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3 years ago

When is the net force on an object equal to zero?

2 answers:

4 0

There are all kinds of conditions that could bring that about.

1. The most common one is friction. An object is moving that it is not accelerating. That means the frictional force is equal to the force causing motion.

2. The downward gravitational force is being negated by a rope suspending an object. The upward pull by the rope cancels the downward pull of gravity.

3. The rules of the game prohibit motion. A ballplayer would be stupid to leave second base while the pitcher has the ball in his hand (a physics joke. Please ignore).

4. A normal force upwards by the earth's surface is balanced by the weight of an object laying on the ground.

5. Speaking of baseball, there is no acceleration in the horizontal direction, so the net force on the ball horizontally is 0.

6. An elastic band held and aimed horizontally. The force of the elastic to launch a spitball vs the hand holding the elastic back waiting for the best time to let the spitball go.

That should get you going.

Rufina [12.5K]3 years ago

3 0

When ALL the forces are equal, or in general when <span>forces in the opposite direction are equal

so we can have newton example:

10 newtons X -10 newtons

netforce = 0 newtons</span>

You might be interested in

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

The mass of the block is $m_b = 1.45 kg$

The spring constant is $k = 860 N/m$

The coefficient of static friction is $\mu = 0.36$

For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

$F_y = m_b*g$

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

$F_x = k *x * \mu$

where x is the minimum compression to keep the block from slipping

Now equating this two formulas and making x the subject

$x = \frac{m_b * g}{k * \mu}$

substituting values we have

$x = \frac{1.45 * 9.8}{860 *0.36}$

$x= 0.046m$

3 0

3 years ago

A 60-watt light bulb has a voltage of 120 volts applied across it and a current of 0.5 amperes flows through the bulb. What is t

Naddik [55]

Answer: 240ohms

Explanation: P=IV

P=60watts

I=0.5A

V=120volts

From ohms law

V=IR

R=V/I

R=120/0.5

R= 240ohms

7 0

3 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

3 0

3 years ago

What is the direction of a vector with an x component of -12 m and a y component of -15 m?

dusya [7]

Answer:

$51.34^{\circ}$

Explanation:

Given that,

x component of a vector = -12 m

The y component of a vector = -15 m

We need to find the direction of a vector. The direction of a vector is given by :

$\tan\theta=\dfrac{y}{x}$

Put all the values,

$\tan\theta=\dfrac{-15}{-12}\\\\\theta=\tan^{-1}(\dfrac{-15}{-12})\\\\\theta=51.34^{\circ}$

So, the direction of vector is $51.34^{\circ}$ to x component.

3 0

3 years ago

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

7.5

×m

...........................(

)

in Above equation in x and t. Separating the variables and integrating,

∫

/7.5

×=

∫

−

4.7619

Here C =constant of integration.

0 at t

, we get: C

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

−

4.7619

(

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

−

4.7619

= −

3.7619

or x

7.15

(b) To find the jogger's acceleration in m

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

7.5×

Now differentiating above equation w.r.t time we get:

−

0.675

the joggers acceleration is :

−

0.675

−

4.34

6 in equation (2). We get:

−

4.7619

(

−

(

0.04

6 )

10=

−

4.7619

0.832

6 0

3 years ago