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Rashid [163]
2 years ago
10

What force (in N) must be applied to a 500.0 kg crate on a frictionless plane inclined at 30° to cause an acceleration of 5.0 m/

s^2 up the plane? (Enter the magnitude.)
Physics
1 answer:
Goryan [66]2 years ago
5 0

The force must be applied to the crate on a frictionless plane to cause the acceleration is is 4,243.4 N

The given parameters;

  • <em>mass of the crate, m = 500 kg</em>
  • <em>inclination of the angle, θ = 30⁰</em>
  • <em>acceleration of the crate, a = 5 m/s²</em>

The force that must be applied to the crate to cause the acceleration is calculated as follows;

F = ma

∑Fₓ = 0

mgcosθ - F_k = F

where;

  • <em>F</em>_k<em> is the frictional force on crate </em>

<em />

mgcosθ - 0 = F

mgcosθ = F

500 x 9.8 x cos(30) = F

4,243.4 N

Thus, the force must be applied to the crate on a frictionless plane to cause the acceleration is is 4,243.4 N.

Learn more here:brainly.com/question/19887955

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In this photograph, a soccer player is about to kick the ball. Use the situation to explain that when two objects interact, the
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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:

<u>Step 1:</u> 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

<u>Step 2:</u> 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

<u>Step 3: </u>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}

<u>Step 4:</u> 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,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
3 years ago
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