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2 years ago

What is the force on a 250 kg elevator that is falling freely at 9.8 m/sec2?

Physics

2 answers:

Anettt [7]2 years ago

7 0

Answer:

$\boxed {\boxed {\sf 2450 \ Newtons }}$

Explanation:

The elevator is falling freely and the only force acting on it is weight. So, calculate the weight.

We can do this by multiplying the mass by the acceleration of gravity.

$W=m*g$

The mass of the elevator is 250 kilograms and the acceleration of gravity is 9.8 meters per second squared.

$m= 250 \ kg \\g= 9.8 \ m/s^2$

Substitute the values into the formula.

$W= 250 \ kg * 9.8 \ m/s^2$

Multiply.

$W=2450 \ kg*m/s^2$

1 kilogram meter per second squared is equal to 1 Newton
So, our answer of 2450 kg*m/s² is equal to 2450 N

$W= 2450 \ N$

The force of weight on the elevator is 2450 Newtons

Viefleur [7K]2 years ago

6 0

Answer:

Since it is falling freely, the only force on it is its weight, w.

w = m × g = 250 kg × 9.8 m/s^2 = 2450 Newton/N

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2 years ago

A 0.90-kg falcon is diving at 28.0 m/s at a downward angle of 35° . It catches a 0.325-kg pigeon from behind in midair. What is

pashok25 [27]

Answer:

22.11 m / s

Explanation:

The falcon catches the prey from behind means both are flying in the same direction ( suppose towards the left )

initial velocity of falcon = 28 cos 35 i - 28 sin 35 j

( falcon was flying in south east direction making 35 degree from the east )

momentum = .9 ( 28 cos 35 i - 28 sin 35 j )

= 20.64 i - 14.45 j

initial velocity of pigeon

= 7 i

initial momentum = .325 x 7i

= 2.275 i

If final velocity of composite mass of falcon and pigeon be V

Applying law of conservation of momentum

( .9 + .325) V = 20.64 i - 14.45 j +2.275 i

V = ( 22.915 i - 14.45 j ) / 1.225

= 18.70 i - 11.8 j

magnitude of V

= √ [ (18.7 )² + ( 11.8 )²]

= 22.11 m / s

6 0

3 years ago

Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5

lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0

2 years ago

To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i

ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

$m_1v_1+m_2v_2=(m_1+m_2)v$

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

$m_1(-v_1)+m_2v_2=(m1+m2)(-v2)$

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

$-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s$

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

$v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s$