.

Assume your mass is 60 kg. The acceleration due to gravity is 9.8 m/s 2 . How much work against gravity do you do when you climb

Andre45 [30]

Answer:

W=1705.2 J

Explanation:

Given that

mass ,m= 60 kg

Acceleration due to gravity ,g= 9.8 m/s²

Height ,h= 2.9 m

As we know that work done by a force given as

W = F . d

F=force

d=Displacement

W=work done by force

Now by putting the values

F= m g (Acting downward )

d= h (Upward)

W= m g h ( work done against the force)

W= 60 x 9.8 x 2.9 J

W=1705.2 J

Therefore the answer will be 1705.2 J.

8 0

3 years ago

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

2 years ago

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed. The energies of atoms a

lawyer [7]

Answer:

This is because The energies of atoms are quantized.

Electrons are allowed "in between" quantized energy levels, and, thus, only specific lines are observed

5 0

2 years ago

I NEED HELP ASAP!!! 100 points if you answer these 7 questions:

netineya [11]

Answer:

#1. B. Alfred Wegener

#2. C. Volcanic activity at the ocean bottom.

#3. B, C. Continental, Sea floor (oceanic).

#4. B. Folding

#5. B. Tectonics

#6. C. Point underground where the earthquake starts

#7. C. Flood waters carrying away soil

Explanation:

#1. Alfred Lothar Wegener was a German polar researcher, geophysicist and meteorologist. During his lifetime he was primarily known for his achievements in meteorology and as a pioneer of polar research, and is also known as the father of plate tectonics.

#2. The most prominent feature of ocean topography discovered in the 1960s was: Volcanic activity at the ocean bottom. The continents have always been in their current positions.

#3. Tectonic plates are composed of oceanic lithosphere and thicker continental lithosphere, each topped by its own kind of crust. The two types of tectonic plates are continental and oceanic tectonic plates.

#4. Tremendous pushing forces exerted by two of Earth's plates moving together squeezed rock layers from opposite sides. This caused the rock layers to buckle and fold, forming folded mountains. Folded mountains are mountains formed by the folding of rock layers caused by compression forces.

#5. Plate tectonics is the scientific theory explaining the movement of the earth's crust. It is widely accepted by scientists today. Recall that both continental landmasses and the ocean floor are part of the earth's crust, and that the crust is broken into individual pieces called tectonic plates.

#6. The hypocenter is the point within the earth where an earthquake rupture starts. The epicenter is the point directly above it at the surface of the Earth. Also commonly termed the focus. See also epicenter.

#7. This process is known as erosion. In earth science, erosion is the action of surface processes that removes soil, rock, or dissolved material from one location on the Earth's crust, and then transports it to another location.

Hope this helps!

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3 0

2 years ago

A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

5 0

3 years ago