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lozanna [386]
3 years ago
5

Who was the first scientist to propose that an object could emit only certain amounts of energy?

Physics
1 answer:
Rzqust [24]3 years ago
3 0
It was Niels Bohr who proposed it
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The​ time, t, required to drive a fixed distance varies inversely as the​ speed, r. It takes 3 hr at a speed of 14 ​km/h to driv
kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

\frac{t}{3}=\frac{k/23}{k/14}

or

t=\frac{14\times3}{23}

or

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0
2 years ago
A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider
shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

d=v_x t

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

y(t) = h + v_y t - \frac{1}{2}gt^2

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s

So now we can find the magnitude of the initial velocity:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s

4 0
3 years ago
How long did the trip from camp wood to the pacific ocean and back again take?
maria [59]

The trip from Camp Wood to the Pacific Ocean and back again took 1.5 years to complete.<span>

</span>

<span>The </span>Lewis<span> and Clark </span>Expedition<span> from May 1804 to September 1806, also known as the Corps of Discovery </span>Expedition, was the first American expedition<span> to cross what is now the western portion of the United States.</span>
6 0
3 years ago
NEED HELP FAST ILL MARK BRAINLIEST AND RATE 5 STARS AND SAY THANK YOU
Vitek1552 [10]
It would be letter C


3 0
3 years ago
Read 2 more answers
A child uses a rubber band to launch a bottle cap at an angle of 37.0° above the horizontal. The cap travels a horizontal distan
zavuch27 [327]

Answer:

Initial velocity will be 1.356 m/sec      

Explanation:

Let the initial speed = u

Angle at which rubber band is launched = 37°

Horizontal component of initial velocity u_x=ucos\Theta =ucos37^{\circ}=0.7986u

Time is given as t = 1.20 sec

Distance in horizontal direction = 1.30 m

We know that distance = speed × time

So time t=\frac{distance}{speed}

1.20=\frac{1.3}{0.7986u}

u=1.356m/sec

So initial velocity will be 1.356 m/sec

3 0
2 years ago
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