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3 years ago

A mass accelerates uniformly when the resultant force on it:

1 answer:

3 0

F = ma
So if you want the "a" to stay constant (as it is then uniform acceleration), the F must also be constant to apply a constant force.

2 is the answer

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Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi

Oksanka [162]

<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>

6 0

3 years ago

A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

6 0

3 years ago

Alfredo makes a diagram to organize his notes about charging objects.

konstantin123 [22]

The region marked X in the diagram shows that the objects have the same charge.

<h3>What is conduction?</h3>

The term conduction has to do with the manner of charging in which charge is passed from one object to another. Induction involves charging objects without the objects touching each other.

The region marked X in the diagram shows that the objects have the same charge.

Learn ore about conduction:brainly.com/question/15306642

#SPJ1

8 0

1 year ago

At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the

san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0

3 years ago

Janice has just measured the density of an object. Which value is possible? (Density: D = )

Sunny_sXe [5.5K]

By definition, the density of an object is given by:

$D = \frac{M}{V}$

Where,

M: mass of the object

V: volume of the object

Since the mass and volume of an object are numerical values greater than zero, then it follows that:

$D> 0$

It is important to respect the units of each measure.

For this case we can use the grams for the mass and cubic centimeters for the volume.

Answer:

A possible value for density is given by:

$6 \frac{g}{cm^3}$

7 0

3 years ago

Read 2 more answers