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Semmy [17]
2 years ago
6

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what

is the resistance of the wire?
Physics
1 answer:
Alexxx [7]2 years ago
4 0
<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
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An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance
bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 \frac{m}{sec}

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion V^{2} = u^{2} - 2 a s

Put all the values in above formula we get,

⇒ 0 = 0.8^{2} - 2 × a ×0.2667

⇒ a = 1.2 \frac{m}{sec^{2} }

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0
2 years ago
An amusement park ride spins you around in a circle of radius 2.5 m with a speed of 8.5 m/s. If your mass is 75 kg, what is the
zhannawk [14.2K]
B is the answer to this truly did the math
3 0
2 years ago
Read 2 more answers
What factors affect attractive force
Inga [223]

Two Factors That Affect How Much Gravity Is on an Object. Gravity is the force that gives weight to objects and causes them to fall to the ground when dropped. Two major factors, mass and distance, affect the strength of gravitational force on an object.

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2 years ago
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A rabbit is moving in the positive x-direction at 1.10 m/s when it spots a predator and accelerates to a velocity of 10.9 m/s al
anzhelika [568]

Answer:

aₓ = 0 ,       ay = -6.8125 m / s²

Explanation:

This is an exercise that we can solve with kinematics equations.

Initially the rabbit moves on the x axis with a speed of 1.10 m / s and after seeing the predator acceleration on the y axis, therefore its speed on the x axis remains constant.

x axis

          vₓ = v₀ₓ = 1.10 m / s

          aₓ = 0

y axis

initially it has no speed, so v₀_y = 0 and when I see the predator it accelerates, until it reaches the speed of 10.6 m / s in a time of t = 1.60 s. let's calculate the acceleration

         v_{y}= v_{oy} -ay t

          ay = (v_{oy} -v_{y}) / t

          ay = (0 -10.9) / 1.6

          ay = -6.8125 m / s²

the sign indicates that the acceleration goes in the negative direction of the y axis

8 0
2 years ago
A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume
dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

Initial speed of car 1, u_1=15i\ m/s (east)

Mass of the car 2, m_2=750\ kg

Initial speed of car 2, u_1=20j\ m/s (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

900\times 15i +750\times 20j=(900+750)V

13500i+15000j=1650V

V=(8.18i+9.09j)\ m/s

The magnitude of speed,

|V|=\sqrt{8.18^2+9.09^2}

V = 12.22 m/s

(b) Let \theta is the direction the wreckage move just after the collision. It is given by :

tan\theta=\dfrac{v_y}{v_x}

tan\theta=\dfrac{9.09}{8.18}

\theta=48.01^{\circ}

Hence, this is the required solution.

4 0
2 years ago
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