1.8 × 103 seconds. During operation, the dryer
draws a current of
(1) 10. A (3) 9.0 × 102 A
(2) 15 A (4) 3.3 × 103 A
Answer : The dryer draws a current of 15 A.
Explanation :
Given that,
Energy consumed,
Voltage, V = 220 Volts
Time,
Power dissipated is given by :
Also, P = VI
So,
So, the correct option is (2)
Hence, this is the required solution.
The dryer draws a current of 15 A.Electric currents is the ratio of charge per unit of time.
The pace at which electrons travel through a conductor is known as electric current. The ampere is the SI unit for electric current.
Electrons are tiny particles that reside within a substance's molecular structure.
The given data in the problem is;
V is the voltage= 220 volts
E is the electric energy= 6.0 × 10⁶
t is the time =1.8 × 10³ seconds.
I is the electric current=?
The electric current is found as;
Hence the dryer draws a current of 15 A.Electric currents is the ratio of charge per unit of time.
To learn more about the electric current refer to the link;
b. twice the voltage and the same current will flow through each.
c. the same voltage and different current will flow through each.
d. twice the voltage and different currents will flwo through each.
Answer: option B
Explanation: each cell is suppose to have an emf and an internal resistance.
Let E1 be the emf of the first battery and E2 be the emf of the second battery.
Et is the total emf.
Going by the connection described above, it is a series arrangement.
Hence the total emf is calculated as sum of individual emf
Et = E1 + E2
But each cell are identical, hence E1 = E2 = E
Et = E + E = 2E.
Hence the total voltage is twice the voltage.
Since the connection is a series arrangement, the current passing through each cell is the same.
These points have made option B the correct answer.
Answer:
b) Twice the voltage and the same current
Explanation:
The two batteries are identical, and are connected in series. When elements in a circuit are connected in series, the same current flow through them but the voltage drops are additive.
Since the two batteries are identical, the possess the same voltage of magnitude V.
Therefore, the total voltage,
b.) Find the density of the rock dropped into the graduated cylinder.
The volume of the rock dropped into the graduated cylinder is 10 ml. The density of the rock is 2.3 g/ml.
We first determine the volume of the rock dropped into the graduated cylinder. The initial volume of water in the cylinder is 20 ml, and after placing the rock, the total volume increases to 30 ml. Therefore, the volume of the rock is the total volume minus the initial volume of the water, which is 30 ml - 20 ml = 10 ml.
Next, we find the density of the rock. The formula for density is mass/volume. Given that the mass of the rock is 23 grams and the volume is 10 ml, the density of the rock would be mass/volume = 23 grams / 10 ml = 2.3 g/ml.
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Answer:NOPE you need more
Explanation:
The power required to lift a weight depends on the force needed and the speed at which the weight is lifted. The force required to lift a weight is given by the equation F = m * g, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth).
For example, if a 1000 kg weight is lifted 10 m in 10 seconds, the work done can be calculated as W = (1000 kg) * (9.81 m/s²) * (10 m) = 98100 J (Joules). The power required is work done per unit time P = (98100 J) / (10 s) = 9810 W (Watts), which is approximately 9.8 kW¹.
In terms of horsepower, since 1 horsepower is approximately equal to 746 Watts¹, the power required would be about 13.15 horsepower. So, it does not require 100 horsepower to lift a 1000 kg weight from the ground under these conditions. However, these calculations assume ideal conditions and do not take into account factors such as air resistance or mechanical inefficiencies. In real-world applications, more power might be needed.