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Tanya [424]
3 years ago
7

In a chemical equation, where do the products appear

Physics
1 answer:
tester [92]3 years ago
4 0

Answer:

products would appear after the raw materials

Explanation:

raw material + raw material = product (anything deriving from combining two materials)

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Six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then
Nataly_w [17]

Answer:

The option (b) is correct.

Explanation:

The expression for the power in terms of work done is as follows;

P=\frac{W}{t}

Here, W is the work done, t is the time taken and P is the power.

According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.

The expression for the equivalent resistance in the series combination is as follows;

R_{eq}=R_{1}+R_{2}......+R_{6}

Put R_{1},R_{2}......,R_{6}=R.

R_{eq}=R+R+R+R+R+R

R_{eq}=6R

It is given in the problem that a seventh resistor is added (in series).

R_{eq}=R_{1}+R_{2}......+R_{7}

Put R_{1},R_{2}......,R_{7}=R.

R_{eq}=R+R+R+R+R+R+R

R_{eq}=7R

The expression for the power in terms of voltage and resistance is as follows;

P=\frac{V^{2}}{R}

Here, R is the resistance.

From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.

If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.

Therefore, the option (b) is correct.

5 0
3 years ago
Why is the physics of hssc2 so difficult?
Ivan
Could be easy for some people and hard for some people.
8 0
3 years ago
Read 2 more answers
An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its i
zmey [24]

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

5 0
3 years ago
How much work must you do to push a 13 kg block of steel across a steel table at a steady speed of 1.0 m/s for 8.9 s ? the coeff
maria [59]
Do you have any answer choices

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5 0
3 years ago
Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b
sergeinik [125]

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched

<u>Explanation:</u>

Work, W = 2 J

Initial distance, x_{1} = 30 cm

Final distance,  = 42 cm

Force, F = 30 N

Stretched length, x = ?

We know,

W = 1/2 kΔx²

Δx = 42-30 cm = 12 cm = 0.12 m

2 J = 1/2 k X (0.12)²

k = 277.77 N/m

According to Hooke's law,

F = kx

30 N = 277.77 X x

x = 0.108 m

x = 10.8  cm

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.

7 0
3 years ago
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