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2 years ago

2. A carpenter tosses a shingle off a 9.4 m high roof, giving it an initial horizontal

1 answer:

8 0

Sounds like the shingle/ball is thrown from the roof horizontally, so that the distance it travels x after time t horizontally is

x = (7.2 m/s) t

The object's height y at time t is

y = 9.4 m - 1/2 gt²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity, and its vertical velocity is

v = -gt

(a) The object hits the ground when y = 0:

0 = 9.4 m - 1/2 gt²

t² = 2 * (9.4 m) / (9.80 m/s²)

t ≈ 1.92 s

at which time the object's vertical velocity is

v = -g (1.92 s) = -18.8 m/s ≈ -19 m/s

(b) See part (a); it takes the object about 1.9 s to reach the ground.

(c) The object travels a horizontal distance of

x = (7.2 m/s) * (1.92 s) ≈ 13.8 m ≈ 14 m

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A light wave has a 670 {\rm nm} wavelength in air. Its wavelength in a transparent solid is420 {\rm nm} .a)What is the speed of

tia_tia [17]

Answer:

Explanation:

given,

Light wave wavelength in Air, λ_(vac) = 670 nm

wavelength in transparent solid, λ= 420 nm

Refractive index of the solid

$n = \dfrac{\lambda_{vac}}{\lambda}$

$n = \dfrac{670}{420}$

n = 1.6

a) speed of light in the solid

$v = \dfrac{c}{n}$

$v = \dfrac{c}{1.6}$

v = 0.625 c

b) light's frequency in the solid

$f = \dfrac{c}{\lambda_{vac}}$

frequency of light remain same when light move from one medium to another.

f is the frequency of light

$f = \dfrac{3\times 10^8}{670\times 10^{-9}}$

f = 4.47 x 10⁻¹⁴ Hz

the light's frequency in the solid is equal to 4.47 x 10⁻¹⁴ Hz

5 0

3 years ago

A book that weighs 20 N sits on a table. How big and in what direction does the force of gravity from the Earth act on the book?

taurus [48]

Answer:

A. 20 N down

Explanation:

It's asking how much force of gravity is acting on it. The book weighs 20 Newtons so that's how much gravity is being applied. Hope this helps

7 0

2 years ago

Read 2 more answers

a family drives from boston (100 miles away) to new york (500 miles away) in 10 hours . How fast were they were traveling?

Fynjy0 [20]

It should be 60 mph. Because if you divide 600 by 10 it’s 60

8 0

2 years ago

A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

Mass of the pitcher, m₁ = 50 kg
Mass of the baseball, m₂ = 0.15 kg
Velocity of the ball, u₂ = 35 m/s

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

$u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s$

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0

2 years ago

May you help me answer this

Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$