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3 years ago

How do I solve such problem???

Physics

1 answer:

labwork [276]3 years ago

6 0

1). Calculate how long it takes an object to fall 4,000 m after it's dropped. (Use D = (1/2) (g) (T²) . D is 4,000 m. g = 9.8 m/s². Find T .)

2). Calculate how far the object will move HORIZONTALLY in that length of time, if it's moving at 75 m/s. (Distance = (75 m/s) x (time) . )

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Is 45 m/s2 a scalar or a vector quantity and how do you know?

Umnica [9.8K]

Answer:

Explanation:

it is a vector as it has a magnitude and a direction. If it was a scalar quantity it would just have a magnitude and would be 45 m. Acceleration is an example of a vector quantity

5 0

2 years ago

If the mass of an object increases, how is its acceleration affected, assuming the net force acting on the object remains the sa

vovikov84 [41]

Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,

F = ma

If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.

7 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

1 year ago

What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?

IrinaK [193]

Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m( $\frac{v_{2} - v_{1} }{t}$ )

⇒ Ft = m( $v_{2}$ - $v_{1}$ )

where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, $v_{1}$ is its initialvelocity and $v_{2}$ is the final velocity of the object.

Therefore,

impulse = Ft = m( $v_{2}$ - $v_{1}$ )

From the question, m = 3kg, $v_{1}$ = 0m/s and $v_{2}$ = 20m/s.

So that,

Impulse = 3 (20 - 0)

= 3(20)

= 60Ns

The impulse on the object is 60Ns.

8 0

2 years ago

Which of these factors determines soil texture?

STatiana [176]

It is B particle sizes

8 0

3 years ago

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