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topjm [15]
3 years ago
8

Which determines the reactivity of an alkali metal? A. its boiling and melting points B. the shininess of its surface C. the num

ber of protons it has D. its ability to lose electrons
Physics
2 answers:
Naddik [55]3 years ago
7 0

Answer:

D

Ex plane

I just took the test

Gekata [30.6K]3 years ago
4 0

Answer:

D

Please let me know if this helped! Please rate it the brainlist if possible, thanks

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3 hours into minutes
otez555 [7]

Answer:

180

60 minutes in a hour

60x3 is 180minutes

6 0
2 years ago
A 4,000 kg truck is moving at +10 m/s hits a 1500 kg parked car which moves off at +10 m/s. What is the velocity of the truck
Igoryamba

Answer:

10m/s

Explanation:

Using the law of conservation of momentums

M1u1+m2u2 = (m1+m2)v

Substitute.

4000(10)+1500(10) = (4000+1500)v

40,000+15,000 = 5,500v

55000 = 5500v

v = 55000/5500

v= 10m/s

Hence the velocity of the truck after Collision is 10m/s

6 0
3 years ago
In Thomson experiment, why was the glowing beam repelled by a negatively charged plate
algol13
<span>In Thomson experiment, why was the glowing beam repelled by a negatively charged plate, because the glowing beam was negatively charged. The glowing beam particles were attracted to the positive plate.

</span><span>J.JThomson proved that the cathode rays produced a stream of negatively charged particles called electrons. </span>
6 0
2 years ago
Read 2 more answers
What property of matter do we consider when deciding whether to buy a half-gallon of milk at the store versus a gallon of milk?
Mrac [35]
Liquified matter maybe.
7 0
2 years ago
In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie
Evgesh-ka [11]

Answer:

8.4 V

Explanation:

induced emf, e1 = 5.8 V

Magnetic field, B1 = 0.38 T

magnetic field, B2 = 0.55 T

induced emf, e2 = ?

As we know that the induced emf is directly proportional to the magnetic field strength.

When the other parameters remains constant then

\frac{e_{1}}{e_{2}}=\frac{B_{1}}{B_{2}}

\frac{5.8}{e_{2}}=\frac{0.38}{0.55}

e2 = 8.4 V

Thus, the induced emf is 8.4 V.

4 0
2 years ago
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