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3 years ago

Give two examples of energy transforming from type to another. Make sure to tell the

Physics

1 answer:

pentagon [3]3 years ago

4 0

Well one is from potential energy to ascensive moving energy . And another is from electrical to circuit energy.

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What is the value of x in the equation below 1+2e^x+1=9

GuDViN [60]

<h2>Answer: 1.252</h2>

Explanation:

We are given this equation and we need to find the value of $x$ :

$1+2e^x+1=9$ (1)

Firstly, we have to clear $x$ :

$2e^x=9-1-1$

$2e^x=7$

$e^x=\frac{7}{2}$ (2)

Applying Natural Logarithm on both sides of the equation (2):

$ln(e^x)=ln(\frac{7}{2})$ (3)

$xln(e)=ln(\frac{7}{2})$ (4)

According to the Natural Logarithm rules $xln(e)=x$ , so (4) can be written as:

$x=ln(\frac{7}{2})$ (5)

Finally:

$x=1.252$

3 0

3 years ago

How long would it take a drag racer to increase her speed from 10.m/s to 20 m/s if her car accelerates at a uniform rate of 15 m

garik1379 [7]

It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?

5 0

3 years ago

Hey, I need help can someone help me out, please?

yan [13]

Explanation:

6) newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

6 0

3 years ago

Read 2 more answers

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

3 years ago

A 570 kg elevator accelerates downwards at 1.5 m/s2 for the first 13 m of its motion.

jeka94

Mass of the elevator (m) = 570 Kg
Acceleration = 1.5 m/s^2
Distance (s) = 13 m
Let the force be F.
We know, F = ma,
Therefore, F = (570 × 1.5) N = 855 N
Angle between distance and force (θ) = 0°
We know, work done = F s Cos θ
Therefore, work done by the cable during this part
= (855 × 13 × Cos 0°) J
= (855 × 13 × 1) J
= 11115 J

Answer:

11115 J

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0

2 years ago