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velikii [3]
3 years ago
9

Which safety precaution is of primary interest to ALL laboratory activities?

Physics
1 answer:
mario62 [17]3 years ago
8 0
There’s nothing here to choose from so I’m thinking goggles
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A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of
ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= \frac{\mu_0}{4\pi} \frac{2i}{r}

i is current , r is distance from wire

B =  10⁻⁷ x \frac{2\times49}{r}

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

=  10⁻⁷ x \frac{2\times49}{r} x 33 x 1

= 3234 x \frac{10^{-7}}{r}

This should balance weight of second wire per unit length

3234 x \frac{10^{-7}}{r} = .075

r = \frac{3234}{.075} x 10⁻⁷

= .0043 m

= .43 cm .

5 0
3 years ago
what will the stopping distance be a a 3000-kg car if -3000N of force are applied when the car is traveling 10 m/s
Inga [223]

Answer:

50 m

Explanation:

Acceleration= force/mass

3000/3000=1m/s^-2

Applying equation of motion:

V^2=U^2+2as; V is final velocity, u is initial velocity, a is acceleration and s is the distance covered.

0=10^2 -2*1s;

Solve for s

5 0
3 years ago
An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in
asambeis [7]

Answer:

E=(-17.08 i +7.2 j -7.6 k )N/C

Explanation:

v= (19.0j+18.k)km/s

a=3.0x10^{13}m/s^2

\beta= 400x10{-6}T

Electron information needed to solve the question:

m_e=9.11x10^{-31}kg

q=-1.6x10{-19}C

F=F_E+F_B=q*(E+Vx \beta)

F=m*a

m*a=q*(E+Vx\beta)

E=\frac{m*a}{q}-(Vx\beta )

E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]

E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C

E=(-17.08 i +7.2 j -7.6 k )N/C

6 0
3 years ago
How long does it take a 750-w coffeepot to bring to a boil 0.75 l of water initially at 11°c? assume that the part of the pot wh
Evgen [1.6K]
Just add all of them up and there is your answer I just added it but  I want u to work it out to..


4 0
3 years ago
What is the acceleration of an object with a constant velocity
valina [46]

Zero.

Acceleration is defined as the change in velocity over time.

Since in your case there is no change, there is no acceleration, so it is zero:

Or in formula: <span>a=<span><span>Δv</span>t</span></span>

Where a=acceleration, <span>Δv</span>=change in velocity and t=time

6 0
3 years ago
Read 2 more answers
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