All categories

2 years ago

What is the time required for an object starting at rest to fall freely 500 meters near earth surface

2 answers:

7 0

Without air resistance . . .

Distance = 1/2 g T²

500 m = (1/2) (9.8 m/s²) (T²)

500 m = (4.9 m/s²) (T²)

Divide each side by (4.9 m/s²) :

T² = (500 m) / (4.9 m/s²)

= 102.04 s²

T = √ (102.04 s²)

= 10.1 seconds

Gnom [1K]2 years ago

3 0

Using kinematics, approximately 10.1 seconds

You might be interested in

If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1

ANTONII [103]

Answer:

$\Delta R_{max} = 0.46 ohm$

Explanation:

Resistance is given of different values

now we have

$R_1 = 64 ohm$

$R_2 = 20 ohm$

$R_3 = 8 ohm$

possible error in all resistance is 0.5 %

now we know that

$\Delta R_1 = 0.005\times 64 = 0.32 ohm$

$\Delta R_2 = 0.005 \times 20 = 0.1 ohm$

$\Delta R_2 = 0.005 \times 8 = 0.04 ohm$

Now the maximum possible error when all resistance is connected in series

$\Delta R_{max} = \Delta R_1 + \Delta R_2 + \Delta R_3$

$\Delta R_{max} = 0.32 + 0.1 + 0.04$

$\Delta R_{max} = 0.46 ohm$

7 0

2 years ago

If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb

rusak2 [61]

We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal.

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day.
Here is the equation to calculate that number

Number of days= 17500 / (kcal per day)

If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

4 0

2 years ago

1. If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released. It will oscillate. I

AURORKA [14]

Answer: a) 0.12m; b) 1,6 s; c) 0.625 1/s

Explanation: The simple harmonic movement can be described by a sin or cosine function in time.

This can be in the form:

X(t)= A Sin/Cos (wt+φ) where φ is initial phase o position at t=0

w the angular frequency are related to the frequency (f) as 2Pif

and f=1/T period of oscillating

3 0

2 years ago

The same physics student jumps off the back of her Laser again, but this time the Laser is

soldi70 [24.7K]

a) The speed of the student after the jump is 1.07 m/s

b) The final speed of the laser is 10.4 m/s

Explanation:

We can solve this problem by applying the law of conservation of momentum: if there are no external forces acting on the system, the total momentum of the student+Laser system must be constant. Therefore, we can write:

$p_i = p_f\\0=mv+MV$

where

The initial momentum is zero

m = 42 kg is the mass of the Laser

v = 1.5 m/s is the final velocity of the Laser

M = 59 kg is the mass of the student

V is the final velocity of the student

Solving the equation for V, we find the velocity of the student:

$V=-\frac{mv}{M}=-\frac{(42)(1.5)}{59}=-1.07 m/s$

So, the final speed of the student is 1.07 m/s.

In this case, the laser and the student are travelling at 3.1 m/s before the student jumps off: therefore, the total momentum before the jump is not zero.

So, the equation of the conservation of momentum is

$(m+M)u=mv+MV$