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3 years ago

An object on earth weighs 120 N. What is it’s mass ?

Physics

2 answers:

inna [77]3 years ago

7 0

Weight = (mass) x (gravity)

120 N = (mass) x (9.8 m/s²)

Mass = (120 N) / (9.8 m/s²)

Mass = 12.24 kg (B)

Ainat [17]3 years ago

4 0

Answer:

12.24 kg would be your answer

Explanation:

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Write some interesting facts about isaac newton

Hatshy [7]

Sir Isaac Newton was an English Mathematician,physicist,astronomer,theologian and author who is widely recognized as one of the most influential scientist of all time because of his discovery of gravity(force that attracts a body toward center of earth).

8 0

3 years ago

Multiple questions onlyy!! Please help!! And if u can answer the response questions that would be helpful!!

Iteru [2.4K]

(6) Given that,

Force in east direction = 67 N

Force in west direction = 82 N

Solution,

According to the given figure,

Two forces are acting in opposite direction. Maximum force is acting in west direction. It means the net force is acting in west direction.

Net force = 82-67

= 15 N (Due west)

The forces are unbalanced. It is because the magnitude of force in west is more than in east.

Net force = 15 N, Direction of motion = West and the forces are unbalanced.

4 0

3 years ago

Calculate the pressure of water in a well if the depth of the water is 10m

Kryger [21]

Answer:

P= phg

so:- the height is 10m

density of water 1000 kg/m3

gravity is 9.8m/s2

P=1000 kg/m3*10m*9.8m/s2

=98000Pa

=98KPa

4 0

2 years ago

How do I do this problem?

Aleks04 [339]

Use Charles Law: V1/T1 = V2/T2

0.30 m^3/27 C = V2/127 C

27V2 = 127 * 0.3

V2= 38.1/27 = 1.4 m^3

5 0

3 years ago

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

3 years ago