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A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).

Answers

Answer 1

Answer:

Answer:

time is 32 s and speed is 304.3 m/s

Explanation:

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

$h = u t +0.5 at^2\n\n- 146 =14 sin 43 t - 4.9 t^2\n\n4.9 t^2 - 9.5 t - 146 =0 \n\nt =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\n\nt=(9.5\pm 54.3)/(9.8)\n\nt = 32.05 s, - 22.4 s$

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is

v' = u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

$v=√(v'^2 + v''^2)\n\nv = √(304.1^2 +10.3^2 )\n\nv = 304.3 m/s$

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How do you define 'heat' and 'temperature'

What is an atomic nucleus

Answers

Answer:

The atomic nucleus is the small, dense region consisting of protons and neutrons at the center of an atom, discovered in 1911 by Ernest Rutherford based on the 1909 Geiger–Marsden gold foil experiment.

Explanation:

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 27.0 m/s by a 7850 N braking force acting opposite the car’s motion. What is the car's velocity after 2.52s?
How far does the car move during the 2.52 s?
How long does it take the car to come to a complete stop?

Answers

Answer:

19.1 m/s

58.1 m

8.60 s

Explanation:

Take north to be positive and south to be negative.

Use Newton's second law to find the acceleration.

∑F = ma

-7850 N = (2500 kg) a

a = -3.14 m/s²

Given:

v₀ = 27.0 m/s

a = -3.14 m/s²

Find: v given t = 2.52 s

v = at + v₀

v = (-3.14 m/s²) (2.52 s) + 27.0 m/s

v = 19.1 m/s

Find: Δx given t = 2.52 s

Δx = v₀ t + ½ at²

Δx = (27.0 m/s) (2.52 s) + ½ (-3.14 m/s²) (2.52 s)²

Δx = 58.1 m

Find: t given v = 0 m/s

v = at + v₀

0 m/s = (-3.14 m/s²) t + 27.0 m/s

t = 8.60 s

(a) A woman climbing the Washington Monument metabolizes 6.00×102kJ of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?

Answers

Answer:

492 kJ

Consistent

Explanation:

Q = Heat stored by woman from food = 600 k J

η = Efficiency of woman = 18% = 0.18

Q' = heat transferred to the environment

heat transferred to the environment is given as

Q' = (1 - η) Q

Inserting the values

Q' = (1 - 0.18) (600)

Q' = 492 kJ

Yes the amount of heat transfer is consistent. The process of sweating produces the heat and keeps the body warm

Final answer:

A woman climbing the Washington Monument metabolizes food energy with 18% efficiency, meaning 82% of the energy is lost as heat. When we calculate this value, we find that 492 kJ of energy is released as heat, which is consistent with the fact that people quickly warm up when exercising.

Explanation:

The woman climbing the Washington Monument metabolizes 6.00×10² kJ of food energy with an efficiency of 18%. This implies that only 18% of the energy consumed is used for performing work, while the remaining (82%) is lost as heat to the environment.

To calculate the energy lost as heat:

Determine the total energy metabolized, which is 6.00 × 10² kJ.
Multiply this total energy by the percentage of energy lost as heat (100% - efficiency), which gives: (6.00 × 10² kJ) * (100% - 18%) = 492 kJ.

The released heat of 492 kJ is consistent with the fact that a person quickly warms up when exercising, because a significant portion of the body's metabolic energy is lost as heat due to inefficiencies in converting energy from food into work.

Learn more about Energy Efficiency and Heat Transfer here:

brainly.com/question/18766797

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Why are continental rocks much older than oceanic crust?A. Oceanic crust is continually recycled through convection in the earth's mantle
B. Oceanic crust is made out of much less dense material than continental crust
C. Continental crust is continually renewed through convection in the earth's mantle
D. Continental crust eats oceanic crust for breakfast

Answers

Answer:

A. Oceanic crust is continually recycled through convection in the earth's mantle

Explanation:

The oceanic plate is constantly being recycled through the forces of convection within the earth's mantle.

New oceanic plate are formed mid-oceanic ridge for example. As the magma cools and solidifies, they are moved away continually.

This is not the case for the continental curst.

Two non-conducting slabs of infinite area are given a charge-per-unit area of σ = -16 C/m^2 and σb =+6.0 C/m^2 respectively. A third slab, made of metal, is placed between the first two plates. The charge density σm on the metal slab is 0 (i.e., the slab is uncharged).Required:
Find the magnitude and direction of the electric field.

Answers

Answer:

$E_(C) = -5.65 * 10^(11) \hat{x}$ N/C

$E_(A) = -1.24 * 10^(12) \hat{x}$ N/C

Explanation:

The charge per unit area of the two non-conducting slabs are given by:

$\sigma_(a) = -16 C/m^2$

$\sigma_(b) = 6 C/m^2$

The charge density on the metal $\sigma_(m) = 0$

ε0 = 8.854 x 10-12 C2/N m2

Note that the electric field inside the conductor is zero because it is an equipotential surface.

The diagram attached to this solution typifies the description given in the question:

The electric field in the region C can be calculated by:

$E_(C) = ( |\sigma_(b) |- |\sigma_(a)| )/(2 \epsilon_(0) ) \nE_(C) = (6 - 16 )/(2 * 8.854 * 10^(-12) ) \nE_(C) = -5.65 * 10^(11) \hat{x}$

The electric field in the region A can be calculated by:

$E_(A) = (- |\sigma_(a) |- |\sigma_(b)| )/(2 \epsilon_(0) ) \nE_(A) = (-16 - 6 )/(2 * 8.854 * 10^(-12) ) \nE_(A) = -1.24 * 10^(12) \hat{x}$

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. What could you do to increase the maximum kinetic energy of electrons to 1.5 eV?

Answers

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:

$K_(max)=(hc)/(\lambda)-W$

Here h is the Planck's constant, c is the speed of light, $\lambda$ is the wavelength of the light and W the work function of the element:

$W=(hc)/(\lambda)-K_(max)\nW=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(495*10^(-9)m)-0.5eV\nW=2.01eV$

Now, we calculate the wavelength for the new maximum kinetic energy:

$W+K_(max)=(hc)/(\lambda)\n\lambda=(hc)/(W+K_(max))\n\lambda=((4.14*10^(-15)eV\cdot s)(3*10^8(m)/(s)))/(2.01eV+1.5eV)\n\lambda=3.54*10^(-7)m=354*10^(-9)m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

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