How much force is required to accelerate a 50 kg mass at 2 m/s

You are crouched at the "start" line. Bang! The race begins and 7 seconds later you end up 60 yards down the track. What was you

Nutka1998 [239]

Answer:

a = 7.35 ft / s²

Explanation:

For this exercise we must use the kinematics relations

x = v₀ t + ½ a t²

as the runner leaves the starting line his initial velocity is zero

x = ½ a t²

a = $\frac{2x}{t^2}$

let's reduce the distance to foot

x = 60 yd (3ft / 1yd) = 180 ft

let's calculate

a = 2 180 / 7²

a = 7.35 ft / s²

6 0

3 years ago

A 1400 kg car driving at 25 m/s slams on its brakes. The coefficient of kinetic friction between the tires and the road is 0.7.

tamaranim1 [39]

The acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

The given parameters;

mass of the car, m = 1400 kg
Initial velocity of the car, u = 25 m/s
coefficient of kinetic friction, μ = 0.7

The acceleration of the car is calculated as follows;

a = μg

a = 0.7 x 9.8

a = 6.86 m/s²

The time taken for the car to stop is calculated by using Newton's second law of motion;

F = ma

$F = \frac{mv}{t} \\\\ma = \frac{mv}{t}\\\\a = \frac{v}{t} \\\\t = \frac{v}{a} \\\\t = \frac{25}{6.86} \\\\t = 3.64 \ s$

Thus, the acceleration of the car is 6.86 m/s² and the time taken for the car to stop is 3.64 s.

Learn more here:brainly.com/question/19887955

5 0

3 years ago

Intersystem crossing is:

Semenov [28]

An intersystem crossing (ISC) is a non-radiative process that involves the transition between two electronic states with different spin multiplicity. That is, when an electron is excited in a molecule in a basal singlet state (either by absorption or radiation) into a state of greater energy, an excited singlet or triplet state can be obtained.

Therefore, ISC is understood as an a non radio active transition between states with different spin multiplicity.

Correct answer is C: a radiative transition between states with the same spin.

8 0

4 years ago

PLease Help !!! A.S.A.P!!! will be marked as brainlest !!!!! 16 points

White raven [17]

This one is correct

Jamie is correct, because the mechanical energy is converting to electrical energy.

3 0

3 years ago

Read 2 more answers

A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.

7 0

3 years ago

Read 2 more answers