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2 years ago

What are the two factors in which weight of object depends?

Physics

2 answers:

insens350 [35]2 years ago

5 0

Answer:

An object's weight depends on its mass (the amount of matter it consists of) and the strength of the gravitational pull.

TEA [102]2 years ago

4 0

Two factors that affect the weight of an object is mass and constant gravitational force.

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A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed

oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

$v=30j$

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

$u=475i$

c. The true velocity of the jet as a vector will be represented as w:

$w=u+v$

$w=475i+30j$

d. The true speed of the jet will be calculated as:

$IwI=\sqrt{(475)^2+(30)^2}$

$IwI=\sqrt{225625+900}$

$IwI=\sqrt{226525}$

$IwI=476 mi/h$

e. The direction of the jet will be:

$tita=tan^{-1}\frac{30}{475}$

$tita=tan^{-1}(0.0632)$

$tita=3.62degrees,or,N86.38degreesS$

7 0

2 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )

Also at equilibrium the moment about the point where the string holding the bar (where ( $m_1 +m_2$ ) and $m_3$ are hanged ) is attached is zero

So basically

$(m_1 + m_2 ) * 20 = m_3 * 30$

$(0.42 + 0.47) * 20 = 30 * m_3$

Making $m_3$ subject

$m_3 = \frac{(0.42 + 0.47) * 20 }{30 }$

$m_3 = 0.59 kg$

3 0

3 years ago

you check the weather and find that the winds are coming from the west at 15 milers per hour. this information describes the win

nydimaria [60]

Answer:

Velocity

Explanation:

We finds that the winds are coming from the west at 15 miles per hour. This information shows the velocity of the wind. Since, velocity is a vector quantity. It has both magnitude and direction. 15 miles per hour shows the speed of wind and west shows the direction of wind motion.

Hence, the given information describes wind velocity.

6 0

3 years ago

I. What is the initial velocity of the car?

qaws [65]

Answer:

I. 0 m/s

II. 20 m/s

III. Part BC

Explanation:

I. Determination of the initial velocity.

From the diagram given above,

The motion of the car starts from the origin. This implies that the car start from rest and as such, the initial velocity of the car is 0 m/s

II. Determination of the maximum velocity attained.

From the diagram given above, we can see clearly that the maximum velocity is 20 m/s.

III. Determination of the part of the graph that represents zero acceleration.

It important that we know the meaning of zero acceleration.

Zero acceleration simply means the car is not accelerating. This can only be true when the car is moving with a constant velocity.

From the graph given above, the car has a constant velocity between B and C.

Therefore, part BC illustrates zero acceleration.

6 0

3 years ago

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

$d_1 = v* t$

$d_1 = 18* 0.200 = 3.6 m$

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

$v_f^2 - v_i^2 = 2 a s$

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

$0^2 - 18^2 = 2*(-3.35)(s)$

$s = 48.35 m$

So total distance after which car will stop is

$d = d_1 + d_2$

$d = 48.35 + 3.6 = 51.95 m$

So car will not stop before the intersection as it is at distance 20 m

3 0

3 years ago

What are the two factors in which weight of object depends?​

What are the two factors in which weight of object depends?