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Elan Coil [88]
4 years ago
8

How is velocity affected when balanced forces act on an object?

Physics
1 answer:
MrRissso [65]4 years ago
8 0
When the forces acting on a body are balanced, their effect
\on the body's motion is the same as if no forces at all are
acting on it, and its velocity can't change.  It continues moving
in a straight line at constant speed (which may be zero).
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Alana is skateboarding at 19 km/h and throws a tennis ball at 11 km/h to her friend Oliver who is behind her leaning against a w
NNADVOKAT [17]

The speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

<h3>What is the frame of reference?</h3>

It is a point from where the motion is observed. The speed changes with the frame of reference.

For the person standstill on the sidewalk, If  Alana throw the ball in the front direction, The speed will be,

19+11 = 30 km/h

It Alana just leave the ball, the speed,

19 +0 = 19 km/h

If Alana throws the ball behind her at the speed of 19 km/h. The speed of the ball for a sidewalk observer,

19 +(-19 ) =  0 km/h

Thus, when Aana throw the ball behind the velocity will be negative,

So,

19 + (-11 ) = 8 km/h

Therefore, the speed of the tennis ball is 8 km/h when Alana throws the ball behind at the speed of 11 km/h.

Learn more about Frame of reference:

brainly.com/question/9820962

8 0
2 years ago
Using an example, such as the natural building of sand dunes, the process of DEPOSITION causes?
andre [41]
The correct answer is c. The process of deposition causes rock and soil to be slowly gained. Deposition is a geological process in which soil and rocks are added to a landform.
8 0
3 years ago
Read 2 more answers
The driver or a sports car traveling at 10 m/s steps down hard on the accelerator for 5 seconds. If they accelerate at a rate of
ivanzaharov [21]

Answer:

53.125m

Explanation:

The displacement of the car, denoted by S, can be calculated using the formula:

S = ut + 1/2at²

Where;

u = initial velocity/speed (m/s)

t = time (s)

a = acceleration (m/s²)

According to the information provided in this question, u = 10m/s, t = 5s, a = 0.25m/s², S = ?

S = ut + 1/2at²

S = (10 × 5) + 1/2 (0.25 × 5²)

S = 50 + 1/2 (0.25 × 25)

S = 50 + 1/2(6.25)

S = 50 + 3.125

S = 53.125m

4 0
3 years ago
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
A car moving with a velocity of 20 meters/second has 1.8 × 105 joules of kinetic energy. What is the mass of the car?
PSYCHO15rus [73]
V = 20 m/s 
Ecin = 1.8 x 10^5 J 

Ecin = m*v^2/ 2 
1.8*10^5 = m*400/2 
200m = 1.8*10^5 

I know you can do it :) 

7 0
3 years ago
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