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nalin [4]
3 years ago
7

Please hurry please

Physics
1 answer:
Ghella [55]3 years ago
4 0

Answer:

0.765m

Explanation:

gravitational potential energy GPE = mass * acceleration due to gravity * height

Given

GPE = 30Joules

Mass m = 4kg

acceleration = 9.8m/s²

Required

Height h

From the formula

h = GPE/mg

h = 30/4(9.8)

h = 30/39.2

h = 0.765m\

Hence the height of the counter is 0.765m

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A child tugs on a rope attached to a 0.62 kg toy with a horizontal force of 16.3 N. A puppy pulls the toy in the opposite direct
sattari [20]

Answer:

a=51.77\ m/s^2

Explanation:

Given that,

The mass of a toy, m = 0.62 kg

Force with which a child pull the toy = 16.3 N

The force with which the toy pulled in the opposite direction = -15.8 N

We need to find the acceleration of the toy. Let F be the net force acting on the toy. It is equal to :

F = 16.3 N - (-15.8 N)

= 32.1 N

Let a be the acceleration of the toy. Using Newton' second law of motion to find it.

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{32.1}{0.62}\\\\a=51.77\ m/s^2

So, the acceleration of the toy is 51.77\ m/s^2.

5 0
3 years ago
In lenzs law: If the current flow along the direction of deflection of the galvanometer, predict the direction of current in the
sertanlavr [38]

Answer:

this answer is only for points

Explanation:

Mark me as a brain list please

7 0
3 years ago
Which one the answer to this question
tangare [24]
The second bubble is the answer:)
4 0
3 years ago
A heat engine operates with 70.0 kcal of heat supplied and exhausts 30.0 kcal of heat. How much work did the engine do?
gladu [14]

The work done by the heat engine is 40 kCal.

The given parameters;

  • input heat of the engine, Q₁ = 70 kCal
  • output heat of the engine, Q₂ = 30 kCal

To find:

  • the work done by the heat engine

The work done by the heat engine is the change in the heat energy of the engine;

W = Q₂ - Q₁

Substitute the given parameters and solve work done (W)

W = 70 kCal - 30 kal

W = 40 kCal

Thus, the work done by the heat engine is 40 kCal.

Learn more here: brainly.com/question/4280097

4 0
3 years ago
The tools to distinguish the thickness of two objects of thickness of 1.92mm and the other with thickness of 1.93mm is ?
melomori [17]

Answer:

the instrument that gives this precision is the micrometer screw

Explanation:

The high precision measurements of small parts are the general vernier and the micrometer screw.

In these two instruments the same principle is used: there is a fixed rule and a mobile one that increases precision.

Let's analyze the absolute error or precision of each instrument

* For the vernier, the precision of the fixed rule is 1 mm and there are 20 divisions (the most common); therefore the precision of the instrument is

            Δx = 1 mm / 20

            Δx = 0.05 mm

* For the micrometer screw, the precision of the fida rule is 0.5 mm and the number of divisions is 50, therefore the precision of the screw is

            Δx = 0.5mm / 50

            Δx = 0.01 mm

consequently the instrument that gives this precision is the micrometer screw

8 0
3 years ago
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