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2 years ago

3. A simple way to state Newton’s first law is:

Physics

1 answer:

MrRissso [65]2 years ago

4 0

Explanation:

A simple way to state Newton's first law is:

For every action force, there is a reaction force which is equal in magnitude and opposite in direction.

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What is the velocity of a rocket that goes 700 km north in 25 seconds?

Svet_ta [14]

Use this formula
Displacement= velocity x time
Velocity=Displacement/time
V=700/25
V=28 km/s

5 0

3 years ago

A sinusoidal traveling wave is generated on a string by an oscillating source that completes 116 cycles per minute. What is the

slava [35]

Answer:

$\lambda =6.32\ cm$

Explanation:

given,

number of cycle complete (f) = 116 cycles per minute

wavelength observed at 11 m in 1.5 m.

$v = \dfrac{distance}{time}$

$v = \dfrac{11}{1.5}$

v = 7.33 m/s

$\lambda = \dfrac{v}{f}$

$\lambda = \dfrac{7.33}{116}$

$\lambda =0.0632\ m$

$\lambda =6.32\ cm$

The wavelength of the wave is equal to $\lambda =6.32\ cm$

6 0

3 years ago

The sleigh is being pulled with a force of 800N and has a mass of 200 Kg.

kiruha [24]

The answer is 4 m/s/s

4 0

3 years ago

Read 2 more answers

A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to

jeka57 [31]

Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire $=3.7\ MA$

b) Current at the center of the circular coil $=2.48\times 10^{5}\ A$

c) Current near the center of a solenoid $236.8\ A$

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where $B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}$

$\mu=4\pi \times 10^{-7}\ \frac{henry}{m}$

Plugging the values,

Conversion $1\times 10^6 A = 1\ MA$ ,and $2cm=\frac{2}{100}=0.02\ m$

$I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA$

⇒Magnetic Field at the center due to circular coil (at center) is given by, $B=\frac{\mu\times I (N)}{2(a)}$

So $I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A$

⇒Magnetic field due to the long solenoid, $B=\mu\ nI=\mu (\frac{N}{l})I$

Then $I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A$

So the value of current are $3.7 MA$ , $2.48\times 10^{5} A$ and $236.8\ A$ respectively.

8 0

3 years ago

The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm

Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0

3 years ago