All categories

4 years ago

What value is needed to create a motion diagram? a. position c. time interval b. velocity d. both a and c

1 answer:

5 0

For classical kinematics, you usually graph position versus time, meaning D. This would be a graph where time is on the x-axis and position (can also be treated as displacement if you consider the net distance) is on the y axis.

You might be interested in

The electric field strength is 20,000 N/C inside a parallel-platecapacitor with a 1.0 mm spacing. An electron is released fromre

creativ13 [48]

Explanation:

It is given that,

Electric field strength, E = 20,000 N/C

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

$F=qE$

Also, $ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 20,000}{9.1\times 10^{-31}}$

$a=3.51\times 10^{15}\ m/s^2$

Using third equation of motion as :

$v^2-u^2=2ad$

$v=\sqrt{2ad}$

$v=\sqrt{2\times 3.51\times 10^{15}\times 0.001}$

v = 2649528.2599 m/s

$v=2.64\times 10^6\ m/s$

So, the velocity of the electron when it reaches the positive plate is $2.64\times 10^6\ m/s$ . Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the dis

irakobra [83]

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

4 0

3 years ago

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire

svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance = 20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

$W_{fr} = -F\cdot d$

Where, F = force

d = distance

Put the value into the formula

$W_{fr}=-35.0\times20$

$W_{fr}=−700\ J$

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

$W=fd\cos\theta$

Put the value into the formula

$W=35.0\times20\cos90$

$W=0\ J$

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

$W_{sh}=W_{net}-W_{fr}$

Put the value into the formula

$W_{sh}=0-(-700)$

$W_{sh}=700\ J$

(D). We need to calculate the force the shopper exerts

Using formula of force

$F_{sh}=\dfrac{W_{fr}}{d\cos\theta}$

Put the value into the formula

$F_{sh}=\dfrac{700}{20\cos25}$

$F_{sh}=38.61\ N$

(E). We need to calculate the total work done on the cart

Using formula of work done

$W_{cart}=W_{fr}+W_{sh}$

Put the value into the formula

$W_{cart}=700-(-700)$

$W_{cart}=0\ J$

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0

3 years ago

How do we know if something is moving?

nadya68 [22]

An object is in motion if its distance relative to another object is changing. To tell if an object is moving, you use a reference point. If an object's distance from another object [reference point] is changing. A place or object used for comparison to determine if something is in motion.

5 0

3 years ago

What is the public storm warning signal that carries winds of 30 to 60 km/hr and can cause no damage to very light damage

Shkiper50 [21]

Answer:

I know u need this

Explanation:

You gave me the runaround

I really hate the runaround

You really got me paranoid

I always keep a gun around

You always give me butterflies

When you come around

7 0

3 years ago