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3 years ago

. An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

Physics

1 answer:

Hoochie [10]3 years ago

5 0

The acceleration of the object is $6.7 m/s^2$

Explanation:

We can solve the problem by using Newton's second law, which states that the net force exerted on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass of the object

a is its acceleration

For the object in this problem,

F = 500 N is the applied force

m = 75 kg is the force

Solving the equation for a, we find the acceleration:

$a=\frac{F}{m}=\frac{500}{75}=6.7 m/s^2$

Learn more about Newton's second law:

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I need the answer to this question what has the student plotted on the vertical axis?

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Answer:

The correct option is D

Explanation:

In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.

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3 years ago

A 50-g chunk of 80 degrees C iron is dropped into a cavity in a very large block of ice at 0 degrees C. Show that 5.5 g of ice w

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Answer:

5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

Explanation:

The concept to solve this problem is given by Energy Transferred, the equation is given by,

$Q = mc\Delta T$

Where,

Q= Energy transferred

m = mass of water

c = specific heat capacity

$\Delta T =$ Temperature change (K or °C)

Replacing the values where mass is 50g and temperature is 80°C to 0°C we have,

$Q = mc\Delta T$

$Q = 50*0.11*(80-0)$

$Q = 440cal$

Then we can calculate the heat absorbed by m grams of ice at 0°C, then

$Q_2 = mL = 80*m$

How Q_1=Q_2, so

$80m=440$

$m=\frac{440}{80}$

$m = 5.5g$

Then 5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

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3 years ago

You are making a round trip from City A to City B and back to City A again at constant speed. At what point in the trip is your

MA_775_DIABLO [31]

Answer:

Halfway between B and A on the return leg.

Explanation:

Your average SPEED for the entire trip will equal your constant speed as the time and distance increase at proportionate rates.

Your average VELOCITY will equal your constant speed while you travel from A to B because time and displacement are increasing at proportionate rates.

When you turn around at B to return, your Displacement is now decreasing while your travel time continues to increase, so your average velocity decreases.

Lets say the distance from A to B is 90 km and your constant speed is 30 km/hr.

your average speed is 30 km/hr because you took 6 hrs to travel 180 km

We want to find your position when your average velocity is 30/3 = 10 km/hr

it took 3 hrs to go 90 km from A to B. Let t be the time lapsed since turn around

your displacement is given by d = 90 - 30(t)

and your total time of travel is t + 3 hrs

v = d/t

10 = (90 - 30t) / (t + 3)

10(t + 3) = (90 - 30t)

10t + 30 = 90 - 30t

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2 years ago

The movement of an object at a constant speed around a circular radius is known as

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Answer:

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3 years ago

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