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3 years ago

A heavy object is dropped from a vertical height of 8.0 m. What is its speed when it hits the ground?

1 answer:

5 0

What do you know:
d=-8.0mm (starts higher and ends lower so it is negative)
Vi=0m/s (because it was dropped, not thrown)
A=-9.8m/s^2 (gravity)
Vf=???
Use the equation:
Vf^2=Vi^2+2A(d)
Vf^2=2(-9.8)(-8.0)
Vf^2= 156.8 -->find the square root of this
Final answer: Vf= 12.5m/s

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A hockey puck is hit on a frozen lake and starts moving with a speed of 13.60 m/s. Exactly 6.2 s later, its speed is 7.20 m/s. (

stellarik [79]

Answer:

-1.03 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².

Mathematically, acceleration is expressed as

a = (v-u)/t ........................ Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t = time.

Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.

Substituting into equation 2

a = (7.20-13.60)/6.2

a = -6.4/6.2

a = -1.03 m/s²

Note: a is negative because, the hockey puck is decelerating.

Hence the average acceleration = -1.03 m/s²

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2 years ago

A student pulls a cart across the floor with a force of 45 N. If the cart travels 10 m, how much work does the cart do on the st

bixtya [17]

I'm assuming we're applying the standard Integral form of the calculation of work. The solution is provided in the image.

6 0

3 years ago

Net force is the sum of all forces acting on an object. For example, in a tug of war, when one team is pulling the rope with a f

melisa1 [442]

Answer:

If there are equal forces in both directions and there is no motion, the net force is 0 Newtons. This is because you'd be subtracting 100 from 100 which just equals 0.

5 0

2 years ago

Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected

V125BC [204]

Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e

q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)

=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.

3 0

3 years ago

A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves.

san4es73 [151]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.

Mathematically its definition can be given as:

$d sin\theta =m\lambda$

Where

d = Width of the slit

$\theta =$ Angle between the beam and the source

m = Order (any integer) which represent the number of repetition of the spectrum, at this case 1 (maximum respect the wavelength)

Since the point of the theta angle for which the diffraction becomes maximum will be when it is worth one then we have to:

$\lambda = d$

$\lambda = 20cm = 20*10^{-2}m$

Applying the given relation of frequency, speed and wavelength then we will have that the frequency would be:

$f = \frac{v}{\lambda}$

Here the velocity is equal to the speed of light and the wavelength to the value previously found.

$f = \frac{3*10^8}{0.2}$

$f = 1.5Ghz$

Therefore the smallest microwave frequency for which only the central maximum occurs is 1.5Ghz

7 0

3 years ago