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Shalnov [3]
3 years ago
10

What is an electromagnet ​

Physics
1 answer:
damaskus [11]3 years ago
6 0

Answer:

An electromagnet is a magnet that uses electricity.

Explanation:

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A star with a large luminosity would have a relatively _____ absolute magnitude.
Mademuasel [1]
A star with large luminosity would have a relatively low absolute magnitude. Absolute magnitude is a number that tells how bright a star is from the Earth. However, this scale is backwards and logarithmic, so having a large absolute magnitude value means that the star is faint.
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Put on the ground a shrimp that has just been taken out of water.Now touch the shrimp from a distance by a stick.The shrimp will
Assoli18 [71]

Answer: Yes.

Explanation:

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2 years ago
Gaining neutrons makes an element?
Kobotan [32]

Answer:

C. Have no change in electrical charge

Explanation:

If a element gains neutron it become an Isotope. The electrical charge do not change for this, only the atomic mass changes when an element gains neutrons.

The electrical charge is affected when there is a variation in the number of electrons or protons in the element.

6 0
3 years ago
Let T be a linear transformation from a vector space V with dimension 11 onto a vector space W with dimension 7. What is the dim
erica [24]

Answer:

The dimension of the nullspace of T = 4

Explanation:

The rank/dimension theorem is explains that:

Suppose V and W are vector spaces over F, and T:V → W is linear. If V is finite dimensional, then

nullity(T) + rank(T) = dim(V).

rank(T) = dimension of T = dim(T) = dim(W) = 7

nullity(T) = dimension of the nullspace of T = dim(T) = ?

dim(V) = 11

nullity(T) = dim(V) - dim(T) = 11 - 7

nullity(T) = 4.

8 0
3 years ago
Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same
Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

\underset{V_{A}}{\rightarrow} = velocity of car A before collision = 0 i - V_{A} j

\underset{V_{B}}{\rightarrow} = velocity of car B before collision = V_{B} i + 0 j

\underset{V_{AB}}{\rightarrow} = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

M_{A} = mass of car A = 1750 kg

M_{B} = mass of car B = 1450 kg

Using conservation of momentum

M_{A}  \underset{V_{A}}{\rightarrow} + M_{B}  \underset{V_{B}}{\rightarrow} = (M_{A} + M_{B}) ( \underset{V_{AB}}{\rightarrow} )

(1750) (0 i - V_{A} j) + (1450) (V_{B} i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) V_{B} i - (1750) V_{A} j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) V_{B} = 97600    and - (1750) V_{A} = - 60160

V_{B} = 67.3 km/h        and  V_{A} = 34.4 km/



6 0
3 years ago
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