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Vaselesa [24]
3 years ago
8

Which conversion can take place in a transformer

Physics
1 answer:
Naddik [55]3 years ago
5 0

Answer:

D. A lower Voltage into a higher

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A stone is thrown upward at an angle. what happens to the horizontal component of its velocity as it rises? as it falls?
Ne4ueva [31]
<span>Rising or falling, it does not change.</span>
5 0
3 years ago
A 75 kg astronaut floating in space throws a 5 kg rock at 5 m/sec. How fast does the astronaut move backwards?
JulijaS [17]

Answer:

The astronaut will get a velocity 0.064ms−1 opposite to the direction of the object.

4 0
2 years ago
How far apart would you have to place the poles of a 1. 5 v battery to achieve the same electric field?
Zarrin [17]

To place the poles of a 1. 5 v battery to achieve the same electric field is 1.5×10−2 m

The potential difference is related to the electric field by:

∆V=Ed

where,

∆V is the potential difference

E is the electric field

d is the distance

what is potential difference?

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other.

We want to know the distance the detectors have to be placed in order to achieve an electric field of

E=1v/cm=100v/cm

when connected to a battery with potential difference

∆v=1.5v

Solving the equation,we find

d =  \frac{ \:Δv}{e}

=  \frac{1.5v}{100v/m}

= 1.5 \times 10 {}^{ - 2} m

learn more about potential difference from here: brainly.com/question/28166044

#SPJ4

6 0
9 months ago
A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c
satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0
2 years ago
Please help I’m stuck
Gnoma [55]

The weight of the box is <em>w</em> = <em>mg</em>, where <em>m</em> is the mass. So

<em>m</em> = <em>w</em>/<em>g</em> = (3893.40 N) / (9.80 m/s²) ≈ 397 kg

Then the box has density

(397 kg)/(4.60 m³) ≈ 86.4 kg/m³

which is less than the density of the given liquid, so the box will float.

8 0
2 years ago
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