I need an expert in physics for help

A sphere has______ sides

Alina [70]

Answer:

5738929292

Explanation:

simple math

8 0

3 years ago

Read 2 more answers

Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an

Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

$W = F\cdot \Delta s$

Where:

$W$ - Work, measured in joules.

$F$ - Force exerted on the object, measured in newtons.

$\Delta s$ - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

$W_{net} = W_{Joel} - W_{Jerry}$

$W_{net} = F\cdot \Delta s - F\cdot \Delta s$

$W_{net} = (F-F)\cdot \Delta s$

$W_{net} = 0\cdot \Delta s$

$W_{net} = 0\,J$

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

$W_{net} = W_{Joel}$

$W_{net} = F\cdot \Delta s$

Hence, the work done by Joel is greater than the work done by Jerry.

7 0

3 years ago

1.) Rn-222 decays from 400 grams to 6.25 grams in 240 minutes. How long is one “half-life.

KiRa [710]

Answer:

40

Explanation:

7 0

3 years ago

Interactive Solution 6.39 presents a model for solving this problem. A slingshot fires a pebble from the top of a building at a

mariarad [96]

(a) 29.8 m/s

To solve this problem, we start by analyze the vertical motion first. This is a free fall motion, so we can use the following suvat equation:

$v_y^2 - u_y^2 = 2as$

where, taking upward as positive direction:

$v_y$ is the final vertical velocity

$u_y = 0$ is the initial vertical velocity (zero because the pebble is launched horizontally)

$a=g=-9.8 m/s^2$ is the acceleration of gravity

s = -25.0 m is the displacement

Solving for vy,

$v_y = \sqrt{u^2+2as}=\sqrt{0+2(-9.8)(-25)}=-22.1 m/s$ (downward, so we take the negative solution)

The pebble also have a horizontal component of the velocity, which remains constant during the whole motion, so it is

$v_x = 20.0 m/s$

So, the final speed of the pebble as it strikes the ground is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{20.0^2+(-22.1)^2}=29.8 m/s$

(b) 29.8 m/s

In this case, the pebble is launched straight up, so its initial vertical velocity is

$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

(c) 29.8 m/s

This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

As before, the horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

We notice that the final value of the speed is always the same in all the three parts, so it does not depend on the direction of launching. This is due to the law of conservation of energy: in fact, the initial mechanical energy of the pebble (kinetic+potential) is the same in all three cases (because the height h does not change, and the speed v does not change either), and the kinetic energy gained during the fall is also the same (since the pebble falls the same distance in all 3 cases), therefore the final speed must also be the same.

7 0

3 years ago

Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by