire is to be used

as a resistor that will develop 130 W of power when connected to a 9.00 V battery. What length of wire is required?
Physics
1 answer:
PolarNik [594]3 years ago
6 0

Using  the formula for power:

P = V^2 / R  

130 W = (9.00 V)^2 / R  

Solve for r:

R = 81/130

R = 0.623 ohms

Now solve for the length of wire:

R = rho L / A  

A must be in m^2 - 2.32 mm^2 * 1 m^2/10^6 mm^2 = 2.132x10^-6 m^2  

Now you have:

0.623 = (6.47x10^-8) L / (2.32x10^-6)  

L = 22.34 m (Round answer as needed.)

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A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational
velikii [3]

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

           sin  35 = Wₓ / W

            cos 35 = W_y / W

            Wₓ = W sin 35

            W_y = W cos 35

            Wₓ = 2500 9.8 sin 35

             Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis  

       N-W_y = 0

       N = W_y

X axis  

       F -Wₓ = m a

       F = Wₓ + m a = mg sin 35 + m a

       F = m (a + g sin 35)

   

let's calculate

       F = 2500 (5 + 9.8 sin 35)

       F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

3 0
3 years ago
A 2.0 kg mass weighs 10 Newtons on planet X. what is the acceleration due to gravity on planet X? Show the work.
KengaRu [80]

Answer: The gravitational acceleration on planet X is 5 N/kg

On Earth (with the gravitational accelartion g_E) the mass of 2kg will correspond to

F_E = m\cdot g_E = 2.0 \mbox{kg}\cdot 9.8 \frac{N}{kg} = 19.6 N

On planet X we are told the same measure is only 10N. Since there is a proportional relationship between g and F, we can calculate g_X:

\frac{F_E}{g_E}=\frac{F_X}{g_X} \implies\\g_x = \frac{g_E\cdot F_X}{F_E}=\frac{9.8 N/kg \cdot 10 N }{19.6N}=5 \frac{N}{kg}

8 0
3 years ago
7. Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If th
Ainat [17]

Answer:

Explanation:

Given

mass of children m_1=20\ kg

m_2=30\ kg

distance between two children L=3\ m

suppose  small child is at a distance of x m from  pivot point

so torque of small child and heavier child must be equal

20\times (x)=30\times (3-x)

2x=9-3x

5x=9

x=1.8\ m    

7 0
3 years ago
How does a mathematical model help you understand the science concepts
Dmitrij [34]
It can help with measurements and when you want to add measurements to a cylinder or a beaker so ya
7 0
3 years ago
The speed that a tsunami can travel is modeled by the equation , where s is the speed in kilometers per hour and d is the averag
CaHeK987 [17]

The speed of tsunami is a.0.32 km. 

Steps involved  :

The equation s = 356d models the maximum speed that a tsunami can move at. It reads as follows: s = 200 km/h d =?

Let's now change s to s in the equation to determine d: s = 356√d 200 = 356√d √d = 200 ÷ 356 √d = 0.562 Let's square the equation now by squaring both sides: (√d)² = (0.562) ² d = (0.562)² = 0.316 ≈ 0.32

As a result, 0.32 km is roughly the depth (d) of water for a tsunami moving at 200 km/h.

To learn more about tsunami refer : brainly.com/question/11687903

#SPJ4

6 0
2 years ago
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