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4 years ago

7

You have a large flashlight that takes 4 D-cell batteries, each with a voltage of 1.5 volts. If the current in the flashlight is

2.0 amps, what is the resistance?. Select one of the options below as your answer:. . A.. 12 ohms. B.. 0.75 ohms. C.. 3.0 ohms. D.. 0.5 ohms. E.. 6.0 ohms.

2 answers:

Sever21 [200]4 years ago

5 0

B: 0.75 ohms on plato

Naddik [55]4 years ago

3 0

Ohm's law states that,
V = I x R
where V, I, and R are voltage, current, and resistance, respectively. Rearranging the equation gives,
R = V / I
Replacing the variables with the given values,
R = 1.5 volts / 2.0 amp
The answer is letter B. 0.75 ohms.

You might be interested in

What minimum speed does a 170 gg puck need to make it to the top of a frictionless ramp that is 3.6 mm long and inclined at 27 ∘

amid [387]

Answer:

0.176 m/s

Explanation:

given,

mass of the puck, m = 170 g

Length of the ramp, L = 3.6 mm

angle of inclination, θ = 26°

the minimum speed require to reach at the top of the ramp

using equation of motion

v² = u² + 2 a s

final speed of the puck is zero

0² = u² - 2 g s

$u = \sqrt{2gh}$

height of the pluck

$sin \theta = \dfrac{h}{L}$

$sin 26^0= \dfrac{h}{3.6}$

h = 1.58 mm

h = 0.00158 m

now,

$u = \sqrt{2\times 9.8\times 0.00158}$

u = 0.176 m/s

hence, the speed required by the pluck to reach at the top is equal to 0.176 m/s

3 0

4 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

3 0

4 years ago

A 1000-kg car traveling north at 15 m/s collides with a2000-kg truck traveling east at 10 m/s. The occupants, wearing seat belts

Aleksandr-060686 [28]

Answer:

8.33 m/s, 36.87° North of East

Explanation:

$m_n$ = Mass of car = 1000 kg

$v_n$ = Velocity of car = 15 m/s

$m_e$ = Mass of truck = 2000 kg

$v_e$ = Velocity of truck = 10 m/s

M = Combined mass = 1000+2000 = 3000 kg

Momentum

$p_n=m_nv_n\\\Rightarrow p_n=1000\times 15\\\Rightarrow p_n=15000\ kgm/s$

Momentum of car traveling East is 15000 kgm/s

$p_e=m_ev_e\\\Rightarrow p_n=2000\times 10\\\Rightarrow p_n=20000\ kgm/s$

Momentum of truck traveling North is 20000 kgm/s

Angle

$\theta=tan^{-1}\frac{p_n}{p_e}\\\Rightarrow \theta=tan^{-1}\frac{15000}{20000}\\\Rightarrow \theta=36.87^{\circ}$

As the two vehicles are vectors, the resultant velocity is

$(Mv)^2=p_n^2+p_e^2\\\Rightarrow v=\sqrt{\frac{p_n^2+p_e^2}{M^2}}\\\Rightarrow v=\sqrt{\frac{15000^2+20000^2}{3000^2}}\\\Rightarrow v=8.33\ m/s$

Velocity of the two vehicles when they are locked together is 8.33 m/s and direction is 36.87° North of East

5 0

3 years ago

You drive 773 miles east, then turn around and drive 773 miles west. What is your displacement, to the nearest mile?

Oliga [24]

Answer:0miles

Explanation:

Since they are moving opposite direction we say

Displacement =773-773

Displacement =0miles

8 0

3 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>

<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = <em>0.62 m/s</em> <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago