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zalisa [80]
4 years ago
15

Explain to me an example of forces on an object you have used recently. Include all forces acting on the object and explain if t

he motion is balanced or unbalanced
Physics
1 answer:
masha68 [24]4 years ago
4 0

Answer:

a toy car being pushed by a kid, a force acts on the car

Explanation:

when the boy is pushing the car a force acts on the car which is the effort that is applied on the car and the motion is unbalanced because there will be constant stopping of the boy and also the force acting on the car

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A force of 400.0N is exerted on a 1,250N car while moving it at a distance of 3.0m. How much work was done on the car?
S_A_V [24]

Answer:

1200 J

Explanation:

Work=force * displacement/distance

Work=400 * 3

Work=1200 J

8 0
3 years ago
Acartravelingat15m/sstartstodeceleratesteadily.Itcomestoacompletestopin10seconds.Whatisit'sacceleration? a. 1500 m/s b. -1.5 m/s
Amiraneli [1.4K]

Answer:

b. -1.5 m/s²

Explanation:

Given the following data;

Initial velocity = 15m/s

Time = 10 seconds.

Since the car came to rest, final velocity = 0m/s

To find acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{initial \; speed  -  final \; speed}{time}

a = \frac{v  -  u}{t}

Where,

a is acceleration measured in ms^{-2}

v and u is initial and final speed respectively, measured in ms^{-1}

t is time measured in seconds.

Substituting into the equation, we have;

Acceleration, a = \frac{0 - 15}{10}

Acceleration, a = \frac{-15}{10}

Acceleration = -1.5 m/s²

6 0
3 years ago
For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo
Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}

so that its period is T=3.644\,\frac{\mathrm s}{\mathrm{rev}} (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration \Delta t. Denote by \theta the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ

\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ

\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ

We can then compute the magnitude of the velocity vector differences \Delta\vec v for each time interval by using the law of cosines:

|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta

\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}

and in turn we find the magnitude of the average acceleration vectors to be

\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0
3 years ago
What does activation energy has to do with a chemical reaction.
asambeis [7]

Explanation:

Activation energy and reaction rate

The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be. ... The released energy helps other fuel molecules get over the energy barrier as well, leading to a chain reaction.

6 0
3 years ago
High energy waves have short wave lengths, true or false?
Anna71 [15]
True, the wavelength dies down due to high frequency and low amptitude. 
5 0
3 years ago
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