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marusya05 [52]
3 years ago
11

Please help urgent i need help easy queston

Physics
1 answer:
mihalych1998 [28]3 years ago
3 0
I believe it’s B but i’m not sure
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How much heat is required to raise the temperature of 0.25 kg of water from 20°C to 30°C
pentagon [3]

Answer:

10500 J/kg/*C

Explanation:

Quantity of heat required=mass of substance x specific heat capacity x change in temperature

Quantity of heat required=0.25 x 4200 x [30-20]

Quantity of heat required=0.25 x 4200 x 10

Quantity of heat required=10500 J/kg/*C

3 0
3 years ago
A charged particle is surrounded by an electric field and a magnetic field
zlopas [31]

Answer:

yeah

Explanation:

electric fields help the charged particles interact

and isn't magnetic field the same as electric field.

7 0
3 years ago
Read 2 more answers
_____ replacement involves one element replacing another element in a compound.
Scorpion4ik [409]

Answer:

Single replacement

Explanation:

A single replacement reaction is a reaction in which the less reactive elements are replaced by the more reactive element in a compound. This is so because the more reactive element are higher in the electrochemical series element than the less reactive elements

3 0
3 years ago
Earth Space Science:
Leya [2.2K]
The answer is C. Chemistry because it is fact based and has the characteristics of a real science.
8 0
3 years ago
A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

Mass of object = 2.00 kg

Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

W=F\cdot d

W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0
3 years ago
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