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NeTakaya
3 years ago
15

How does a single fix pully help you do work

Physics
1 answer:
Flauer [41]3 years ago
4 0
It makes things easier because if it is broken it would take longer to fix the main problem
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the very high voltage needed to create a spark across the spark plug is produced at the a. transformer's primary winding. b. tra
Karo-lina-s [1.5K]
I think the correct answer from the choices listed above is option B. The very high voltage needed to create a spark across the spark plug is produced at the  transformer's secondary winding. <span>The secondary coil is engulfed by a powerful and changing magnetic field. This field induces a current in the coils -- a very high-voltage current.</span>
5 0
3 years ago
Read 2 more answers
A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision
Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is a=2,800m/s^{2}

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, v_{i}=-30m/s, final speed has a positive direction, v_{f}=26m/s, \Delta t=20ms=0.020s and mass m_{b}.

<u>Initial momentum</u>

p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s

<u>final momentum</u>

p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s

<u>Impulse</u>

I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s

<u>Average Force</u>

F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}

<u>Average acceleration</u>

F=ma, so a=\frac{F}{m_{b}}.

Therefore, a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}

8 0
3 years ago
What is displacement ​
omeli [17]

The word displacement implies that an object has moved, or has been displaced. Displacement is defined to be the change in position of an object.

Displacement is defined as the act of moving someone or something from one position to another or the measurement of the volume replaced by something else

8 0
4 years ago
A.Substitute in the units for each one and combine like terms.
storchak [24]
IDK     ghjfnhgfjmrmhjgfhgfmmfh
8 0
3 years ago
You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm. You have a helium-neon las
topjm [15]

Answer:

0.776 m far Pinhole should be placed before the viewing screen

Explanation:

For circular aperture of diameter D will have a bright central maximum of diameter, width is given by

w=\frac{2.44 \lambda L}{D}

where \lambda is wavelength of helium neon laser = 633 nm, D=10.cm, w=0.12 mm

Pinhole should be placed before the viewing screen is

L=\frac{wD}{2.44\lambda}\\L=\frac{0.12\times 10^{-3}\times 0 .01}{2.44\times 633 \times 10^{-9}}\\L=0.776 m

4 0
3 years ago
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