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Alex787 [66]
3 years ago
8

An object was released from rest at height of 1.65 m with respect to ground. Determine the time it takes the object to reach the

ground. (g = 9.80 m/s
Physics
1 answer:
vivado [14]3 years ago
8 0

Answer:

The time taken by the object to reach the ground is 0.58 seconds.

Explanation:

Given that,

An object was released from rest at height of 1.65 m with respect to ground. We need to find the time taken by the object to reach the ground. Initial speed of the object is 0 as it is at rest. It will move downward under the action of gravity such that, the distance covered by the object is given by :

d=ut+\dfrac{1}{2}gt^2

d=\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2d}{g}}

t=\sqrt{\dfrac{2\times 1.65}{9.8}}

t = 0.58 seconds

So, the time taken by the object to reach the ground is 0.58 seconds. Hence, this is the required solution.

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3. A certain horizontal east-west lined wire has a mass of 0.2kg per meter of length and carries a current I. Impressed on the w
erica [24]

Answer:

i = 4.9 A

Explanation:

The expression for the magnetic force in a wire carrying a current is

          F = i L x B

bold letters indicate vectors.

The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition

            F - W = 0

            i L B = m g

They indicate the linear density of the cable λ = 0.2 kg / m

           λ = m / L

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we substitute

           i B = λ g

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let's calculate

          i = 0.2 9.8 / 0.4

          i = 4.9 A

3 0
3 years ago
A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

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