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3 years ago

When there is more speed does that mean there is more or less force?

2 answers:

7 0

If a heavy (more massive) object is in motion, more force must be applied to get the object moving faster. If the same force is applied to two objects, the object with the smaller mass will change speeds more quickly.

Mashcka [7]3 years ago

5 0

Answer:

1. The answer is that there will be more force with more speed.

2. less speed means less force.

3. And last, but certainly not the least, the less the mass has, the less the force will be, unless the object is moving at a high speed.

Explanation:

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Which of the following situations would cause the greatest decrease in the motion of molecules in a system? A. Q = -50 J, W = -5

amid [387]

To reduce the internal energy U of an ideal gas you need the gas to lose heat and/or do work.

6 0

3 years ago

A space rover weighs less on Mars than it does on Earth. Which statement explains this difference? A. The gravitational constant

Maksim231197 [3]

Answer:

B. The mass of Mars is less than the mass of Earth.

Explanation:

Mass of an object is the constant anywhere in the universe.

The weight of an object is equal to the gravitational force acting on it.

Weight is given by

$W=\dfrac{GMm}{R^2}\\\Rightarrow W=\dfrac{GM}{R^2}m\\\Rightarrow W=mg$

where

G = Gravitational constant

M = Mass of Planet

R = Radius of planet

m = Mass of object

g = Acceleration due to gravity

So weight of an object depends on the acceleration due to gravity on that planet. The acceleration due to gravity depends on the mass and radius of the planet.

The weight of the object is less on Mars because mars has less mass compared to Earth.

6 0

2 years ago

Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0

2 years ago

What is the magnitude of electrical force of attraction between an copper nucleus (29 protons) and its innermost electron if the

Agata [3.3K]

The charge of the copper nucleus is 29 times the charge of one proton:
$Q=29 q= 29 \cdot 1.6 \cdot 10^{-19}C=4.64 \cdot 10^{-18}C$
the charge of the electron is
$e=-1.6 \cdot 10^{-19}C$
and their separation is
$r=1.0 \cdot 10^{-12} m$

The magnitude of the electrostatic force between them is given by:
$F=k_e \frac{Qe}{r^2}$
where $k_e$ is the Coulomb's constant. If we substitute the numbers, we find (we can ignore the negative sign of the electron charge, since we are interested only in the magnitude of the force)
$F=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(4.64 \cdot 10^{-18}C)(1.6 \cdot 10^{-19}C)}{(1.0 \cdot 10^{-12} m)^2}=6.68 \cdot 10^{-3} N$

3 0

3 years ago

The battleship and enemy ships 1 and 2 lie along a straight line. Neglect air friction. battleship 1 2 Consider the motion of th

DaniilM [7]

Answer:

$\frac{t_1}{t_2} = \frac{sin\theta_1}{sin\theta_2}$

Explanation:

The vertical component of the initial velocities are

$v_v = v_0sin\theta$

If we ignore air resistance, and let g = -9.81 m/s2. The the time it takes for the projectiles to travel, vertically speaking, can be calculated in the following motion equation

$v_vt - gt^2/2 = s = 0$

$t(v_v - gt/2) = 0$

$v_v - gt/2 = 0$

$t = 2v_v/g = 2v_0sin\theta/g$

So the ratio of the times of the flights is

$t_1 / t_2 = \frac{2v_0sin\theta_1/g}{2v_0sin\theta_2/g} = \frac{sin\theta_1}{sin\theta_2}$

8 0

2 years ago