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allochka39001 [22]
3 years ago
6

1. Is The larger the mass the greater the acceleration is on a smaller object?

Physics
1 answer:
zheka24 [161]3 years ago
7 0

Answer:No,a smaller object would have more acceleration because it would take less force to move.

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Un movil aumenta su velocidad de 10m/s a 20m/s acelerando uniformemente a razon de 5m/s2 ¿que distancia logro en dicha operacion
Kruka [31]

v₀ = initial velocity of the mobile = 10 m/s

v = final velocity of the mobile = 20 m/s

a = acceleration of the mobile = 5 m/s²

d = distance traveled during this operation = ?

Using the kinematics equation

v² = v²₀ + 2 a d

inserting the above values in the equation

20² = 10² + 2 (5) d

400 = 100 + 10 d

subtracting 100 both side

400 - 100 = 100 - 100 + 10 d

300 = 10 d

dividing both side by 10

300/10 = 10 d/10

d = 30 m

hence mobile travels 30 m.

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Energy is measured in units called what?
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They are measured in joules, calories, and kilocalories
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Un cuerpo suspendido en el aire es capaz de:
Rama09 [41]

Answer:

Explanation:

d

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What happens to the resistance of a wire as it gets wider
Alchen [17]

Answer:

it snaps

Explanation:

the more force you put on it, the wired out it gets than it snaps. I think

5 0
3 years ago
Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N
nadezda [96]
First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say F_{1}, and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
F_{1} = 7i, where i = Unit vector in the x-direction.

2) Take the second force, say F_{2}, and assume that it is making an angle \alpha with the first force F_{1}.

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

F_{2} = (15*cos \alpha)i + (15*sin \alpha )j

Now from simple vector addition, we know that,
F_{R} = F_{1} + F_{2} --- (A)

Where F_{R} = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find F_{1}+F_{2} first!
F_{1}+F_{2} =  7i+(15*cos \alpha)i + (15*sin \alpha )j

=> F_{1}+F_{2} =  (7+15*cos \alpha)i + (15*sin \alpha )j

Now the magnitude of F_{1}+F_{2} is,
| F_{1}+F_{2}| = \sqrt{ (7+ 15*cos \alpha)^{2} +  (15*sin \alpha )^{2}}

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}

Since (sin \alpha)^{2} + (cos \alpha)^{2} = 1, therefore,

=> | F_{1}+F_{2}| = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Since  | F_{1}+F_{2}| = |F_{R}|, and the magnitude of the resultant force is 20N, therefore,

 |F_{R}| = | F_{1}+F_{2}|
20 = \sqrt{ 49 + 225 + 210*(cos \alpha)}

Take square on both sides,
400 = 49 + 225 + 210*(cos \alpha)
(cos \alpha) =  \frac{3}{5}

\alpha = 53.13^{o}

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0
2 years ago
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