Question 12 of 20

6. An excited dog runs full-speed toward his owner who has just returned home from

mixas84 [53]

Answer:

v = 8.65 m/s

Explanation:

Given that,

Distance covered by the doge, d = 45 m

Time taken, t = 5.2 s

We need to find its average speed. The total distance covered divided by the total time taken is called the average speed of an object. So,

$v=\dfrac{45\ m}{5.2\ s}\\\\=8.65\ m/s$

So, the average speed is 8.65 m/s.

4 0

3 years ago

32. Only a small percentage of the energy

nordsb [41]

Only a small Percentage of the energy emitted by the sun strikes earth because, since the earth is going in circles round and round, the sun only hits part of the earth and not fully. So for example if the earth is going around and the sun hits Africa then in a couple minutes it will go to the next country and that country will have sunlight. Let me know if you need anything else or if this isn't partially correct. GLAD TO HELP! :)

5 0

3 years ago

A 59kg child starting from rest slides down a water slide with a vertical height of 5.0m. what is the child's speed halfway down

KIM [24]

<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>

6 0

3 years ago

When a certain element is excited with electricity, we see three main lines in its emission spectrum: two red lines and one oran

Eddi Din [679]

The absorption spectrum would have all the wavelengths of the light source but would have black lines where the two red and one orange lines were in the emission spectrum

4 0

3 years ago

A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .

rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

$-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0$

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

$v=u+at$

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

$v = 15.5 +(-9.8)(2.64)=-10.4 m/s$

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

$s=ut+\frac{1}{2}at^2$

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

$-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0$

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

5 0

3 years ago

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