1) 15 / 12 = 1.25 ratio
2) to increase acceleration 1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg
choose A
Answer:
It will take the plant
days or 4.44 days to grow to a height of 200 inches tall.
Explanation:
From the question, the rate at which the species of the bamboo tree grows is 36 inches per day.
To determine how long it would take a plant 40 inches tall initially to grow at this rate (that is, 36 inches per day) to a height of 200 inches.
This means we will calculate the number of days it will take the plant to grow additional 160 inches ( 200 inches - 40 inches) at this rate.
Now,
If the plant grows 36 inches in 1 day
then it will grow 160 inches in x days
x = (160 inches × 1 day) / 36 inches
x = 160 / 36
x =
days or 4.44 days
Hence, it will take the plant
days or 4.44 days to grow to a height of 200 inches tall.
False. Radio waves<span> have much longer </span>wavelengths<span> and lower frequencies </span>than<span> </span><span>visible light waves</span>
The correct graph is <u>D</u>.
The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.
The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.
The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.
The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.
Answer:
which corresponds to the second option shown: "voltage times amperage"
Explanation:
The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.
Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.
This corresponds to the second option shown in the question: "Voltage times amperage".