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Pepsi [2]
3 years ago
12

I need help Mr or ms tutor​

Physics
1 answer:
Ronch [10]3 years ago
7 0

Explanation:

Height is the x-axis, and gravitational potential energy is the y-axis.  As the height increases, the gravitational potential energy increases linearly.

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When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot
gayaneshka [121]

Answer:

The value is E =  1.35 *10^{14} \ J

Explanation:

From the question we are told that

    The mass of matter converted to energy on first test is  m  =  1 \  g  = 0.001 \  kg

    The mass of matter converted to energy on second test m_1 =  1.5 \  g = 1.5 *10^{-3} \ kg

    Generally the amount of energy that was released by  the explosion is  mathematically  represented as  

         E =  m * c^2

=>       E =  1.5 *10^{-3}  * [ 3.0 *10^{8}]^2

=>       E =  1.35 *10^{14} \ J

7 0
3 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
A particle moving on a plane curve what is the degree of freedom
kirill115 [55]

A particle confined to move along a curved path has only one degree of freedom. inclined plane are some examples of constrained motion. Every condition of constraint reduces the number of degree of freedom by one.

I hope this helps!

7 0
3 years ago
PLEASE HELP!!
Anarel [89]

Answer:

C. Bacteria

Explanation:

Bacteria and archaea are the two domains which consist of organisms with prokaryotic cells. Prokaryotic cells are cells without nuclei or membrane-bound organelles.

8 0
3 years ago
Read 2 more answers
How to use kinetic energy to calculate velocity
BartSMP [9]

As we know the formula of kinetic energy is

KE = \frac{1}{2} mv^2

here given that

KE = 150,000 J

mass = 120 kg

we can use this to find speed

150,000 = \frac{1}{2} * 120 * v^2

v^2 = 2500

v = 50 m/s

So speed of above object is 50 m/s

7 0
3 years ago
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