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valkas [14]
3 years ago
12

In order to do work on an object, the force you exert must be

Physics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0
The force must be made over a distance
You might be interested in
Using the Bohr model, what is the ratio of the energy of the nth orbit of a triply ionized beryllium atom (Be3+, Z = 4) to the e
emmainna [20.7K]

Answer:

16

Explanation:

solution:

by taking the ratio of the energy E_n,be of the nth orbit of a beryllium atom

Z_be=4 to the energy E_n,h of the nth orbit of a hydrogen (Z_h=1) atom  gives

E_n,B/E_n,H=-(2.18*10^-18)*Z^2_BE/-(2.18*10^-18)*Zh^2/n^2

                    =Z^2_BE+/Z^2_H

                    =(4)^2/(1)^2

                    = 16

8 0
4 years ago
7 Calculate the weight of an apple of mass 100 grams
Leni [432]

Answer:

See the anwers below

Explanation:

To solve this problem we must understand that the weight of anyW = 0.1*10\\W = 1 [N]body is defined as the product of mass by gravitational acceleration. Gravitational acceleration depends on the place in space where a certain body is located.

W = m*g\\

where:

W = weight [N]

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration [m/s²] or [N/kg]

<u>Now for the Earth</u>

<u />W =m*g\\W =0.1*10\\W = 1[N]<u />

<u>For the Moon</u>

<u />W = m*g\\W =0.1*1.6\\W = 0.16 [N]<u />

4 0
3 years ago
A hockey puck slides off the edge of a horizontal platform with an initial velocity of 28.0 m/shorizontally in a city where the
kozerog [31]

Answer:

θ = 12.60°

Explanation:

In order to calculate the angle below the horizontal for the velocity of the hockey puck, you need to calculate both x and y component of the velocity of the puck, and also you need to use the following formula:

\theta=tan^{-1}(\frac{v_y}{v_x})       (1)

θ: angle below he horizontal

vy: y component of the velocity just after the puck hits the ground

vx: x component of the velocity

The x component of the velocity is constant in the complete trajectory and is calculated by using the following formula:

v_x=v_o

vo: initial velocity = 28.0 m/s

The y component is calculated with the following equation:

v_y^2=v_{oy}^2+2gy         (2)

voy: vertical component of the initial velocity = 0m/s

g: gravitational acceleration = 9.8 m/s^2

y: height

You solve the equation (2) for vy and replace the values of the parameters:

v_y=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(2.00m)}=6.26\frac{m}{s}

Finally, you use the equation (1) to find the angle:

\theta=tan^{-1}(\frac{6.26m/s}{28.0m/s})=12.60\°

The angle below the horizontal is 12.60°

7 0
3 years ago
The ends of a magnet where thee forces are the strongest are called magnets
ladessa [460]
They are called magnets poles.
7 0
4 years ago
What type of foods do bodyguards eat give 5 examples
Rainbow [258]
They eat rice, chicken, beef, cereal they eat what we eat in a daily basis I guess
8 0
3 years ago
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