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3 years ago

What’s the potential difference across a 5.0 ohms resistor that carries a current of 5.0A

Physics

1 answer:

vivado [14]3 years ago

3 0

Answer:

The correct answer is B-25 V

Explanation:

We apply Ohm's Law, according to which:

V = i x R

V = 5A x 5Ω

V= 25 V

Being V the potential difference whose unit is the VOLT, i the current intensity (Ampere) and R the electrical resistance (ohm)

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3 years ago

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A concave mirror has a radius of curvature of 1.6m. Find the focal length

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Given,

Radius of curvature of concave mirror = 1.6m

We know that ,

Focal length = radius/2

Hence ,

Focal length of concave mirror = radius of concave mirror /2

=> F = 1.6/2

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2 years ago

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The low areas created as a sound wave propagates are called rarefactions

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3 years ago

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Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) $x=\frac{1}{2}at^2+v_0t+x_0$

The equation for velocity v and a constant acceleration a is:

(ii) $v=at+v_0$

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) $a = \frac{v -v_0}{t}$

(iv) $x = \frac{v-v_0}{2t}t^2+v_0t + x_0$

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) $x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t$

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v: $v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}$

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

$f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}$

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3 years ago

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its

maxonik [38]

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

$KE=0.5mv^{2}$ where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

$KE=0.5*800*(20 m/s)^{2}=160,000$

Frictional force by vehicle pads is given by

$Fr=\frac {KE}{d}$ where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

$Fr=\frac {160000}{50}=3200 N$

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

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3 years ago