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3 years ago

PLEASE HELP

Physics

1 answer:

gulaghasi [49]3 years ago

7 0

Answer:

The answer should be Playing Area

Explanation:

I HOPE THIS HELPS.....GOOD LUCK!!!

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1. What is the new kinetic energy of the 1900kg ship on the right moving at 4 m/s?

neonofarm [45]

Explanation:

That`s is the answer, just check

6 0

2 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

2 years ago

Which of the following would produce the most power?

Fantom [35]

Answer:

A mass of 10 kilograms lifted 10 meters in 5 seconds.

Explanation:

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

But Energy = mgh

Substituting into the equation, we have

$Power = \frac {mgh}{time}$

Given the following data;

Mass = 10kg

Height = 10m

Time = 5 seconds

We know that acceleration due to gravity is equal to 9.8 m/s²

$Power = \frac {10*9.8*10}{5} = 490 Watts$

Hence, a mass of 10 kilograms lifted 10 meters in 5 seconds would produce the most power.

6 0

2 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

2 years ago

PLEASE HELPPPPP!!!! :)

GuDViN [60]

Answer:

F=X.F=mxq. 3. 1 N/kg=0.5kg g=9.80

4 0

2 years ago