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Allushta [10]
1 year ago
15

could you help with question 5. Your solutions to the word problems in volving Newton's Laws should have the following features:

Each solution should begin with one or more free body diagrams with a sign convention.The solution process should follow the GRASP process.Formulae should always be used in a three-step process: formula with letter variables, substitution of values, and then the answer.Answers should be expressed with the correct number of significant digits and with the direction if required.

Physics
1 answer:
Ganezh [65]1 year ago
3 0

ANSWER:

(a) 20 m/s^2

(b) 250 m

STEP-BY-STEP EXPLANATION:

Given:

Mass (m) = 250 g = 0.25 kg

Force (F) = 20 N

Frictional force (Ff) = 15 N

(a)

The net force on the block is:

\begin{gathered} F_{net}=F-F_f \\ F_{net}=20-15 \\ F_{net}=5\text{ N} \end{gathered}

We know that:

\begin{gathered} F=m\cdot a \\ \text{ We replacing} \\ 5=0.25\cdot a \\ a=\frac{5}{0.25} \\ a=20\text{ m/s}^2 \end{gathered}

(b)

We can calculate the partial distance from the following equation:

initial velocity (u) = 0 (start from rest)

\begin{gathered} s=ut+\frac{1}{2}at^2 \\ \text{ we replacing} \\ s=0\cdot5+\frac{1}{2}\cdot20\cdot5^2 \\ s=250\text{ m} \end{gathered}

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The process of sediments being compacted and cemented to form sedimentary rocks is called
crimeas [40]

Answer:

Lithification is the answer.

8 0
2 years ago
A 1250 kg car, driving 7.39 m/s, runs into the back of a stationary 5380 kg truck. After the collision, the truck moves forward
Leno4ka [110]

Explanation:

Given that,

Mass of the car, m₁ = 1250 kg

Initial speed of the car, u₁ = 7.39 m/s

Mass of the truck, m₂ = 5380 kg

It is stationary, u₂ = 0

Final speed of the truck, v₂ = 2.3 m/s

Let v₁ is the final velocity of the car. Using the conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

1250\times 7.39+5380\times 0=1250\times v_1+5380\times 2.3

v_1=-2.5\ m/s

So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.

3 0
3 years ago
What is a distraction that a pedestrian may engage in while crossing the street?
goldenfox [79]

Answer:

a dog walking or their phone rings or heard a neighbor talking to them

6 0
3 years ago
A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

T∞ be the temperature of the room = 70°F

T₀ be the initial temperature of the thermometer = 212°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

- kt = In (55/142) = In (0.3873)

- k(2) = - 0.9485

k = 0.4742 /min

At time t = 4 mins

kt = 0.4742 × 4 = 1.897

(T - 70) = 142 e⁻ᵏᵗ

e^(-1.897) = 0.15

T - 70 = 142 × 0.15 = 21.3

T = 91.3°F

7 0
3 years ago
Why is velocity and not just speed important to these pilots
Alborosie
Velocity is a vector. Therefore, it depends on the direction. Pilots need to know the direction of wind, not just the speed. If the pilot is going South, and there's 5 mph wind going South, they'll be happy, but if the wind is going 5 mph North, they'll be going against the wind.
7 0
3 years ago
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