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2 years ago

A+20 N force acts on a car and at the same time, a -30 N force acts on the

Physics

1 answer:

katovenus [111]2 years ago

5 0

Answer:

-10 N, not balanced

Explanation:

Force is a vector quantity, so in order to find the net force on an object, we must use vector addition rule.

This means that if two forces acting on an object along the same line, we have to choose one direction as positive, and write the two forces with the correct sign.

In this problem, we have two forces acting along a line on the car:

- A first force of

$F_1 = +20 N$

- A second force of

$F_2=-30 N$

Therefore, the net force on the car is:

$\sum F=F_1+F_2=+20 +(-30)=-10 N$

Moreover, the net force on an object is said to be "balanced" if it is zero: in this case, it is not zero, so it is not balanced.

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A body of volume 100cc immersed completely in water contained in a jar. The weight of water and jar before immersion of the body

BlackZzzverrR [31]

<h2>Answer:</h2>

800gm

<h2>Explanation:</h2>

Archimedes principle states that when an object is immersed in a liquid there is an apparent loss of weight of the object. This apparent loss of weight is also the upthrust experienced by the liquid. The upthrust is equal to the weight of the liquid displaced.

Following from the above statement, when the body of volume 100c.c is immersed in the water contained in the jar, the upthrust experienced is equal to the weight of the water displaced.

<em>Note: In the question, weight is measured just using the mass.</em>

Mass (m) is the product of density (ρ) of liquid (which is water in this case) and volume (v) of body immersed. i.e

m = ρ x v

Where;

ρ = 1 gm/cm³

v = 100c.c = 100cm³

=> m = 1 gm/cm³ x 100cm³

=> m = 100gm

Therefore the weight of water displaced is 100gm

Now, the weight of the water and jar after immersion is the sum of the weight of water and jar before immersion, and the weight of the water displaced. i.e

Weight of water and jar after immersion = 700gm + 100gm = 800gm

8 0

2 years ago

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end when it

Tems11 [23]

Answer:

5.4 ms⁻¹

Explanation:

Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.

$L$ = length of the meter stick = 1 m

$m$ = mass of the meter stick

$w$ = angular speed of the meter stick as it hits the floor

$v$ = speed of the other end of the stick

we know that, linear speed and angular speed are related as

$v = r w\\w = \frac{v}{r}$

$h$ = height of center of mass of meter stick above the floor = $\frac{L}{2} = \frac{1}{2} = 0.5 m$

$I$ = Moment of inertia of the stick about one end

For a stick, momentof inertia about one end has the formula as

$I = \frac{mL^{2} }{3}$

Using conservation of energy

Rotational kinetic energy of the stick = gravitational potential energy

$(0.5) I w^{2} = mgh\\(0.5)(\frac{mL^{2} }{3}) (\frac{v}{L} )^{2} = mgh\\(0.5)(\frac{v^{2} }{3}) = gh\\(0.5)(\frac{v^{2} }{3}) = (9.8)(0.5)\\v = 5.4 ms^{-1}$

7 0

3 years ago

What is the value of acceleration due to gravity at heght h=4re

Norma-Jean [14]

Answer: Formula for Acceleration Due to Gravity

These two laws lead to the most useful form of the formula for calculating acceleration due to gravity: g = G*M/R^2, where g is the acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is distance.please mark as brainliest

Explanation:

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3 years ago

A mass on a spring A oscillates at twice the frequency of the same mass on spring B. Which statement is correct?A.The spring con

Nataliya [291]

Answer:

A.The spring constant for B is one quarter of the spring constant for A.

Explanation:

If spring A oscillates at twice the frequency of spring B, and period is frequency inverted. It means spring B has a period twice of spring A's.

$T_B = 2T_A$