You can increase the gravitational potential between a car and the road by

A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho

fiasKO [112]

Answer:

$v_f=8.17\frac{m}{s}$

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

$W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m$

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

$W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$

The box is initially at rest, so $v_i=0$ . Solving for $v_f$ :

$v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}$

5 0

2 years ago

How can the density of irregularly shaped objects can be calculated?

Aleks04 [339]

1.Use the balance to find the mass of the object. Record the value on the "Density Data Chart."
2.Pour water into a graduated cylinder up to an easily-read value, such as 50 milliliters and record the number.
3.Drop the object into the cylinder and record the new value in millimeters.
4.The difference between the two numbers is the object's volume. Remember that 1 milliliter is equal to 1 cubic centimeter. Record the volume on the data chart.
5.Compute the density of the object by dividing the mass value by the volume value. Record the density on the data chart.

6 0

2 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

ELEN [110]

A person who is not wearing a seatbelt during a collision will be thrown forward because it maintains forward motion

<h3>Further explanation </h3>

In Newton's law, it is stated that if the resultant force acting on an object of magnitude is zero, it can be formulated :

$\large{\boxed{\bold{\Sigma F = 0}}}$

then the object tends to defend itself from its state. So for objects in a state of movement, objects tend to move forever. Likewise, for objects in a state of rest, they tend to remain forever. The tendency of objects like this is called<em> inertia </em>

The size of inertia is proportional to mass, the greater the mass of the object, the greater the inertia of the object.

In objects with mass m that move translatively, the object will maintain its linear velocity

When we are in a vehicle that moves forward, then we will still maintain a state of forwarding motion. If our vehicle stops suddenly, then we keep moving forward so we will be pushed forward. From this point, the use of a safety belt serves to hold back our movements so that there are no fatal accidents or collisions.

<h3>Learn more </h3>

Newton's law of inertia

brainly.com/question/1412777

example of Newton's First Law of inertia

brainly.com/question/1090504

law of motion

brainly.com/question/75210

Keywords: inertia, Newton's First Law

7 0

2 years ago

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Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv

deff fn [24]

Answer:

electric field E = - k Q (1 /r(r-a)), force F = - k Q qo / r (r-a) and force for r>>a F ≈ - k Q qo / r²

Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

E = k ∫dp /r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C², r is the distance between the load distribution and the test charge, in this case everything is on the X axis.

We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

λ = q / x

As it is constant, we can write it based on differentials

λ = dq / dx

dq = λ dx

We already have all the terms, let's integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

E = k ∫ λ dx / x²

E = k la (- 1 / x)

Let's get the negative sign from the parentheses

E = - k λ (1 / x)

E = - k λ (1 /(r-a) -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

E = - k Q/a [a / r (r-a)]

E = - k Q (1 /r(r-a))

b) We calculate the force.

F = E qo

F = - k Q qo / r (r-a)

c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

F = - k Qqo / r² (1 - a/r)

F ≈ - k Q qo / r²

6 0

3 years ago

eXCUSE ME!!! EXTRA POINTS PLUS BRAINLIEST!!! SUPER DESPERATE HERE!!! The position of an object does not move relative to a refer

Sedbober [7]

It depends whether or not the object in moving relative or not to the reference point, for instance imagine a void where there were no dimensions, but there was an object. You can’t tell if the object is moving or not without a reference point.

5 0

3 years ago

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