Answer:
(1) P = 100 kg-m/s (2) M = 40 kg.
Explanation:
We need to solve the missing variable :
1. P=?, M=20 kg, V=5 m/s
If M is mass, V is velocity, then momentum P is given by :
P = MV
P = 20 kg × 5 m/s
P = 100 kg-m/s
(2) P= 3000 kg(m/s), M=?, V=75 m/s
Hence, P = 100 kg-m/s and M = 40 kg.
In order to answer this, we will set up a simple ratio as such:
1 calorie = 4.184 joules
1 kilocalorie = 1000 calories
1 kilocalorie = 4,184 joules
250 kilocalories = x joules
Cross multiplying the second and third equations, we get:
x joules = 4,184 * 250
250 kilocalories are equivalent to 1,046 kJ
Answer:
The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q.
Explanation:
Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge on the test charge.
I’m pretty sure one increases just straight up velocity and the other is kind of a deceleration
The stereo uses an energy of
in a time
, therefore the power of the stereo is given by
We also know that the power of an electrical device is related to its voltage, V, and its resistance, R, by the following equation
therefore, we can rearrange the equation to calculate the resistance of the stereo:
