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3 years ago

Help me PLEASE! 25 points!!!!! :)

Physics

1 answer:

NeTakaya3 years ago

7 0

Answer:

Explanation:

All the arrows align to the point right

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An object with a mass of 7.60 kg is moving to the right and experiences an applied force of 50 N to the right. The friction forc

horrorfan [7]

Answer:

Explanation:

We will use the equation F - f = ma, which is a fancy way of stating Newton's 2nd Law.

F = +50.0,

f = -30.0,

m = 7.60 kg. Therefore:

50.0 - 30.0 = 7.60a and

20.0 = 7.60a and

a = 20.0/7.60 so

a = 2.63 m/s/s to the right

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3 years ago

Identify the smallest unit of an element

marysya [2.9K]

atom ................................................... sorry for the periods

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3 years ago

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A car has a mass of 1000 kg. What is the acceleration produced by a force of 2000 N?

EastWind [94]

F=ma
a=F/m
a=2000/1000
a=2 m/s^2

6 0

2 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

PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST The actual subject is Science but they dont have that as a option in pick

Sophie [7]

Explanation:

speed : • how fast an object changes position

• miles per hour.

• distance/time.

velocity: • speed in a direction

• miles per hour North

• distance/ time in a direction

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3 years ago