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kodGreya [7K]
2 years ago
7

A body accelerate uniformly from rest at 2m/s square.Calculate its velocity after traveling 9m

Physics
1 answer:
Thepotemich [5.8K]2 years ago
5 0
The correct answer for this question is 6m/s. I hope this helps.
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What force is affected when the distance between two objects remains the same and the mass of each object is doubled?.
rusak2 [61]

Answer:

Gravitational force between masses there is Gravitational force .

3 0
3 years ago
Find the resultant force of the following forces :
bezimeni [28]

The resultant of the given forces is; 6√2 N

<h3>How to find the resultant of forces</h3>

We are given the forces as;

10 N along the x-axis which is +10 N in the x-direction

6 N along the y-axis which is +6N in the y-direction

4 N along the negative x-axis which is -4N

Thus;

Resultant force in the x-direction is; 10 - 4 = 6N

Resultant force in the y-direction is; 6N

Thus;

Total resultant force = √(6² + 6²)

Total resultant force = 6√2 N

Read more about finding resultant of a force at; brainly.com/question/14626208

4 0
2 years ago
Read 2 more answers
What is the force on a person’s hand, which is using a rope to accelerate a 5 kg block upward with an acceleration of 2.2 m/s 2
slavikrds [6]

Answer:

F = 11 N

Explanation:

Given,

Mass of a block, m = 5 kg

Acceleration of the block, a = 2.2 m/s²

We need to find the force on the person's hand. Let it is F. We know that force is given by the product of mass and acceleration as follows :

F = ma

F = 5 kg × 2.2 m/s²

F = 11 N

So, the force on a person's hand is 11 N.

5 0
3 years ago
xperiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are gi
creativ13 [48]

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)\neq0

d) 2+2+(-4)=0

e) 2+2+(-2)\neq0

f) 2+2+(-3) \neq0; 2+(-2)+3\neq0

g) 2+2+(-5)\neq0; 2+(-2)+5\neq0

h)200 + 200 +(-5)\neq0; 200+(-200)+5\neq0

6 0
3 years ago
Find the final temperature of 375 grams of tea (c = 4.184 J/g°C) if its initial temperature is 95°C just before it is placed in
Minchanka [31]

Answer:

the final temperature of the tea is 7.39⁰C.

Explanation:

Given;

mass of the tea, m = 375 g

specific heat capacity of the tea, C = 4.184 JJ/g°C

initial temperature of the tea, t₁ = 95°C

the final temperature of the tea, t₂ = ?

Energy lost by the refrigerator, Q = 137,460 J

The energy lost by the refrigerator is given by the following formula;

-Q = mc(t₂ - t₁)

-137,460 =375 x 4.184(t₂ - 95°C)

-137,460 = 1569(t₂ - 95°C)

t_2-95 = \frac{-137,460}{1569} \\\\t_2-95 = -87.61\\\\t_2 = -87.61 + 95\\\\t_2 = 7.39 \ ^0 C

Therefore, the final temperature of the tea is 7.39⁰C.

4 0
3 years ago
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