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3 years ago

Exit ticket: A lamp is plugged into a 110 Volt electrical outlet. There is a 9 Watt LED bulb in the lamp. a. What is

Physics

1 answer:

notka56 [123]3 years ago

5 0

Answer:

(a) 0.081 A (b) 1358.02 ohms

Explanation:

Voltage, V = 110 volt

Power of a LED bulb, P = 9 Watt

(a) Let the current is I. The formula for ower in terms of voltage and power is given by :

P = VI

$I=\dfrac{P}{V}\\\\I=\dfrac{9}{110}\\\\=0.081\ A$

(b) Let R is the resistance of the bulb. Using Ohm's law as follows :

V = IR

$R=\dfrac{V}{I}\\\\R=\dfrac{110}{0.081}\\\\=1358.02\ \Omega$

Hence, this is the required solution.

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Explanation:

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3 years ago

What is the total resistance in the circuit? (include unit in answer - ohms)

solmaris [256]

Answer:60 ohms

Explanation:

R1=30 ohms

R2=15 ohms

R3=15 ohms

Let the total resistance be R

R=R1 + R2 + R3

R=30 + 15 +15

R=60

Total resistance is 60 ohms

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3 years ago

one mole of water is equivalent to 18 grams of water. a glass of water has a mass of 200 g. how many moles of water is in this?

ehidna [41]

1 mole = 18 g
200 g = glass of water
200 ÷ 18 = 11.1
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2 years ago

A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s

Romashka [77]

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

p₀ = m v₀

celestines que clash case

p_f = (m + M) v

po = pf

m v₀ = (n + M) v

v = $\frac{m}{m+M}$

calculemos

v= $\frac{0.200}{0.200+M} 0.450$

v= 0.09 m/s

elastic shock case

p₀ = m v₀

p_f = m v₁ +M v₂

p₀ = p_f

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6 0

3 years ago

A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:

madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take $g = 10\; \rm m \cdot s^{-2}$ .

a. The momentum of the ball would be approximately $60\;\rm kg \cdot m \cdot s^{-1}$ two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately $\rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum $p$ of an object is equal its mass $m$ times its velocity $v$ . That is: $\vec{p} = m \cdot \vec{v}$ .

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . In other words, its velocity would become approximately $10\; \rm m \cdot s^{-1}$ more negative every second.

The initial velocity of the ball is $60\; \rm m \cdot s^{-1}$ . After two seconds, its velocity would have become $60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}$ . The momentum of the ball at that time would be around $p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}$ .

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant $g \approx -10\; \rm m \cdot s^{-2}$ . That's how the ball's velocity becomes negative.

After three more seconds, the velocity of the ball would be $0\; \rm m \cdot s^{-1} + 3\; \rm s \times \left(-10\; \rm m \cdot s^{-2}\right) = -30 \; \rm m \cdot s^{-1}$ . Accordingly, the ball's momentum at that moment would be $p = m \cdot v \approx \left(-45\; \rm kg \cdot m \cdot s^{-1}\right)$ .

3 0

3 years ago