A go-kart accelerates at a rate

"In the diagram below, point P is located in the electric field between two oppositely charged parallel plates.

olya-2409 [2.1K]

As proton is heavier then electron, and electrostatic force is directly proportional to mass, so it would be greater and direction would be changed 'cause of different charges.

In short, Your Answer would be Option 4

Hope this helps!

5 0

4 years ago

Read 2 more answers

When Royce was 10 years old, he had a mass of

coldgirl [10]

Answer:

The gravitational force between Royce and Earth would be doubled at 16 years.

Explanation:

"Newton's law of universal gravitation states that gravitation force between two masses is proportional to the magnitude of their masses and inverse-squared of their distance".

Royce Scenario

At the age of 10 Royce's mass = 30kg

At the age of 16 Royce's mass = 60kg

From Newton's law of universal gravitation, an Increase in the mass of a body would amount to a corresponding increase in the gravitational force.

In the case of Royce, the mass double between the age of 10 and 16, so there would be an increase of the gravitation force by double.

6 0

3 years ago

An object (mass 12.5 kg) slides on a frictionless surface at 1.32 m/s. How much work must be done to bring the object to rest

mote1985 [20]

Answer:

10.89 J.

Explanation:

The following data were obtained from the question:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Work done =?

To obtain the workdone, we shall determine the kinetic energy of the object since work and energy has the same unit of measurement. This is illustrated below:

Mass (m) = 12.5 kg

Velocity (v) = 1.32 m/s

Kinetic energy (K.E) =?

K.E = ½mv²

K.E = ½ × 12.5 × 1.32²

K.E = 6.25 × 1.7424

K.E = 10.89 J

The kinetic energy of the object is 10.89 J. Hence, the workdone in bringing the object to rest is 10.89 J.

8 0

3 years ago

The tallest sequoia sempervirens tree in California’s redwood national parks is 111 m tall. Suppose an object is thrown downward

ch4aika [34]

The object's initial velocity is equal to $-10.59\frac{m}{s}$

Why?

From the statement we know the height of the tree and the time it takes to reach the ground, so, if we need to calculate its initial velocity, we can use the following formula:

$y=y_o-v_{o}*t-\frac{1}{2}g*t^{2}$

Where,

y, is the final height (0 meters in this case)

yo, is the initial height (111 meters in this case)

t, is the time elapsed (3.8 seconds in this case)

vo, is the initial speed.

g, is the acceleration due to gravity (-9.81 m/s2)

Now, let's set the origin at the top of the tree, so, rewriting the formula, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

So, isolating the initial velocity, we have:

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}$

$y=y_o+v_{o}*t+\frac{1}{2}g*t^{2}\\\\v_{o}*t=y-y_o-\frac{1}{2}g*t^{2}\\\\v_{o}=\frac{y-y_o-\frac{1}{2}g*t^{2}}{t}$

Finally, substituting and calculating, we have:

$v_{o}=\frac{-111m-0-\frac{1}{2}(-9.8\frac{m}{s^{2}}) *(3.8s)^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-9.8\frac{m}{s^{2}})*(14.44s^{2})}{3.8s}\\\\v_{o}=\frac{-111m-\frac{1}{2}(-141.51m)}{3.8s}=\frac{-111m+70.75m}{3.8s}\\\\v_{o}=\frac{-40.25m}{3.8s}=-10.59\frac{m}{s}$

Hence, we have that the initial velocity of the object is $-10.59\frac{m}{s}$

Have a nice day!

7 0

4 years ago

A skateboarder shoots off a ramp with a velocity of 6.6 m/s, directed at an angle of 58 above the horizontal. The end of the ram

sweet-ann [11.9K]

Answer:

a) Maximum height reached above ground = 2.8 m

b) When he reaches maximum height he is 2 m far from end of the ramp.

Explanation:

a) We have equation of motion v²=u²+2as

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0²=5.6² + 2 x -9.81 x s

s = 1.60 m

Height above ground = 1.2 + 1.6 = 2.8 m

b) We have equation of motion v= u+at

Considering vertical motion of skateboarder.

When he reaches maximum height,

u = 6.6sin58 = 5.6 m/s

a = -9.81 m/s²

v = 0 m/s

Substituting

0= 5.6 - 9.81 x t

t = 0.57s

Now considering horizontal motion of skateboarder.

We have equation of motion s =ut + 0.5 at²

u = 6.6cos58 = 3.50 m/s

a = 0 m/s²

t = 0.57

Substituting

s =3.5 x 0.57 + 0.5 x 0 x 0.57²

s = 2 m

When he reaches maximum height he is 2 m far from end of the ramp.

8 0

4 years ago