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slavikrds [6]
2 years ago
5

0.3050 has significant figures ?

Physics
1 answer:
marshall27 [118]2 years ago
6 0

Answer:

NO

Explanation:

because it just doesn't.

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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
2 years ago
1. Describe what happens to the gravitational when you increase mass versus when you
Rudik [331]

Answer: if the mass is doubled, the force of gravity is doubled, meaning it decreases. If the distance is doubled, the force of gravity is 1/4 as strong as before

Explanation:

7 0
3 years ago
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How is the radiating electric field (or electromagnetic signal) produced when radio stations broadcast
Natali5045456 [20]

Answer:

Radio stations have dipole type antennas

this field increases in intensity and propagates outwards,

Explanation:

Radio stations have dipole type antennas, that is, all sides are isolated from each other, when the AC signal from the radio station arrives, the lcharge begins at times and by the Lens law a field appears that opposes this movement, this field increases in intensity and propagates outwards, when the voltage reaches a maximum, the generated wave also reaches the maximum, now the incident wave begins to decrease, an electric hand appears to oppose this prisoner, and in this way a cap is created. electric .

5 0
3 years ago
The type of bond formed by shared electrons is covalent, ionic, or metallic
loris [4]
I believe the answer in Covalent Bond.
7 0
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A force acting on a body causes a change in the momentum of the from 12 kgms-1 to 16 kgms-1 in 0.2 s. calculate the magnitude of
Nonamiya [84]

Answer: Impulse = 4 kgm/s

Explanation:

From the question, you're given the following parameters:

Momentum P1 = 12 kgm/s

Momentum P2 = 16 kgm/s

Time t = 0.2 s

According to second law of motion,

Force F = change in momentum ÷ time

That is

F = (P2 - P1)/t

Cross multiply

Ft = P2 - P1

Where Ft = impulse

Substitute P1 and P2 into the formula

Impulse = 16 - 12 = 4 kgm/s

The magnitude of the impulse is therefore 4 kgm/s.

6 0
3 years ago
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