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3 years ago

The thrust from a car is 2000 N. The air resistance and friction together total 680 N. What happens to the speed of the car?

1 answer:

6 0

The car will speed up because if the air resistance and friction forces are reacting in the direction opposite to the Applied force, The Net force will still be in the direction of the car's motion and since F=ma a will cause your velocity to increase

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Compare and contrast the energy transfer of a roller coaster to that of a pendulum

Softa [21]

When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

<h3>Compare and contrast the energy transfer of a roller coaster to that of a pendulum:</h3><h3>What is the transfer of energy in a roller coaster?</h3>

The transfer of potential energy to kinetic energy occur when the roller coaster move along the track. As the motor pulls the cars to the top, the body has more potential energy whereas when the body comes to the bottom , it has kinetic energy in the object.

<h3>What is the energy transfer in a pendulum?</h3>

As a pendulum swings, its potential energy changes to kinetic energy and kinetic energy changes into potential energy. At the top more potential energy is present.

So we can conclude that When the pendulum and roller coaster move to the top, its has more potential energy whereas when comes to the bottom has more kinetic energy.

Learn more about energy here: brainly.com/question/13881533

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8 0

1 year ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$

5 0

2 years ago

A girl playing tug-of-war with her dog pulls the dog a distance of 8.0m by exerting a force at an angle of 18° with the horizont

AnnZ [28]

Answer:

25 N

Explanation:

Work is a product of force and perpendicular distance moved.

W=Fd where F is force exerted and d is perpendicular distance.

However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as $dcos\theta$

Therefore, $W=Fdcos\theta$

Making F the subject of the formula then

$F=\frac {W}{dcos\theta}$ where $\theta$ is the angle of inclination. Substituting 190 J for W then 18 degrees for $\theta$ and 8 m for d then

$F=\frac {190}{8cos18^{\circ}}\approx 25N$

3 0

2 years ago

A very humble bumble bee is flying horizontally due North at a constant speed of 3.11 m/s. At the current location of the bumble

Reil [10]

To solve this problem we will apply the concepts of the Magnetic Force. This expression will be expressed in both the vector and the scalar ways. Through this second we can directly use the presented values and replace them to obtain the value of the magnitude. Mathematically this can be described as,

$\vec{F_B} = q(\vec{V}\times \vec{B})$