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vivado [14]
3 years ago
7

HALP me!! This is a physics question.

Physics
2 answers:
zepelin [54]3 years ago
5 0
Distance 6.9 m

Work done

345 kJ = 345000

Force 5 x 10 4 = 50000 N
emmasim [6.3K]3 years ago
4 0
<h2>Answer:</h2>

<u>Distance covered is 6.9 meters</u>


<h2>Explanation:</h2>

Data given:

Work Done = 345 kJ = 345000 J

Force = 5 x 10 ^ 4 =  50000 N

Distance = ?


Solution:

As we know that

Work Done = Force applied x Distance covered

By arranging the equation we get

Work / Force = Distance covered

By putting the values

345000 / 50000 = 6.9

So distance covered is 6.9 meters

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What runs all the way around the house but never moves?
amm1812
<span>A fence runs all the way around the house but never moves.</span>
6 0
3 years ago
Read 2 more answers
A truck of mass 200kg rests on an inclined plane hindered from rolling down the surface by a storing sprint whose force constant
mixas84 [53]

Answer:

1.92 J

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 200 Kg

Spring constant (K) = 10⁶ N/m

Workdone =?

Next, we shall determine the force exerted on the spring. This can be obtained as follow:

Mass (m) = 200 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = m × g

F = 200 × 9.8

F = 1960 N

Next we shall determine the extent to which the spring stretches. This can be obtained as follow:

Spring constant (K) = 10⁶ N/m

Force (F) = 1960 N

Extention (e) =?

F = Ke

1960 = 10⁶ × e

Divide both side by 10⁶

e = 1960 / 10⁶

e = 0.00196 m

Finally, we shall determine energy (Workdone) on the spring as follow:

Spring constant (K) = 10⁶ N/m

Extention (e) = 0.00196 m

Energy (E) =?

E = ½Ke²

E = ½ × 10⁶ × (0.00196)²

E = 1.92 J

Therefore, the Workdone on the spring is 1.92 J

3 0
3 years ago
An object with a mass of 20 kg has a net force of 80 N acting on it. What is the acceleration of the object?
Archy [21]

Answer:

4m/s^2

Explanation:

mass(m)=20 kg

force=80 N

acceleration (a)=?

Therefore,

Force = mass * acceleration

80 = 20*a

a=80/20

=4m/s^2

7 0
2 years ago
To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3
hodyreva [135]

Answer : The correct option is, (c) 3.7\times 10^2J/^oC

Explanation :

First we have to calculate the energy or heat.

Formula used :

E=V\times I\times t

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

E=(3.6V)\times (2.6A)\times (350s)

E=3276J

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}

where,

C = heat capacity of the calorimeter

T_{initial} = initial temperature = 20.3^oC

T_{final} = final temperature = 29.1^oC

Now put all the given values in this formula, we get:

C=\frac{3276J}{(29.1-20.3)^oC}

C=372.27J/^oC=3.7\times 10^2J/^oC

Therefore, the heat capacity of the calorimeter is, 3.7\times 10^2J/^oC

7 0
3 years ago
Need help!
tatuchka [14]
الجواب هو الأول الجواب هو الأول
5 0
3 years ago
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