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Dennis_Churaev [7]

2 years ago

11

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

1 answer:

irakobra [83]2 years ago

6 0

Answer:

1.2 amps :)

Explanation:

A heater has a resistance of 10.0 Ω. It operates on a 12.0 V. What is the current through the resistor?

Known:

R = 10.0 Ω

V = 12.0 V

Unknown:

I = ???

I = V/R

= 12.0 V / 10.0 Ω

= 1.2 amps

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A train moves from rest to a speed of 25 m/s in 30.0 seconds. What is the acceleration?

fredd [130]

Answer:

$a = 0.83\ m/s^2$

Explanation:

<u>Uniform Acceleration </u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting:

$\displaystyle a=\frac{25-0}{30}$

$\boxed{a = 0.83\ m/s^2}$

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3 years ago

stellarik [79]

Answer:

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5 0

2 years ago

Read 2 more answers

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find

True [87]

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion $s=ut+\frac{1}{2} at^2$ , where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = $-9.8m/s^2$ .

$-20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0$

t = 6.53 seconds or t = -0.63 seconds

So time = 6.53 seconds.

Considering the horizontal motion of arrow

u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = $0m/s^2$

$s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m$

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

7 0

3 years ago

A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou

prisoha [69]

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

$tan\theta =\frac{v^{2}}{rg}$

$tan\theta =\frac{18.61^{2}}{53.1\times 9.8}$

tan θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

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3 years ago

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Maurinko [17]

Answer: Baking Soda

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7 0

3 years ago