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umka2103 [35]
2 years ago
13

Choice are 1.3 1.0 17​

Physics
1 answer:
viva [34]2 years ago
8 0

The current flowing in resistor 1 in the given series circuit is determined as 1.0 A.

<h3>Equivalent resistance of the circuit</h3>

The equivalent resistance of the circuit is determined as follows;

Rt = R1 + R2

Rt = 15 + 5

Rt = 20ohms

<h3>Current flowing in the circuit</h3>

The current flowing in the circuit is determined from Ohm's law.

V = IR

I = V/R

I = 20/20

I = 1.0 A

since it is series circuit, same current will flow in R1 and R2.

Thus, the current flowing in resistor 1 in the given series circuit is determined as 1.0 A.

Learn more about current here: brainly.com/question/24858512

#SPJ1

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A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at In a few mi
Nataliya [291]

Answer:

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

m_{m} = mass of metal = 400 g

c_{m} = specific heat of metal = ?

T_{mi} = initial temperature of metal = 100 °C

m_{a} = mass of aluminum cup = 100 g

c_{a} = specific heat of aluminum cup = 900.0 J/kg ∙ K

T_{ai} = initial temperature of aluminum cup = 15 °C

m_{w} = mass of water = 500 g

c_{w} = specific heat of water = 4186 J/kg ∙ K

T_{wi} = initial temperature of water = 15 °C

T = Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

m_{m} c_{m} (T_{mi} - T) = m_{a} c_{a} (T - T_{ai}) + m_{w} c_{w} (T - T_{wi} ) \\(400) (100 - 40) c_{m} = (100) (900) (40- 15) + (500) (4186) (40 - 15)\\ c_{m} = 2274 Jkg^{-1}K^{-1}

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3 years ago
Two tiny beads are 25 cm apart with no other charges or fields present. Bead A carries 10 µC of charge and bead B carries 1 µC.
steposvetlana [31]
The electric forces on both beads are equal.
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3 years ago
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Friction always acts____ and in the ____ direction of the motion of an object ​
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Answer:

Friction force always acts tangent to the surface at points of contact. Friction force acts opposite to the direction of motion. There are 2 types of friction: Static friction: If the two surfaces in contact do not move relative to each other, one has static friction.

5 0
2 years ago
One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what
Inessa [10]

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

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F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

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F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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Answer:

Explanation:

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